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<?xml version="1.0" encoding="utf-8"?><echo xmlns="http://www.mpiwg-berlin.mpg.de/ns/echo/1.0/" xmlns:de="http://www.mpiwg-berlin.mpg.de/ns/de/1.0/" xmlns:dcterms="http://purl.org/dc/terms" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:echo="http://www.mpiwg-berlin.mpg.de/ns/echo/1.0/" xmlns:xhtml="http://www.w3.org/1999/xhtml" xmlns:mml="http://www.w3.org/1998/Math/MathML" xmlns:xlink="http://www.w3.org/1999/xlink" version="1.0RC">
  <metadata>
    <dcterms:identifier>ECHO:FDWQ9FD5.xml</dcterms:identifier>
    <dcterms:creator identifier="GND:11864548X">Appolonius Pergaeus</dcterms:creator>
    <dcterms:contributor>Lawson, John</dcterms:contributor>
    <dcterms:title xml:lang="en">The two books of Apollonius Pergaeus, concerning tangencies, as they have been restored by Franciscus Vieta and Marinus Ghetaldus : with a supplement to which is now added, a second supplement, being Mons. Fermat's Treatise on spherical tangencies</dcterms:title>
    <dcterms:date xsi:type="dcterms:W3CDTF">1771</dcterms:date>
    <dcterms:language xsi:type="dcterms:ISO639-3">eng</dcterms:language>
    <dcterms:rights>CC-BY-SA</dcterms:rights>
    <dcterms:license xlink:href="http://creativecommons.org/licenses/by-sa/3.0/">CC-BY-SA</dcterms:license>
    <dcterms:rightsHolder xlink:href="http://www.mpiwg-berlin.mpg.de">Max Planck Institute for the History of Science, Library</dcterms:rightsHolder>
    <parameters>despecs = 1.1.2</parameters>
    <log>removed curly braces: Simpſon’s Algebra, 8vo. {3d Edit. \\ 1ſt Edit.} # _Lond_. 1767 \\ _ibid_. 1745
      removed "/tb" and "tb" in the index </log>
  </metadata>
  <text xml:lang="en" type="free">
<div xml:id="echoid-div1" type="section" level="1" n="1"><pb file="0001" n="1"/>
<pb file="0002" n="2"/>
<pb file="0003" n="3"/>
<pb file="0004" n="4"/>
<pb file="0005" n="5"/>
</div>
<div xml:id="echoid-div2" type="section" level="1" n="2">
<head xml:id="echoid-head1" xml:space="preserve">THE <lb/>TWO BOOKS <lb/>OF <lb/>APOLLONIUS PERGÆUS, <lb/>CONCERNING <lb/>TANGENCIES, <lb/>As they have been Reſtored by <lb/>FRANCISCUSVIET A and MARINUSGHETALDUS. <lb/>WITH A <lb/>SUPPLEMENT.</head>
<p>
  <s xml:id="echoid-s1" xml:space="preserve">By JOHN LAWSON, B. </s>
  <s xml:id="echoid-s2" xml:space="preserve">D. </s>
  <s xml:id="echoid-s3" xml:space="preserve">Rector of Swanſcombe in Kent.</s>
  <s xml:id="echoid-s4" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div3" type="section" level="1" n="3">
<head xml:id="echoid-head2" xml:space="preserve">THE SECOND EDITION.</head>
<head xml:id="echoid-head3" xml:space="preserve">TO WHICH IS NOW ADDED, <lb/>A SECOND SUPPLEMENT, <lb/>BEING <lb/>Monſ. FERMAT’S Treatiſe on Spherical Tangencies.</head>
<head xml:id="echoid-head4" xml:space="preserve">LONDON: <lb/>Printed by G. BIGG, Succeſſor to D. LEACH.</head>
<head xml:id="echoid-head5" xml:space="preserve">And ſold by B. <emph style="sc">White</emph>, in Fleet-Street; L. <emph style="sc">Davis</emph>, in Holborne, J. <emph style="sc">Nourse</emph>, in the <lb/>Strand; and T. <emph style="sc">Payne</emph>, near the Mews-Gate.</head>
<head xml:id="echoid-head6" xml:space="preserve">MDCCLXXI.</head>
<pb file="0006" n="6"/>
<pb o="(iii)" file="0007" n="7"/>
</div>
<div xml:id="echoid-div4" type="section" level="1" n="4">
<head xml:id="echoid-head7" xml:space="preserve">PREFACE.</head>
<p>
  <s xml:id="echoid-s5" xml:space="preserve">OF the twelve Analytical Treatiſes recited by Pappus in <lb/>his Preface to the 7th Book of his Mathematical Col-<lb/>lections, we have very little of the Originals remaining, viz. <lb/></s>
  <s xml:id="echoid-s6" xml:space="preserve">only Euclid’s Data, and part of Apollonius’s Conics. </s>
  <s xml:id="echoid-s7" xml:space="preserve">The <lb/>loſs of the reſt is very much to be lamented by all Lovers of <lb/>the Mathematics. </s>
  <s xml:id="echoid-s8" xml:space="preserve">“ Valde quidem dolendum eſt quod re-<lb/>liqui tractatus Veterum Analytici, a Pappo memorati, <lb/>aut perierint, aut nondum lucem conſpexerint. </s>
  <s xml:id="echoid-s9" xml:space="preserve">Nam <lb/>minime dubito quin eorum nonnulli, Arabicè ſaltem <lb/>verſi, alicubi terrarum lateant, pulvere magis quam tene-<lb/>bris ſuis involuti. </s>
  <s xml:id="echoid-s10" xml:space="preserve">” Dr. </s>
  <s xml:id="echoid-s11" xml:space="preserve"><emph style="sc">Halley’s</emph> <emph style="sc">Preface</emph> to his <emph style="sc">Apol</emph>-<lb/><emph style="sc">LONIUS DE</emph> <emph style="sc">Sectione</emph> <emph style="sc">Rationis et</emph> <emph style="sc">Spatii</emph>.</s>
  <s xml:id="echoid-s12" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s13" xml:space="preserve"><emph style="sc">Some</emph> ingenious men have attempted, from the account of <lb/>them given by Pappus, to reſtore ſome of theſe loſt Trea-<lb/>tiſes. </s>
  <s xml:id="echoid-s14" xml:space="preserve">Snellius has endeavoured to give us the Books <emph style="sc">De</emph> <lb/><emph style="sc">Sectione</emph> <emph style="sc">Rationis</emph>, <emph style="sc">De</emph> <emph style="sc">Sectione</emph> <emph style="sc">Spatii</emph>, and <emph style="sc">De</emph> <emph style="sc">Sec</emph>-<lb/><emph style="sc">TIONE</emph> <emph style="sc">Determinata</emph>. </s>
  <s xml:id="echoid-s15" xml:space="preserve">Fermat and Schooten have laboured <lb/>in the Treatiſe <emph style="sc">De</emph> <emph style="sc">Locis</emph> <emph style="sc">Planis</emph>; </s>
  <s xml:id="echoid-s16" xml:space="preserve">and Marinus Ghetaldus <lb/>in that <emph style="sc">De</emph> <emph style="sc">Inclinationibus</emph>. </s>
  <s xml:id="echoid-s17" xml:space="preserve">But thoſe who have ſuc-<lb/>ceeded beſt, and done the moſt this way, are two incom-
<pb o="(iv)" file="0008" n="8"/>
parable Mathematicians of our own Country, Dr. </s>
  <s xml:id="echoid-s18" xml:space="preserve">Halley <lb/>and Dr. </s>
  <s xml:id="echoid-s19" xml:space="preserve">Simſon, to whom the World is very much obliged <lb/>for their Geometrical Labours. </s>
  <s xml:id="echoid-s20" xml:space="preserve">The firſt of theſe, from an <lb/>Arabic MS in the Bodleian Library, has reſtored the Books <lb/><emph style="sc">De</emph> <emph style="sc">Sectione</emph> <emph style="sc">Rationis</emph>; </s>
  <s xml:id="echoid-s21" xml:space="preserve">and from his own Sagacity ſup-<lb/>plied thoſe <emph style="sc">De</emph> <emph style="sc">Sectione</emph> <emph style="sc">Spatii</emph>: </s>
  <s xml:id="echoid-s22" xml:space="preserve">and the other has with <lb/>equal pains and ingenuity completed thoſe <emph style="sc">De</emph> <emph style="sc">Loc<unsure/>is</emph> <emph style="sc">Planis</emph>.</s>
  <s xml:id="echoid-s23" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s24" xml:space="preserve">As to the Treatiſe <emph style="sc">De</emph> <emph style="sc">Tactionibus</emph>, which I now give <lb/>the Engliſh Reader, it has been reſtored by Vieta under the <lb/>Title of Apollonius Gallus, and his Deficiencies ſupplied by <lb/>Marinus Ghetaldus. </s>
  <s xml:id="echoid-s25" xml:space="preserve">I have endeavoured to do Juſtice to my <lb/>Authors by all poſſible Care both in the Text and in the <lb/>Figures; </s>
  <s xml:id="echoid-s26" xml:space="preserve">and have added a few Propoſitions of my own, by <lb/>way of Supplement, in which I have propoſed Ghetaldus’s <lb/>Problems over again without a Determination, and have <lb/>found the <emph style="sc">Locus</emph> of the center of the circle required, which <lb/>I have not ſeen done before in any Author.</s>
  <s xml:id="echoid-s27" xml:space="preserve"/>
</p>
<pb o="(v)" file="0009" n="9"/>
</div>
<div xml:id="echoid-div5" type="section" level="1" n="5">
<head xml:id="echoid-head8" xml:space="preserve">EXTRACT from PAPPUS’s Preſace to his Seventh Book <lb/>in Dr. HALLEY’s Tranſlation.</head>
<head xml:id="echoid-head9" xml:space="preserve">DE TACTIONIBUS II.</head>
<p>
  <s xml:id="echoid-s28" xml:space="preserve">HIS ordine ſubnexi ſunt libri duo <emph style="sc">DE</emph> <emph style="sc">Tactionibus</emph>, in <lb/>quibus plures ineſſe propoſitiones videntur; </s>
  <s xml:id="echoid-s29" xml:space="preserve">ſed &amp; </s>
  <s xml:id="echoid-s30" xml:space="preserve">ex <lb/>his unam etiam faciemus, ad hunc modum ſe habentem. </s>
  <s xml:id="echoid-s31" xml:space="preserve">“ E <lb/>punctis rectis &amp; </s>
  <s xml:id="echoid-s32" xml:space="preserve">circulis, quibuſcunque tribus poſitione <lb/>datis, circulum ducere per ſingula data puncta, qui, ſi fieri <lb/>poſſit, contingat etiam datas lineas.</s>
  <s xml:id="echoid-s33" xml:space="preserve">” Ex hac autem ob mul-<lb/>titudinem in Hypotheſibus datorum, tam ſimilium quam diſſi-<lb/>milium <emph style="sc">GENERUM</emph>, fiunt neceſſario decem propoſitiones di-<lb/>verſæ; </s>
  <s xml:id="echoid-s34" xml:space="preserve">quia ex tribus diſſimilibus generibus fiunt diverſæ <lb/>triades inordinatæ numero decem. </s>
  <s xml:id="echoid-s35" xml:space="preserve">Data etenim eſſe poſſunt <lb/>vel tria puncta; </s>
  <s xml:id="echoid-s36" xml:space="preserve">vel tres rectæ; </s>
  <s xml:id="echoid-s37" xml:space="preserve">vel duo puncta &amp; </s>
  <s xml:id="echoid-s38" xml:space="preserve">recta; </s>
  <s xml:id="echoid-s39" xml:space="preserve">vel <lb/>duæ rectæ &amp; </s>
  <s xml:id="echoid-s40" xml:space="preserve">punctum; </s>
  <s xml:id="echoid-s41" xml:space="preserve">vel duo puncta &amp; </s>
  <s xml:id="echoid-s42" xml:space="preserve">circulus; </s>
  <s xml:id="echoid-s43" xml:space="preserve">vel duo <lb/>circuli &amp; </s>
  <s xml:id="echoid-s44" xml:space="preserve">punctum; </s>
  <s xml:id="echoid-s45" xml:space="preserve">vel duo circuli &amp; </s>
  <s xml:id="echoid-s46" xml:space="preserve">recta; </s>
  <s xml:id="echoid-s47" xml:space="preserve">vel punctum, <lb/>recta &amp; </s>
  <s xml:id="echoid-s48" xml:space="preserve">circulus; </s>
  <s xml:id="echoid-s49" xml:space="preserve">vel duæ rectæ &amp; </s>
  <s xml:id="echoid-s50" xml:space="preserve">circulus; </s>
  <s xml:id="echoid-s51" xml:space="preserve">vel tres circuli. <lb/></s>
  <s xml:id="echoid-s52" xml:space="preserve">Horum duo quidem prima problemata oſtenduntur in libro <lb/>quarto primorum Elementorum. </s>
  <s xml:id="echoid-s53" xml:space="preserve">Nam per tria data puncta, <lb/>quæ non ſint in linea recta, circulum ducere, idem eſt ac <lb/>circa datum triangulum circumſcribere. </s>
  <s xml:id="echoid-s54" xml:space="preserve">Problema autem in <lb/>tribus datis rectis non parallelis, ſed inter ſe occurrentibus, <lb/>idem eſt ac dato triangulo circulum inſcribere. </s>
  <s xml:id="echoid-s55" xml:space="preserve">Caſus vero <lb/>duarum rectarum parallelarum cum tertiâ occurrente, quaſi
<pb o="(vi)" file="0010" n="10"/>
pars eſſet ſecundæ ſubdiviſionis, cæteris permittitur. </s>
  <s xml:id="echoid-s56" xml:space="preserve">Deinde <lb/>proxima ſex problemata continentur in primo libro. </s>
  <s xml:id="echoid-s57" xml:space="preserve">Reliqua <lb/>duo, nempe de duabus rectis datis &amp; </s>
  <s xml:id="echoid-s58" xml:space="preserve">circulo, &amp; </s>
  <s xml:id="echoid-s59" xml:space="preserve">de tribus <lb/>datis circulis, ſola habentur in ſecundo libro; </s>
  <s xml:id="echoid-s60" xml:space="preserve">ob multas di-<lb/>verſaſque poſitiones circulorum &amp; </s>
  <s xml:id="echoid-s61" xml:space="preserve">rectarum inter ſe, quibus <lb/>fit ut etiam plurium determinationum opus ſit. </s>
  <s xml:id="echoid-s62" xml:space="preserve">Prædictis his <lb/>Tactionibus congener eſt ordo problematum, quæ ab edito-<lb/>ribus omiſſa fuerant. </s>
  <s xml:id="echoid-s63" xml:space="preserve">Nonnulli autem priori horum librorum <lb/>illa prefixerunt: </s>
  <s xml:id="echoid-s64" xml:space="preserve">Compendioſus enim &amp; </s>
  <s xml:id="echoid-s65" xml:space="preserve">introductorius erat <lb/>tractatus ille, &amp; </s>
  <s xml:id="echoid-s66" xml:space="preserve">ad plenam de Tactionibus doctrinam abſol-<lb/>vendam maxime idoneus. </s>
  <s xml:id="echoid-s67" xml:space="preserve">Hæc omnia rurſus una propoſitio <lb/>complectitur, quæ quidem quoad Hypotheſim magis quam <lb/>præcedentia contracta eſt, ſuperaddita autem eſt conditio ad <lb/>conſtructionem: </s>
  <s xml:id="echoid-s68" xml:space="preserve">eſtque hujuſmodi. </s>
  <s xml:id="echoid-s69" xml:space="preserve">“ E punctis, rectis, vel <lb/>circulis, datis duobus quibuſcunque, deſcribere circulum <lb/>magnitudine datum, qui tranſeat per punctum vel puncta <lb/>data, ac, ſi fieri poſſit, contingat etiam lineas datas. </s>
  <s xml:id="echoid-s70" xml:space="preserve">” Con-<lb/>tinet autem hæc propoſitio ſex problemata: </s>
  <s xml:id="echoid-s71" xml:space="preserve">ex tribus enim <lb/>quibuſcunque diverſis generibus fiunt Duades inordinatæ di-<lb/>verſæ numero ſex. </s>
  <s xml:id="echoid-s72" xml:space="preserve">Vel enim datis duobus punctis, vel duabus <lb/>rectis, vel duobus circulis, vel puncto &amp; </s>
  <s xml:id="echoid-s73" xml:space="preserve">rectâ, vel puncto &amp; </s>
  <s xml:id="echoid-s74" xml:space="preserve"><lb/>circuìo, vel rectâ &amp; </s>
  <s xml:id="echoid-s75" xml:space="preserve">circulo, opportet circulum magnitudine <lb/>datum deſcribere, <emph style="sc">QUI DATA CONTINGAT</emph>; </s>
  <s xml:id="echoid-s76" xml:space="preserve">hæc autem re-<lb/>ſolvenda ſunt &amp; </s>
  <s xml:id="echoid-s77" xml:space="preserve">componenda ut &amp; </s>
  <s xml:id="echoid-s78" xml:space="preserve">determinanda juxta Caſus. <lb/></s>
  <s xml:id="echoid-s79" xml:space="preserve">Liber primus <emph style="sc">Tactionum</emph> problemata habet ſeptem; </s>
  <s xml:id="echoid-s80" xml:space="preserve">ſe-<lb/>cundus vero quatuor. </s>
  <s xml:id="echoid-s81" xml:space="preserve">Lemmata autem ad utrumque librum <lb/>ſunt XXI; </s>
  <s xml:id="echoid-s82" xml:space="preserve">Theoremata LX.</s>
  <s xml:id="echoid-s83" xml:space="preserve"/>
</p>
<pb o="(vii)" file="0011" n="11"/>
</div>
<div xml:id="echoid-div6" type="section" level="1" n="6">
<head xml:id="echoid-head10" xml:space="preserve"><emph style="sc">Synopsis</emph> of the PROBLEMS.</head>
<note position="right" xml:space="preserve"> <lb/>Prob. 1. ● ● # Prob. 2. | | # Prob. 3. ○ ○ <lb/>4. ● | # 6. | ● <lb/>5. ● ○ <lb/>Euclid ● ● ● # Euclid | | | # Prob. 14. ○ ○ ○ <lb/>Prob. 7. ● ● | # Prob. 9. | | ○ <lb/>12. ● ● ○ # 11. | ○ ○ <lb/>8. ● | | <lb/>13. ● ○ ○ <lb/>10. ● ○ ○ <lb/></note>
<pb file="0012" n="12"/>
<pb file="0013" n="13"/>
</div>
<div xml:id="echoid-div7" type="section" level="1" n="7">
<head xml:id="echoid-head11" xml:space="preserve">PROBLEMS <lb/>CONCERNING <lb/>TANGENCIES.</head>
<head xml:id="echoid-head12" xml:space="preserve">PROBLEM I.</head>
<p>
  <s xml:id="echoid-s84" xml:space="preserve">THROUGH two given points A and B to deſcribe a circle whoſe radius <lb/>ſhall be equal to a given line Z.</s>
  <s xml:id="echoid-s85" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s86" xml:space="preserve"><emph style="sc">Limitation</emph>. </s>
  <s xml:id="echoid-s87" xml:space="preserve">2 Z muſt not be leſs than the diſtance of the points A and B.</s>
  <s xml:id="echoid-s88" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s89" xml:space="preserve"><emph style="sc">Construction</emph>. </s>
  <s xml:id="echoid-s90" xml:space="preserve">With the centers A and B, and diſtance Z, deſcribe two <lb/>arcs cutting or touching one another in the point E, ( which they will neceſſarily <lb/>do by the Limitation ) and E will be the center of the circle required.</s>
  <s xml:id="echoid-s91" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div8" type="section" level="1" n="8">
<head xml:id="echoid-head13" xml:space="preserve">PROBLEM II.</head>
<p>
  <s xml:id="echoid-s92" xml:space="preserve"><emph style="sc">Having</emph> two right lines AB CD given in poſition, it is required to draw 2 <lb/>circle, whoſe Radius ſhall be equal to the given line Z, which ſhall alſo touch <lb/>both the given lines.</s>
  <s xml:id="echoid-s93" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s94" xml:space="preserve"><emph style="sc">Case</emph> 1ſt. </s>
  <s xml:id="echoid-s95" xml:space="preserve">Suppoſe AB and CD to be parallel.</s>
  <s xml:id="echoid-s96" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s97" xml:space="preserve"><emph style="sc">Limitation</emph>. </s>
  <s xml:id="echoid-s98" xml:space="preserve">2Z muſt be equal to the diſtance of the parallels, and the <lb/>conſtruction is evident.</s>
  <s xml:id="echoid-s99" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s100" xml:space="preserve"><emph style="sc">Case</emph> 2d. </s>
  <s xml:id="echoid-s101" xml:space="preserve">Suppoſe AB and CD to be inclined to each other, let them be <lb/>produced till they meet in E, and let the angle BED be biſected by EH, and <lb/>through E draw EF perpendicular to ED, and equal to the given line Z; </s>
  <s xml:id="echoid-s102" xml:space="preserve">through <lb/>F draw FG parallel to EH, meeting ED in G, and through G draw GH paral-<lb/>lel to EF. </s>
  <s xml:id="echoid-s103" xml:space="preserve">I ſay that the circle deſcribed with H center, and HG radius, <lb/>touches the two given lines: </s>
  <s xml:id="echoid-s104" xml:space="preserve">it touches CD, becauſe EFGH is a Parallelogram,
<pb o="(2)" file="0014" n="14"/>
and hence the angle FEG is equal to EGH, but FEG is a right one by Con-<lb/>ſtruction. </s>
  <s xml:id="echoid-s105" xml:space="preserve">Let now HI be drawn from H perpendicular to AB: </s>
  <s xml:id="echoid-s106" xml:space="preserve">then the two <lb/>triangles EHI and EHG having two angles in one HEI and EIH reſpectively <lb/>equal to two angles in the other HEG and EGH, and alſo the ſide EH com-<lb/>mon, by Euc. </s>
  <s xml:id="echoid-s107" xml:space="preserve">I. </s>
  <s xml:id="echoid-s108" xml:space="preserve">26. </s>
  <s xml:id="echoid-s109" xml:space="preserve">HI will be equal to HG, and therefore the circle will <lb/>touch alſo the other line AB: </s>
  <s xml:id="echoid-s110" xml:space="preserve">and HG or HI equals the given line Z, becauſe <lb/>EF was made equal to Z, and HG and EF are oppoſite ſides of a paral-<lb/>lelogram.</s>
  <s xml:id="echoid-s111" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div9" type="section" level="1" n="9">
<head xml:id="echoid-head14" xml:space="preserve">PROBLEM III.</head>
<p>
  <s xml:id="echoid-s112" xml:space="preserve"><emph style="sc">Having</emph> two circles given whoſe centers are A and B, it is required to draw <lb/>another, whoſe Radius ſhall be equal to a given line Z, which ſhall alſo touch <lb/>the two given ones.</s>
  <s xml:id="echoid-s113" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s114" xml:space="preserve"><emph style="sc">This</emph> Problem has various Caſes, according to the various poſition of the <lb/>given circles, and the various manner of deſcribing the circle required: </s>
  <s xml:id="echoid-s115" xml:space="preserve">but there <lb/>are ſix principal ones, and to the conditions of theſe all the reſt are ſubject.</s>
  <s xml:id="echoid-s116" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s117" xml:space="preserve"><emph style="sc">Case</emph> 1ſt. </s>
  <s xml:id="echoid-s118" xml:space="preserve">Let the circle to be deſcribed be required to be touched outwardly <lb/>by the given circles.</s>
  <s xml:id="echoid-s119" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s120" xml:space="preserve"><emph style="sc">Limitation</emph>. </s>
  <s xml:id="echoid-s121" xml:space="preserve">Then it is neceſſary that 2Z, or the given Diameter, ſhould <lb/>not be leſs than the ſegment of the line joining the centers of the given circles <lb/>which is intercepted between their convex circumferences, viz. </s>
  <s xml:id="echoid-s122" xml:space="preserve">not leſs than CD <lb/>in the Figure belonging to Caſe 1ſt.</s>
  <s xml:id="echoid-s123" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s124" xml:space="preserve"><emph style="sc">Case</emph> 2d. </s>
  <s xml:id="echoid-s125" xml:space="preserve">Let the circle to be deſcribed be required to be touched inwardly by <lb/>the given circles.</s>
  <s xml:id="echoid-s126" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s127" xml:space="preserve"><emph style="sc">Limitation</emph>. </s>
  <s xml:id="echoid-s128" xml:space="preserve">Then it’s Diameter muſt not be given leſs than the right line, <lb/>which drawn through the centers of the given circles, is contained between their <lb/>concave circumferences; </s>
  <s xml:id="echoid-s129" xml:space="preserve">viz. </s>
  <s xml:id="echoid-s130" xml:space="preserve">not leſs than CD.</s>
  <s xml:id="echoid-s131" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s132" xml:space="preserve"><emph style="sc">Case</emph> 3d. </s>
  <s xml:id="echoid-s133" xml:space="preserve">Let the circle to be deſcribed be required to be touched outwardly <lb/>by one of the given circles, and inwardly by the other.</s>
  <s xml:id="echoid-s134" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s135" xml:space="preserve"><emph style="sc">Limitation</emph>. </s>
  <s xml:id="echoid-s136" xml:space="preserve">Then it’s Diameter muſt not be given leſs than the ſegment <lb/>of the right line, joining the centers of the given circles, which is intercepted <lb/>between the convex circumference of one and the concave circumference of the <lb/>other; </s>
  <s xml:id="echoid-s137" xml:space="preserve">viz. </s>
  <s xml:id="echoid-s138" xml:space="preserve">not leſs than CD.</s>
  <s xml:id="echoid-s139" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s140" xml:space="preserve"><emph style="sc">Case</emph> 4th. </s>
  <s xml:id="echoid-s141" xml:space="preserve">Let one of the given circles include the other, and let it be re-<lb/>quired that the circle to be deſcribed be touched outwardly by them both.</s>
  <s xml:id="echoid-s142" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s143" xml:space="preserve"><emph style="sc">Limitation</emph>. </s>
  <s xml:id="echoid-s144" xml:space="preserve">Then it’s Diameter muſt not be given greater than the greater <lb/>ſegment of the right line, joining the centers of the given circles, which is in-<lb/>tercepted between the concave circumference of one and the convex circumference <lb/>of the other; </s>
  <s xml:id="echoid-s145" xml:space="preserve">nor leſs than the leſſer ſegment; </s>
  <s xml:id="echoid-s146" xml:space="preserve">viz. </s>
  <s xml:id="echoid-s147" xml:space="preserve">not greater than CD, nor <lb/>leſs than MN.</s>
  <s xml:id="echoid-s148" xml:space="preserve"/>
</p>
<pb o="(3)" file="0015" n="15"/>
<p>
  <s xml:id="echoid-s149" xml:space="preserve"><emph style="sc">Case</emph> 5th. </s>
  <s xml:id="echoid-s150" xml:space="preserve">Let one of the given circles include the other, and let be required <lb/>that the circle to be deſcribed be touched outwardly by one of the given circles, <lb/>and inwardly by the other.</s>
  <s xml:id="echoid-s151" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s152" xml:space="preserve"><emph style="sc">Limitation</emph>. </s>
  <s xml:id="echoid-s153" xml:space="preserve">Then it’s Diameter muſt not be given greater than the greater <lb/>ſegment of the right line, joining the centers of the given circles, which is in-<lb/>tercepted between the two concave circumferences of the ſaid circles, nor leſs <lb/>than the leſſer ſegment; </s>
  <s xml:id="echoid-s154" xml:space="preserve">viz. </s>
  <s xml:id="echoid-s155" xml:space="preserve">not greater than CD, nor leſs than MN.</s>
  <s xml:id="echoid-s156" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s157" xml:space="preserve"><emph style="sc">Case</emph> 6th. </s>
  <s xml:id="echoid-s158" xml:space="preserve">Let the two given circles cut each other, and let it be required <lb/>that the circle to be deſcribed, and to be touched by them both, ſhall alſo be <lb/>included in each of them.</s>
  <s xml:id="echoid-s159" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s160" xml:space="preserve"><emph style="sc">Limitation</emph>. </s>
  <s xml:id="echoid-s161" xml:space="preserve">Then it’s Diameter muſt not be given greater than the ſeg-<lb/>ment of the right line, joining the centers of the given circles, intercepted by <lb/>their concave circumferences, which lies in the ſpace common to both the given <lb/>circles; </s>
  <s xml:id="echoid-s162" xml:space="preserve">viz. </s>
  <s xml:id="echoid-s163" xml:space="preserve">not greater than CD.</s>
  <s xml:id="echoid-s164" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s165" xml:space="preserve"><emph style="sc">There</emph> may be alſo three other Caſes of this Problem, when the given circles <lb/>cut each other; </s>
  <s xml:id="echoid-s166" xml:space="preserve">but becauſe they are ſimilar to the 1ſt, 2d, and 4th Caſes already <lb/>propoſed, and ſubject to juſt the ſame Limitations; </s>
  <s xml:id="echoid-s167" xml:space="preserve">except that which is ſimilar <lb/>to the 1ſt, which is ſubject to no Limitation at all, they are here omitted; </s>
  <s xml:id="echoid-s168" xml:space="preserve">as are <lb/>likewiſe thoſe Caſes where the given circles touch each other; </s>
  <s xml:id="echoid-s169" xml:space="preserve">becauſe they are <lb/>the ſame as the preceding, and ſolved in the ſame manner.</s>
  <s xml:id="echoid-s170" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div10" type="section" level="1" n="10">
<head xml:id="echoid-head15" xml:space="preserve"><emph style="sc">The</emph> <emph style="sc">GENERAL</emph> <emph style="sc">Solution</emph>.</head>
<p>
  <s xml:id="echoid-s171" xml:space="preserve">Join the given centers A and B, and where the Caſe requires, let AB be pro-<lb/>duced to meet the given circumferences in C and D: </s>
  <s xml:id="echoid-s172" xml:space="preserve">and let CI and DH be <lb/>taken equal to the given line Z: </s>
  <s xml:id="echoid-s173" xml:space="preserve">and let two circles be deſcribed; </s>
  <s xml:id="echoid-s174" xml:space="preserve">one with cen-<lb/>ter A and diſtance AI, and the other with center B and diſtance BH: </s>
  <s xml:id="echoid-s175" xml:space="preserve">and theſe <lb/>two circles will neceſſarily cut or touch each other by the Limitations given. </s>
  <s xml:id="echoid-s176" xml:space="preserve">Let <lb/>the point of concourſe be E: </s>
  <s xml:id="echoid-s177" xml:space="preserve">from E draw the right line EAF cutting the circle <lb/>whoſe center is A in F; </s>
  <s xml:id="echoid-s178" xml:space="preserve">as alſo EBG cutting the circle whoſe center is B in G: <lb/></s>
  <s xml:id="echoid-s179" xml:space="preserve">then with center E and diſtance EF deſcribe a circle FK, this will be the circle <lb/>required: </s>
  <s xml:id="echoid-s180" xml:space="preserve">becauſe AF and AC are equal as alſo AI and AE; </s>
  <s xml:id="echoid-s181" xml:space="preserve">therefore FE <lb/>and CI are alſo equal: </s>
  <s xml:id="echoid-s182" xml:space="preserve">but CI was made equal to Z, therefore FE is equal to <lb/>Z. </s>
  <s xml:id="echoid-s183" xml:space="preserve">Again, becauſe BD and BG are equal, as alſo BH and BE, therefore DH <lb/>and EG are alſo equal: </s>
  <s xml:id="echoid-s184" xml:space="preserve">but DH was made equal to Z, therefore EG is equal to <lb/>Z. </s>
  <s xml:id="echoid-s185" xml:space="preserve">Hence it appears that the circle FK, paſſing through F will alſo paſs thro’ <lb/>G, and likewiſe that it will alſo touch the given circles in F and G, becauſe EAF <lb/>and EBG are right lines paſſing through the centers.</s>
  <s xml:id="echoid-s186" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div11" type="section" level="1" n="11">
<head xml:id="echoid-head16" xml:space="preserve">PROBLEM IV.</head>
<p>
  <s xml:id="echoid-s187" xml:space="preserve"><emph style="sc">Having</emph> a given point A, and a given right line BC, it is required to draw a <lb/>circle, whoſe Radius ſhall be equal to a given line Z, which ſhall paſs through <lb/>the given point, and alſo touch the given line.</s>
  <s xml:id="echoid-s188" xml:space="preserve"/>
</p>
<pb o="(4)" file="0016" n="16"/>
<p>
  <s xml:id="echoid-s189" xml:space="preserve"><emph style="sc">Limitation</emph>. </s>
  <s xml:id="echoid-s190" xml:space="preserve">2Z muſt not be given leſs than the perpendicular let fall from <lb/>the given point A upon the given line BC.</s>
  <s xml:id="echoid-s191" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s192" xml:space="preserve"><emph style="sc">From</emph> the point A let AD be drawn perpendicular to BC, and in this perpen-<lb/>dicular take DE equal to the given line Z: </s>
  <s xml:id="echoid-s193" xml:space="preserve">and through E draw EF parallel to <lb/>BC, and from A upon this line EF ſet off AF equal to Z, which may be done, <lb/>for by the Limitation Z is not leſs than AE: </s>
  <s xml:id="echoid-s194" xml:space="preserve">then with center F and diſtance <lb/>FA deſcribe a circle, and I ſay it will touch the line BC: </s>
  <s xml:id="echoid-s195" xml:space="preserve">for through F drawing <lb/>FG parallel to AD, FGDE will be a Parallelogram, and FG will be equal to <lb/>DE, that is to Z, and at right angles to BC.</s>
  <s xml:id="echoid-s196" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div12" type="section" level="1" n="12">
<head xml:id="echoid-head17" xml:space="preserve">PROBLEM V.</head>
<p>
  <s xml:id="echoid-s197" xml:space="preserve"><emph style="sc">Having</emph> a given point A, and alſo a given circle whoſe center is B, it is re-<lb/>quired to draw a circle whoſe Radius ſhall be equal to a given line Z, which ſhall <lb/>paſs through the given point, and alſo touch the given circle.</s>
  <s xml:id="echoid-s198" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s199" xml:space="preserve"><emph style="sc">This</emph> Problem has three Caſes, each of which is ſubject to a Limitation.</s>
  <s xml:id="echoid-s200" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s201" xml:space="preserve"><emph style="sc">Case</emph> Iſt. </s>
  <s xml:id="echoid-s202" xml:space="preserve">Let the circle to be doſcribed be required to be touched outwardly <lb/>by the given circle.</s>
  <s xml:id="echoid-s203" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s204" xml:space="preserve"><emph style="sc">Limitation</emph>. </s>
  <s xml:id="echoid-s205" xml:space="preserve">Then the Diameter muſt not be given leſs than the ſegment <lb/>of the right line, joining the given point and the center of the given circle, <lb/>which is intercepted between the given point and the convex circumference; </s>
  <s xml:id="echoid-s206" xml:space="preserve">viz. <lb/></s>
  <s xml:id="echoid-s207" xml:space="preserve">not leſs than AC.</s>
  <s xml:id="echoid-s208" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s209" xml:space="preserve"><emph style="sc">Case</emph> 2d. </s>
  <s xml:id="echoid-s210" xml:space="preserve">Let the circle to be deſcribed be required to be touched inwardly <lb/>by the given circle.</s>
  <s xml:id="echoid-s211" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s212" xml:space="preserve"><emph style="sc">Limitation</emph>. </s>
  <s xml:id="echoid-s213" xml:space="preserve">Then the Diameter muſt not be given leſs than the right line <lb/>which, drawn from the given point through the center of the given circle, is con-<lb/>tained between the given point and the concave circumſerence; </s>
  <s xml:id="echoid-s214" xml:space="preserve">viz. </s>
  <s xml:id="echoid-s215" xml:space="preserve">not leſs <lb/>than AC.</s>
  <s xml:id="echoid-s216" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s217" xml:space="preserve"><emph style="sc">Case</emph> 3d. </s>
  <s xml:id="echoid-s218" xml:space="preserve">Let the given point lie in the given circle.</s>
  <s xml:id="echoid-s219" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s220" xml:space="preserve"><emph style="sc">Limitation</emph>. </s>
  <s xml:id="echoid-s221" xml:space="preserve">Then a diameter of the given circle being drawn through the <lb/>given point, it is divided into two ſegments by the ſaid point, and the Diameter <lb/>of the circle required muſt not be given greater than the greater of them, nor <lb/>leſs than the leſſer; </s>
  <s xml:id="echoid-s222" xml:space="preserve">viz. </s>
  <s xml:id="echoid-s223" xml:space="preserve">not greater than AC, nor leſs than AG.</s>
  <s xml:id="echoid-s224" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div13" type="section" level="1" n="13">
<head xml:id="echoid-head18" xml:space="preserve"><emph style="sc">The general</emph> <emph style="sc">Solution</emph>.</head>
<p>
  <s xml:id="echoid-s225" xml:space="preserve"><emph style="sc">Let</emph> A and B be joined, and in the line AB take CF equal to Z, and then <lb/>with center A and diſtance Z, let an arc be drawn, and with center B, and <lb/>diſtance BF let another be drawn, which by the Limitations will neceſſarily <lb/>either touch or cut the former; </s>
  <s xml:id="echoid-s226" xml:space="preserve">let the point of their concourſe be D; </s>
  <s xml:id="echoid-s227" xml:space="preserve">then with <lb/>D center and DA diſtance let a circle be drawn, and I ſay it will touch the <lb/>given circle whoſe center is B: </s>
  <s xml:id="echoid-s228" xml:space="preserve">for DB being drawn meeting the circumference <lb/>of the circle whoſe center is B in E, BC is equal to BE, and hence CF equals <lb/>ED, and they are both equal to the given line Z.</s>
  <s xml:id="echoid-s229" xml:space="preserve"/>
</p>
<pb o="(5)" file="0017" n="17"/>
</div>
<div xml:id="echoid-div14" type="section" level="1" n="14">
<head xml:id="echoid-head19" xml:space="preserve">PROBLEM VI.</head>
<p>
  <s xml:id="echoid-s230" xml:space="preserve"><emph style="sc">Having</emph> a right line given BC, and alſo a circle whoſe center is A, it is re-<lb/>quired to draw another circle, whoſe Radius ſhall be equal to a given right line <lb/>Z, and which ſhall touch both the given line and alſo the given circle.</s>
  <s xml:id="echoid-s231" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s232" xml:space="preserve"><emph style="sc">This</emph> Problem has alſo three caſes, each of which is ſubject to a Limitation.</s>
  <s xml:id="echoid-s233" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s234" xml:space="preserve"><emph style="sc">Case</emph> Iſt, Let the circle to be deſcribed be required to be touched outwardly <lb/>by the given circle.</s>
  <s xml:id="echoid-s235" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s236" xml:space="preserve"><emph style="sc">Limitation</emph>. </s>
  <s xml:id="echoid-s237" xml:space="preserve">Then the Diameter of the circle required muſt not be given <lb/>leſs than the ſegment of a line, drawn from the center of the given circle, per-<lb/>pendicular to the given line, which is intercepted between the ſaid line and the <lb/>convex circumference; </s>
  <s xml:id="echoid-s238" xml:space="preserve">viz. </s>
  <s xml:id="echoid-s239" xml:space="preserve">not leſs than BD.</s>
  <s xml:id="echoid-s240" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s241" xml:space="preserve"><emph style="sc">Case</emph> 2d. </s>
  <s xml:id="echoid-s242" xml:space="preserve">Let the circle to be deſcribed be required to be touched inwardly by <lb/>the given circle.</s>
  <s xml:id="echoid-s243" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s244" xml:space="preserve"><emph style="sc">Limitation</emph>. </s>
  <s xml:id="echoid-s245" xml:space="preserve">Then the given line muſt not be in the given circle, neither <lb/>muſt the Diameter of the circle required be given leſs than that portion of the <lb/>perpendicular, drawn from the center of the given circle to the given line, which <lb/>is intercepted between the ſaid line and the concave circumference; </s>
  <s xml:id="echoid-s246" xml:space="preserve">viz. </s>
  <s xml:id="echoid-s247" xml:space="preserve">not leſs <lb/>than BD.</s>
  <s xml:id="echoid-s248" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s249" xml:space="preserve"><emph style="sc">Case</emph> 3d. </s>
  <s xml:id="echoid-s250" xml:space="preserve">Let the circle to be deſcribed be required to be both touched and <lb/>included in the given circle.</s>
  <s xml:id="echoid-s251" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s252" xml:space="preserve"><emph style="sc">Limitation</emph>. </s>
  <s xml:id="echoid-s253" xml:space="preserve">Then the given right line muſt be in the given circle, and <lb/>when a Diameter of this given circle is drawn cutting the given line at right an-<lb/>gles, the Diameter of the circle required muſt not be given greater than the <lb/>greater ſegment ; </s>
  <s xml:id="echoid-s254" xml:space="preserve">viz. </s>
  <s xml:id="echoid-s255" xml:space="preserve">not greater than BD.</s>
  <s xml:id="echoid-s256" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div15" type="section" level="1" n="15">
<head xml:id="echoid-head20" xml:space="preserve"><emph style="sc">The general</emph> <emph style="sc">Solution</emph>.</head>
<p>
  <s xml:id="echoid-s257" xml:space="preserve"><emph style="sc">From</emph> A draw AB perpendicular to BC, cutting the given circumference in D; <lb/></s>
  <s xml:id="echoid-s258" xml:space="preserve">and in this perpendicular let BG and DF be taken each equal to the given line <lb/>Z; </s>
  <s xml:id="echoid-s259" xml:space="preserve">and through G draw GE parallel to BC; </s>
  <s xml:id="echoid-s260" xml:space="preserve">and with center A and diſtance <lb/>AF let an arc be ſtruck, which by the Limitations will neceſſarily either touch <lb/>or cut GE; </s>
  <s xml:id="echoid-s261" xml:space="preserve">let the point of concourſe be E, let AE be joined, and, if neceſſary, <lb/>be produced to meet the given circumference in H; </s>
  <s xml:id="echoid-s262" xml:space="preserve">then with E center and <lb/>EH diſtance deſcribe a circle, and I ſay it will be the required circle; </s>
  <s xml:id="echoid-s263" xml:space="preserve">it is evi-<lb/>dent it will touch the given circle: </s>
  <s xml:id="echoid-s264" xml:space="preserve">and becauſe AD and AH are equal, as alſo <lb/>AF and AE, therefore DF (which was made equal to Z) will be equal to HE: </s>
  <s xml:id="echoid-s265" xml:space="preserve"><lb/>let now EC be drawn perpendicular to BC, then GBCE will be a Parallelogram, <lb/>and EC will be equal to GB, which was alſo made equal to Z: </s>
  <s xml:id="echoid-s266" xml:space="preserve">hence the <lb/>circle will alſo touch the given line BC, becauſe the angle ECB is a right <lb/>one.</s>
  <s xml:id="echoid-s267" xml:space="preserve"/>
</p>
<pb o="(6)" file="0018" n="18"/>
</div>
<div xml:id="echoid-div16" type="section" level="1" n="16">
<head xml:id="echoid-head21" xml:space="preserve">PROBLEM VII.</head>
<p>
  <s xml:id="echoid-s268" xml:space="preserve"><emph style="sc">Having</emph> two points A and B given in poſition, and likewiſe a right line EF <lb/>given in poſition, it is required to find the center of a circle, which ſhall paſs <lb/>through the given points and touch the given line.</s>
  <s xml:id="echoid-s269" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s270" xml:space="preserve"><emph style="sc">Case</emph> Iſt. </s>
  <s xml:id="echoid-s271" xml:space="preserve">When the points A and B are joined, ſuppoſe AB to be parallel to <lb/>EF: </s>
  <s xml:id="echoid-s272" xml:space="preserve">then biſecting AB in D, and through D drawing DC perpendicular to it, <lb/>DC will alſo be perpendicular to EF: </s>
  <s xml:id="echoid-s273" xml:space="preserve">draw a circle therefore which will paſs <lb/>through the three points A, B, andC, (by Euc. </s>
  <s xml:id="echoid-s274" xml:space="preserve">IV. </s>
  <s xml:id="echoid-s275" xml:space="preserve">5.) </s>
  <s xml:id="echoid-s276" xml:space="preserve">and it will be the circle <lb/>required: </s>
  <s xml:id="echoid-s277" xml:space="preserve">(by a Corollary from Euc. </s>
  <s xml:id="echoid-s278" xml:space="preserve">III. </s>
  <s xml:id="echoid-s279" xml:space="preserve">1. </s>
  <s xml:id="echoid-s280" xml:space="preserve">and another from Euc. <lb/></s>
  <s xml:id="echoid-s281" xml:space="preserve">III. </s>
  <s xml:id="echoid-s282" xml:space="preserve">16.)</s>
  <s xml:id="echoid-s283" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s284" xml:space="preserve"><emph style="sc">Case</emph> 2d. </s>
  <s xml:id="echoid-s285" xml:space="preserve">Suppoſe AB not parallel to EF, but being produced meets it in <lb/>E: </s>
  <s xml:id="echoid-s286" xml:space="preserve">then from EF take the line EC ſuch, that its ſquare may be equal to the <lb/>rectangle BEA, and through the points A, B, C, deſcribe a circle, and it will be <lb/>the circle required by Euc. </s>
  <s xml:id="echoid-s287" xml:space="preserve">III. </s>
  <s xml:id="echoid-s288" xml:space="preserve">37.</s>
  <s xml:id="echoid-s289" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s290" xml:space="preserve"><emph style="sc">This</emph> is Vieta’s Solution. </s>
  <s xml:id="echoid-s291" xml:space="preserve">But Mr. </s>
  <s xml:id="echoid-s292" xml:space="preserve">Thomas Simpſon having conſtructed this, <lb/>and ſome of the following, both in the Collection of Problems at the end of his <lb/>Algebra, and alſo among thoſe at the end of his Elements of Geometry, I ſhall <lb/>add one of his Conſtructions.</s>
  <s xml:id="echoid-s293" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s294" xml:space="preserve"><emph style="sc">Let</emph> A and B be the points given, and CD the given line: </s>
  <s xml:id="echoid-s295" xml:space="preserve">drawing AB and <lb/>biſecting it in F, through E let EF be drawn perpendicular to AB and meeting <lb/>CD in F: </s>
  <s xml:id="echoid-s296" xml:space="preserve">and from any point H in EF draw HG perpendicular to CD, and <lb/>having drawn BF, to the ſame apply HI = HG, and parallel thereto draw BK <lb/>meeting EF in K: </s>
  <s xml:id="echoid-s297" xml:space="preserve">then with center K and radius BK let a circle be deſcribed, <lb/>and the thing is done: </s>
  <s xml:id="echoid-s298" xml:space="preserve">join KA, and draw KL perpendicular to CD, then be-<lb/>cauſe of the parallel lines, HG: </s>
  <s xml:id="echoid-s299" xml:space="preserve">HI:</s>
  <s xml:id="echoid-s300" xml:space="preserve">: KL: </s>
  <s xml:id="echoid-s301" xml:space="preserve">KB; </s>
  <s xml:id="echoid-s302" xml:space="preserve">whence as HG and HI are <lb/>equal, KL and KB are likewiſe equal. </s>
  <s xml:id="echoid-s303" xml:space="preserve">But it is evident from the Conſtruction <lb/>that KA = KB, therefore KB = KL = KA.</s>
  <s xml:id="echoid-s304" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s305" xml:space="preserve"><emph style="sc">Because</emph> two equal lines HI and Hi may be applied from H to BF each <lb/>equal to HG, the Problem will therefore admit of two Solutions, as the Figure <lb/>ſhews: </s>
  <s xml:id="echoid-s306" xml:space="preserve">except in the caſe when one of the given points, A for inſtance, is given <lb/>in the line CD, for then the Problem becomes more ſimple, and admits but of <lb/>one conſtruction, as the center of the circle required muſt be in the line EF con-<lb/>tinued, as alſo in the perpendicular raiſed from A to CD, and therefore in their <lb/>common interſection: </s>
  <s xml:id="echoid-s307" xml:space="preserve">and this is the limit of poſſibility; </s>
  <s xml:id="echoid-s308" xml:space="preserve">for ſhould the line CD <lb/>paſs between the given points, the Problem is impoſſible.</s>
  <s xml:id="echoid-s309" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s310" xml:space="preserve">N. </s>
  <s xml:id="echoid-s311" xml:space="preserve">B. </s>
  <s xml:id="echoid-s312" xml:space="preserve"><emph style="sc">Tho</emph>’ Vieta does not take notice that this Problem is capable of two <lb/>anſwers, yet this is as evident from his conſtruction, as from Mr. </s>
  <s xml:id="echoid-s313" xml:space="preserve">Simpſon’s, for <lb/>EC (the mean proportional between E B and EA) may be ſet off upon the given <lb/>line EF either way from the given point E.</s>
  <s xml:id="echoid-s314" xml:space="preserve"/>
</p>
<pb o="(7)" file="0019" n="19"/>
</div>
<div xml:id="echoid-div17" type="section" level="1" n="17">
<head xml:id="echoid-head22" xml:space="preserve">LEMMA I.</head>
<p>
  <s xml:id="echoid-s315" xml:space="preserve"><emph style="sc">Apoint</emph> A being given between the two right lines BC and DE, it is requir-<lb/>ed through the point A to draw a line cutting the two given ones at equal <lb/>angles.</s>
  <s xml:id="echoid-s316" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s317" xml:space="preserve"><emph style="sc">If</emph> the given lines be parallel, then a perpendicular to them through the point <lb/>A is the line required. </s>
  <s xml:id="echoid-s318" xml:space="preserve">But if not, then let them be produced to meet in the <lb/>point F: </s>
  <s xml:id="echoid-s319" xml:space="preserve">and let FG be drawn biſecting the Angle BFD, and through A draw <lb/>a perpendicular to FG, and it will be the line required by Euc. </s>
  <s xml:id="echoid-s320" xml:space="preserve">I. </s>
  <s xml:id="echoid-s321" xml:space="preserve">26.</s>
  <s xml:id="echoid-s322" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div18" type="section" level="1" n="18">
<head xml:id="echoid-head23" xml:space="preserve">PROBLEM VIII.</head>
<p>
  <s xml:id="echoid-s323" xml:space="preserve"><emph style="sc">Having</emph> a point A given, and alſo two right lines BC and DE, to draw a circle <lb/>which ſhall paſs through the given point, and touch both the given right lines.</s>
  <s xml:id="echoid-s324" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s325" xml:space="preserve"><emph style="sc">By</emph> the preceding Lemma draw a line IAH’ through the point A, which ſhall <lb/>make equal angles with the two given lines BC and DE: </s>
  <s xml:id="echoid-s326" xml:space="preserve">biſect IH in K; </s>
  <s xml:id="echoid-s327" xml:space="preserve">and <lb/>taking KL = KA, by means of the preceeding Problem draw a circle which ſhall <lb/>paſs through the points A and L, and likewiſe touch one of the given lines, BC <lb/>for inſtance, in the point M. </s>
  <s xml:id="echoid-s328" xml:space="preserve">I ſay this circle will alſo touch the other given <lb/>line DE: </s>
  <s xml:id="echoid-s329" xml:space="preserve">for from the center N letting fall the perpendicular NO, and joining <lb/>NI, NH, NM; </s>
  <s xml:id="echoid-s330" xml:space="preserve">in the triangles NKH, NKI, NK being common, and HK = <lb/>KI, and the Angles at K right ones, by Euc. </s>
  <s xml:id="echoid-s331" xml:space="preserve">I. </s>
  <s xml:id="echoid-s332" xml:space="preserve">4. </s>
  <s xml:id="echoid-s333" xml:space="preserve">NH = NI likewiſe the <lb/>angle NHK = angle NIK, from hence it follows: </s>
  <s xml:id="echoid-s334" xml:space="preserve">that the angle NHM = angle <lb/>NIO; </s>
  <s xml:id="echoid-s335" xml:space="preserve">and the angles at M and O, being both right, and NH being proved <lb/>equal to NI, NM will be equal to NO by Euc. </s>
  <s xml:id="echoid-s336" xml:space="preserve">I. </s>
  <s xml:id="echoid-s337" xml:space="preserve">26.</s>
  <s xml:id="echoid-s338" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div19" type="section" level="1" n="19">
<head xml:id="echoid-head24" xml:space="preserve">Mr. Simpſon conſtructs the Problem thus.</head>
<p>
  <s xml:id="echoid-s339" xml:space="preserve"><emph style="sc">Let</emph> BD and BC be the given lines meeting in B, and A the given point, <lb/>join AB, and draw BN biſecting the given angle DBC: </s>
  <s xml:id="echoid-s340" xml:space="preserve">and from any point E <lb/>in BN upon BC let fall the perpendicular EF, and to BA apply EG = EF, pa-<lb/>rallel to which draw AH meeting BN in H: </s>
  <s xml:id="echoid-s341" xml:space="preserve">then from center H with Interval <lb/>AH let a circle be deſcribed, and the thing is done. </s>
  <s xml:id="echoid-s342" xml:space="preserve">Upon BC and BD let fall <lb/>the perpendiculars HI, HK, which are manifeſtly equal, becauſe by Conſtruction <lb/>the angle HBI = HBK; </s>
  <s xml:id="echoid-s343" xml:space="preserve">moreover as EF: </s>
  <s xml:id="echoid-s344" xml:space="preserve">EG:</s>
  <s xml:id="echoid-s345" xml:space="preserve">: HI: </s>
  <s xml:id="echoid-s346" xml:space="preserve">HA: </s>
  <s xml:id="echoid-s347" xml:space="preserve">but EF and EG <lb/>are equal, therefore alſo HI and HA.</s>
  <s xml:id="echoid-s348" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div20" type="section" level="1" n="20">
<head xml:id="echoid-head25" xml:space="preserve">PROBLEM IX.</head>
<p>
  <s xml:id="echoid-s349" xml:space="preserve"><emph style="sc">Having</emph> a circle whoſe center is A given in magnitude and poſition, and alſo <lb/>two right lines BD and ZC given in poſition, to draw a circle which ſhall touch <lb/>all three.</s>
  <s xml:id="echoid-s350" xml:space="preserve"/>
</p>
<pb o="(8)" file="0020" n="20"/>
<p>
  <s xml:id="echoid-s351" xml:space="preserve"><emph style="sc">From</emph> A draw two perpendiculars to the right lines DB, ZC; </s>
  <s xml:id="echoid-s352" xml:space="preserve">viz. </s>
  <s xml:id="echoid-s353" xml:space="preserve">ADF and <lb/>AZX; </s>
  <s xml:id="echoid-s354" xml:space="preserve">and in theſe perpendiculars take DF, ZX, on “either ſide of D and Z, <lb/>equal to the Radius of the given circle: </s>
  <s xml:id="echoid-s355" xml:space="preserve">and through F<unsure/> and X draw lines pa-<lb/>rallel to DB, ZC; </s>
  <s xml:id="echoid-s356" xml:space="preserve">viz. </s>
  <s xml:id="echoid-s357" xml:space="preserve">FG, XH; </s>
  <s xml:id="echoid-s358" xml:space="preserve">and then by the preceding Problem draw <lb/>a circle which ſhall paſs through the given center A and touch the two lines <lb/>FG, XH; </s>
  <s xml:id="echoid-s359" xml:space="preserve">and E the center of this circle will alſo be the center of the circle <lb/>required, as appears by ſubtracting equals from equals in Figure 1: </s>
  <s xml:id="echoid-s360" xml:space="preserve">and by adding <lb/>equals to equals in Figure 2.</s>
  <s xml:id="echoid-s361" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div21" type="section" level="1" n="21">
<head xml:id="echoid-head26" xml:space="preserve">LEMMA II.</head>
<p>
  <s xml:id="echoid-s362" xml:space="preserve"><emph style="sc">If</emph> the two circles CEB and CED cut one another C, then I ſay a line drawn <lb/>from the point of ſection CBD cutting both circles, will cut off diſſimilar ſeg-<lb/>ments from thoſe circles.</s>
  <s xml:id="echoid-s363" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s364" xml:space="preserve">1ſt. </s>
  <s xml:id="echoid-s365" xml:space="preserve"><emph style="sc">Suppose</emph> CB to be the Diameter of one of them: </s>
  <s xml:id="echoid-s366" xml:space="preserve">then draw to the <lb/>other point of ſection E the line CE, and joining EB, ED, the angle CEB will <lb/>be a right one, and the angle CED either greater or leſs than a right one, and <lb/>conſequently CD cannot be a Diameter of the other.</s>
  <s xml:id="echoid-s367" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s368" xml:space="preserve">2dly. </s>
  <s xml:id="echoid-s369" xml:space="preserve"><emph style="sc">Suppose</emph> CBD not to paſs through the center of either: </s>
  <s xml:id="echoid-s370" xml:space="preserve">then through <lb/>C draw a Diameter CAG, and continue it to meet the other circle in F, and <lb/>join BG, DF: </s>
  <s xml:id="echoid-s371" xml:space="preserve">then the angle CBG is a right one, and the angle CDF is either <lb/>greater or leſs than a right one: </s>
  <s xml:id="echoid-s372" xml:space="preserve">and therefore the lines BG and DF are not <lb/>parallel: </s>
  <s xml:id="echoid-s373" xml:space="preserve">let H be the center of the other circle, and let a Diameter CHI be <lb/>drawn: </s>
  <s xml:id="echoid-s374" xml:space="preserve">draw DI and continue it meet to meet CG in K: </s>
  <s xml:id="echoid-s375" xml:space="preserve">then DIK will be pa-<lb/>rallel to BG: </s>
  <s xml:id="echoid-s376" xml:space="preserve">hence CB: </s>
  <s xml:id="echoid-s377" xml:space="preserve">CD:</s>
  <s xml:id="echoid-s378" xml:space="preserve">: CG: </s>
  <s xml:id="echoid-s379" xml:space="preserve">CK. </s>
  <s xml:id="echoid-s380" xml:space="preserve">But CI and CK are unequal, <lb/>(being both applied from the ſame point in a right angle) and therefore it cannot <lb/>be 2s CB: </s>
  <s xml:id="echoid-s381" xml:space="preserve">CD:</s>
  <s xml:id="echoid-s382" xml:space="preserve">: CG: </s>
  <s xml:id="echoid-s383" xml:space="preserve">CI: </s>
  <s xml:id="echoid-s384" xml:space="preserve">and hence it appears that the Segments CB and <lb/>CD are diſſimilar.</s>
  <s xml:id="echoid-s385" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div22" type="section" level="1" n="22">
<head xml:id="echoid-head27" xml:space="preserve">LEMMA III.</head>
<p>
  <s xml:id="echoid-s386" xml:space="preserve"><emph style="sc">If</emph> through the legs of any triangle EDF (ſee Figure to Problem 10.) </s>
  <s xml:id="echoid-s387" xml:space="preserve">a line <lb/>BI be<unsure/> drawn parallel to the baſe DF, ſo that there be conſtituted two ſimilar <lb/>triangles about the ſame vertex; </s>
  <s xml:id="echoid-s388" xml:space="preserve">and a circle be circumſcribed about each of <lb/>theſe triangles; </s>
  <s xml:id="echoid-s389" xml:space="preserve">theſe circles will touch one another in the common vertex E.</s>
  <s xml:id="echoid-s390" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s391" xml:space="preserve"><emph style="sc">It</emph> is plain that they will either touch or cut each other in the point E: </s>
  <s xml:id="echoid-s392" xml:space="preserve">if <lb/>they cut each other, then by the preceding Lemma the Segments BE and DE <lb/>would be diſſimilar; </s>
  <s xml:id="echoid-s393" xml:space="preserve">but they are ſimilar, and they muſt therefore touch each <lb/>other.</s>
  <s xml:id="echoid-s394" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div23" type="section" level="1" n="23">
<head xml:id="echoid-head28" xml:space="preserve">PROBLEM X.</head>
<p>
  <s xml:id="echoid-s395" xml:space="preserve"><emph style="sc">Having</emph> a point A, and alſo a right line BC, given in poſition; </s>
  <s xml:id="echoid-s396" xml:space="preserve">together <lb/>with a circle whoſe center is G given both in m<gap/>de and<unsure/> poſition; </s>
  <s xml:id="echoid-s397" xml:space="preserve">to deſcribe
<pb o="(9)" file="0021" n="21"/>
a circle which ſhall paſs through the given point, and likewiſe touch both the <lb/>given line and the given circle.</s>
  <s xml:id="echoid-s398" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s399" xml:space="preserve"><emph style="sc">The</emph> given right line may either 1ſt cut, touch, or be entirely without the <lb/>given circle, and the given point be without the ſame or in the circum-<lb/>ference; </s>
  <s xml:id="echoid-s400" xml:space="preserve">or 2dly, it may cut the given circle, and the given point be within <lb/>the ſame, or in the circumference.</s>
  <s xml:id="echoid-s401" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s402" xml:space="preserve"><emph style="sc">Case</emph> Iſt. </s>
  <s xml:id="echoid-s403" xml:space="preserve">Suppoſe the given line BC to cut, touch, or fall entirely with-<lb/>out the given circle; </s>
  <s xml:id="echoid-s404" xml:space="preserve">and the given point A to be without the ſame, or in the <lb/>circumference: </s>
  <s xml:id="echoid-s405" xml:space="preserve">through G the center of the given circle draw DGFC per-<lb/>pendicular to BC, and joining DA, take DH a 4th proportional to DA, DC, <lb/>DF, ſo that DA X DH = DC X DF: </s>
  <s xml:id="echoid-s406" xml:space="preserve">then through the points A and H draw <lb/>a circle touching the line CB by Problem VII, and I ſay it will alſo touch the <lb/>given circle.</s>
  <s xml:id="echoid-s407" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s408" xml:space="preserve">Draw DB cutting the given circle in E, and join FE. </s>
  <s xml:id="echoid-s409" xml:space="preserve">Now becauſe the <lb/>triangles DEF, DCB are ſimilar, DF: </s>
  <s xml:id="echoid-s410" xml:space="preserve">DE:</s>
  <s xml:id="echoid-s411" xml:space="preserve">: DB: </s>
  <s xml:id="echoid-s412" xml:space="preserve">DC, and therefore DC <lb/>X DF = DB X DE. </s>
  <s xml:id="echoid-s413" xml:space="preserve">But DC X DF = DA X DH by Conſtruction. </s>
  <s xml:id="echoid-s414" xml:space="preserve">Hence <lb/>DB X DE = DA X DH, and therefore the points B, E, H, A, will be alſo in <lb/>a circle: </s>
  <s xml:id="echoid-s415" xml:space="preserve">but the point E is alſo in the given circle; </s>
  <s xml:id="echoid-s416" xml:space="preserve">therefore theſe circles either <lb/>touch or cut one another in that point. </s>
  <s xml:id="echoid-s417" xml:space="preserve">Let now BI be drawn from the point <lb/>of contact B perpendicular to the touching line BC to meet the circumference <lb/>again in I, and it will be a Diameter: </s>
  <s xml:id="echoid-s418" xml:space="preserve">and let EI be joined: </s>
  <s xml:id="echoid-s419" xml:space="preserve">then becauſe the <lb/>angles FED and BEI are vertical and each of them right ones, FEI will be a <lb/>continued ſtraight line: </s>
  <s xml:id="echoid-s420" xml:space="preserve">and it appears that the two circles will touch each <lb/>other by the preceding Lemmas.</s>
  <s xml:id="echoid-s421" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s422" xml:space="preserve"><emph style="sc">Case</emph> 2d. </s>
  <s xml:id="echoid-s423" xml:space="preserve">Suppoſe the given line BC to cut the given circle, and the given <lb/>point to be within the ſame, or in the circumference; </s>
  <s xml:id="echoid-s424" xml:space="preserve">the Conſtruction and De-<lb/>monſtration are exactly the ſame as before, except that the angles FED and <lb/>BEI are not vertical but coincident, and ſo EI is coincident with EF.</s>
  <s xml:id="echoid-s425" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s426" xml:space="preserve">N.</s>
  <s xml:id="echoid-s427" xml:space="preserve">B. </s>
  <s xml:id="echoid-s428" xml:space="preserve">In either of theſe caſes if the point A coincide with E or be given in <lb/>the circumference, draw DEB, and erect BI perpendicular to CB to meet FE in <lb/>I, then upon BI as diameter deſcribe a circle, and the thing will manifeſtly be done.</s>
  <s xml:id="echoid-s429" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div24" type="section" level="1" n="24">
<head xml:id="echoid-head29" xml:space="preserve">PROBLEM XI.</head>
<p>
  <s xml:id="echoid-s430" xml:space="preserve"><emph style="sc">Having</emph> two circles given in magnitude and poſition, whoſe centers are A <lb/>and B, as likewiſe a right line CZ; </s>
  <s xml:id="echoid-s431" xml:space="preserve">to draw a circle which ſhall touch all three.</s>
  <s xml:id="echoid-s432" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s433" xml:space="preserve"><emph style="sc">From</emph> the center of the leſſer circle B let BZ be drawn perpendicular to CZ, <lb/>and in BZ (or in BZ continued as the caſe requires) let be taken ZX = AL the <lb/>Radius of the other circle; </s>
  <s xml:id="echoid-s434" xml:space="preserve">and through X let XH be drawn parallel to CZ, <lb/>and with center B and Radius BG, equal to the difference ( or ſum as the Caſe <lb/>requires) of the Radii of the two given circles, let a circle be deſcribed; </s>
  <s xml:id="echoid-s435" xml:space="preserve">and laſtly
<pb o="(10)" file="0022" n="22"/>
let another circle be deſcribed, touching this laſt, and alſo the line XH and paſſ-<lb/>ing through the point A by Problem X, and I ſay that E the center of this circle <lb/>will alſo be the center of the circle required; </s>
  <s xml:id="echoid-s436" xml:space="preserve">as will appear by taking equals from <lb/>equals, or adding equals to equals, as the aſſigned Caſe and Data ſhall require.</s>
  <s xml:id="echoid-s437" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s438" xml:space="preserve"><emph style="sc">The</emph> Caſes are four, though Vieta makes but three.</s>
  <s xml:id="echoid-s439" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s440" xml:space="preserve"><emph style="sc">Case</emph> Iſt. </s>
  <s xml:id="echoid-s441" xml:space="preserve">If it be required that the circle ſhould touch both the others ex-<lb/>ternally then BG muſt be taken equal to the difference of the Semidiameters of <lb/>the two given circles, and ZX muſt be taken in BZ produced.</s>
  <s xml:id="echoid-s442" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s443" xml:space="preserve"><emph style="sc">Case</emph> 2d. </s>
  <s xml:id="echoid-s444" xml:space="preserve">If it be required that the circle ſhall touch and include both the <lb/>given ones; </s>
  <s xml:id="echoid-s445" xml:space="preserve">then BG muſt be taken equal to the difference, as in Caſe 1ſt, but <lb/>ZX muſt be taken in BZ itſelf.</s>
  <s xml:id="echoid-s446" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s447" xml:space="preserve"><emph style="sc">Case</emph> 3d. </s>
  <s xml:id="echoid-s448" xml:space="preserve">If it be required that the circle ſhould touch and include the greater <lb/>of the given circles, and touch externally the other whoſe center is B; </s>
  <s xml:id="echoid-s449" xml:space="preserve">then BG <lb/>muſt be taken equal to the ſum of the Radii of the given circles, and ZX muſt <lb/>be taken in BZ itſelf.</s>
  <s xml:id="echoid-s450" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s451" xml:space="preserve"><emph style="sc">Case</emph> 4th. </s>
  <s xml:id="echoid-s452" xml:space="preserve">If it be required that the circle ſhould touch the greater of the <lb/>given circles externally, and touch and include the leſſer; </s>
  <s xml:id="echoid-s453" xml:space="preserve">then BG muſt be taken <lb/>equal to the ſum of the Radii, and ZX muſt be taken in BZ produced.</s>
  <s xml:id="echoid-s454" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div25" type="section" level="1" n="25">
<head xml:id="echoid-head30" xml:space="preserve">PROBLEM XII <anchor type="note" xlink:href="" symbol="*"/>.</head>
<p>
  <s xml:id="echoid-s455" xml:space="preserve"><emph style="sc">Having</emph> two points given B and D, and like wiſe a circle whoſe center is A; </s>
  <s xml:id="echoid-s456" xml:space="preserve">to de-<lb/>ſcribe another circle which ſhall paſs through the given points, and touch the <lb/>given circle.</s>
  <s xml:id="echoid-s457" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s458" xml:space="preserve"><emph style="sc">Let</emph> DB be joined, as alſo AB, and let AB be produced to cut the given <lb/>circle in the points I and K, then let BH be taken a 4th proportional to DB, <lb/>BK, BI; </s>
  <s xml:id="echoid-s459" xml:space="preserve">ſo that BD X BH = BI X BK: </s>
  <s xml:id="echoid-s460" xml:space="preserve">from H let a Tangent HF be drawn <lb/>to the given circle; </s>
  <s xml:id="echoid-s461" xml:space="preserve">and BF be joined and cut the circle again in G: </s>
  <s xml:id="echoid-s462" xml:space="preserve">and let DG <lb/>be drawn cutting the given circle again in E, and laſtly through the points D, <lb/>B, G, let a circle be drawn, I ſay it will touch the given circle in G.</s>
  <s xml:id="echoid-s463" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s464" xml:space="preserve"><emph style="sc">For</emph> joining EF; </s>
  <s xml:id="echoid-s465" xml:space="preserve">becauſe the rectangle DBH = the Rectangle KBI, i.</s>
  <s xml:id="echoid-s466" xml:space="preserve">e. </s>
  <s xml:id="echoid-s467" xml:space="preserve">the <lb/>Rectangle GBF, thereforethefourpoints D, H, F, G, are in a circle; </s>
  <s xml:id="echoid-s468" xml:space="preserve">and hence the <lb/>angle HFB = the angle GDB: </s>
  <s xml:id="echoid-s469" xml:space="preserve">(for in the two firſt ſigures one is theexternal angle <lb/>of a quadrilateral figure, and the other is the internal and oppoſite, and in the two <lb/>laſt figures theſe angles are in the ſame ſegment.) </s>
  <s xml:id="echoid-s470" xml:space="preserve">But the angle HFB = the angle <lb/>GEF by Eu. </s>
  <s xml:id="echoid-s471" xml:space="preserve">III. </s>
  <s xml:id="echoid-s472" xml:space="preserve">32. </s>
  <s xml:id="echoid-s473" xml:space="preserve">hence GEF = GDB: </s>
  <s xml:id="echoid-s474" xml:space="preserve">therefore the triangles GEF and <lb/>GDB are ſimilar and under the ſame vertex, and therefore by Lemma 3. </s>
  <s xml:id="echoid-s475" xml:space="preserve">the cir-<lb/>cles deſcribed about them will touch each other in the common vertex G.</s>
  <s xml:id="echoid-s476" xml:space="preserve"/>
</p>
<note symbol="*" position="foot" xml:space="preserve">There are other Conſtructions of this Problem in Hugo de Omerique, Simpſon, and the Mathematician. <lb/>See alſo Monthly Review for Oct. 1764, where the Reviewer is pleaſed to ſpeak favourably of the 1ſt Edi-<lb/>tion of this work, but wiſhes that ſome modern ſolutions of theſe Problems had been inſerted, which, he <lb/>ſays, are more conciſe and elegant than any which are to be met with in the works of the Antients. The <lb/>Editor acknowledges that the conſtruction there given is more ſimple than Vieta’s; but Vieta is not an <lb/>Antient, and he knows of no others that exceed his.</note>
<pb o="(11)" file="0023" n="23"/>
</div>
<div xml:id="echoid-div26" type="section" level="1" n="26">
<head xml:id="echoid-head31" xml:space="preserve">LEMMA IV.</head>
<p>
  <s xml:id="echoid-s477" xml:space="preserve"><emph style="sc">Having</emph> two circles ABCI and EFGH given, it is required to find a point <lb/>M, in the line joining their centers, or in that line continued, ſuch, that any <lb/>line drawn through the ſaid point M, cutting both the circles, ſhall always cut <lb/>off ſimilar ſegments.</s>
  <s xml:id="echoid-s478" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s479" xml:space="preserve"><emph style="sc">Let</emph> the line KL joining the centers be ſo cut or produced, that KM may be <lb/>to LM in the given ratio of the Radii AK to EL [in the caſe of producing <lb/>KL we muſt make it as AK-EL: </s>
  <s xml:id="echoid-s480" xml:space="preserve">EL:</s>
  <s xml:id="echoid-s481" xml:space="preserve">: KL: </s>
  <s xml:id="echoid-s482" xml:space="preserve">LM, for then by compoſition it <lb/>will be AK: </s>
  <s xml:id="echoid-s483" xml:space="preserve">EL:</s>
  <s xml:id="echoid-s484" xml:space="preserve">: KM: </s>
  <s xml:id="echoid-s485" xml:space="preserve">LM] and then I ſay that the point M will be the <lb/>point required. </s>
  <s xml:id="echoid-s486" xml:space="preserve">For from it drawing any line MGFCB cutting the circle <lb/>ABCI in B and C, and the circle EFGH in F and G; </s>
  <s xml:id="echoid-s487" xml:space="preserve">and let B and F be the <lb/>correſpondent points moſt diſtant from M, and C and G the correſpondent <lb/>points that are nearer; </s>
  <s xml:id="echoid-s488" xml:space="preserve">and let be joined BK, CK, FL, GL, and thereby two <lb/>triangles will be formed BKC, FLG. </s>
  <s xml:id="echoid-s489" xml:space="preserve">Now becauſe by Conſtri<unsure/>ction KM: </s>
  <s xml:id="echoid-s490" xml:space="preserve">LM <lb/>:</s>
  <s xml:id="echoid-s491" xml:space="preserve">: KB: </s>
  <s xml:id="echoid-s492" xml:space="preserve">LF, KB and LF will be parallel by Euc. </s>
  <s xml:id="echoid-s493" xml:space="preserve">VI. </s>
  <s xml:id="echoid-s494" xml:space="preserve">7. </s>
  <s xml:id="echoid-s495" xml:space="preserve">and for the ſame rea-<lb/>ſon KC and LG will be parallel; </s>
  <s xml:id="echoid-s496" xml:space="preserve">and therefore the triangles BKC and FLG <lb/>will be equiangled, and hence the ſegment BC will be ſimilar to the ſegment FG.</s>
  <s xml:id="echoid-s497" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div27" type="section" level="1" n="27">
<head xml:id="echoid-head32" xml:space="preserve">LEMMA V.</head>
<p>
  <s xml:id="echoid-s498" xml:space="preserve"><emph style="sc">The</emph> point M being found as in the preceding Lemma, I ſay that it is a pro-<lb/>perty of the ſaid point, that MG x MB = MH x MA: </s>
  <s xml:id="echoid-s499" xml:space="preserve">as alſo that MF x <lb/>MC = ME x MI.</s>
  <s xml:id="echoid-s500" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s501" xml:space="preserve"><emph style="sc">For</emph> joining CI and GH, it is evident that theſe lines will alſo be parallel. <lb/></s>
  <s xml:id="echoid-s502" xml:space="preserve">Hence MI: </s>
  <s xml:id="echoid-s503" xml:space="preserve">MC:</s>
  <s xml:id="echoid-s504" xml:space="preserve">: MH: </s>
  <s xml:id="echoid-s505" xml:space="preserve">MG, but MI: </s>
  <s xml:id="echoid-s506" xml:space="preserve">MC:</s>
  <s xml:id="echoid-s507" xml:space="preserve">: MB: </s>
  <s xml:id="echoid-s508" xml:space="preserve">MA, therefore <lb/>MH: </s>
  <s xml:id="echoid-s509" xml:space="preserve">MG:</s>
  <s xml:id="echoid-s510" xml:space="preserve">: MB: </s>
  <s xml:id="echoid-s511" xml:space="preserve">MA, and MG x MB = MH x MA. </s>
  <s xml:id="echoid-s512" xml:space="preserve">Again MH: </s>
  <s xml:id="echoid-s513" xml:space="preserve">MG <lb/>:</s>
  <s xml:id="echoid-s514" xml:space="preserve">: MF: </s>
  <s xml:id="echoid-s515" xml:space="preserve">ME, but MH: </s>
  <s xml:id="echoid-s516" xml:space="preserve">MG:</s>
  <s xml:id="echoid-s517" xml:space="preserve">: MI: </s>
  <s xml:id="echoid-s518" xml:space="preserve">MC, therefore MF: </s>
  <s xml:id="echoid-s519" xml:space="preserve">ME:</s>
  <s xml:id="echoid-s520" xml:space="preserve">: MI: </s>
  <s xml:id="echoid-s521" xml:space="preserve">MC <lb/>and MF x MC = ME x MI.</s>
  <s xml:id="echoid-s522" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div28" type="section" level="1" n="28">
<head xml:id="echoid-head33" xml:space="preserve">PROBLEM XIII.</head>
<p>
  <s xml:id="echoid-s523" xml:space="preserve"><emph style="sc">Having</emph> two circles given in magnitude and poſition, whoſe centers are K <lb/>and L, and alſo a point D; </s>
  <s xml:id="echoid-s524" xml:space="preserve">to draw a circle which ſhall touch the two given <lb/>ones, and alſo paſs through the point D.</s>
  <s xml:id="echoid-s525" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s526" xml:space="preserve"><emph style="sc">Join</emph> the given centers by drawing KL, and in KL or KL produced ſind the <lb/>point M (by Lemma 4.) </s>
  <s xml:id="echoid-s527" xml:space="preserve">ſuch, that all the lines drawn from it cutting the given <lb/>circles ſhall cut off ſimilar ſegments; </s>
  <s xml:id="echoid-s528" xml:space="preserve">and let KL cut the circumferences, one <lb/>of them in the points A and I, and the other in the points E and H; </s>
  <s xml:id="echoid-s529" xml:space="preserve">and join-<lb/>ing MD, make it as MD: </s>
  <s xml:id="echoid-s530" xml:space="preserve">MA:</s>
  <s xml:id="echoid-s531" xml:space="preserve">: MH: </s>
  <s xml:id="echoid-s532" xml:space="preserve">MN. </s>
  <s xml:id="echoid-s533" xml:space="preserve">Then through the points D and <lb/>N draw a circle which ſhall alſo touch the given circle whoſe center is K, by <lb/>Prob. </s>
  <s xml:id="echoid-s534" xml:space="preserve">XII. </s>
  <s xml:id="echoid-s535" xml:space="preserve">and I ſay that this circle will alſo touch the other given circle whoſe <lb/>center is L. </s>
  <s xml:id="echoid-s536" xml:space="preserve">For let B be the point of contact and BM be drawn cutting the <lb/>circle K in C, and the circle L in F and G; </s>
  <s xml:id="echoid-s537" xml:space="preserve">then by Lemma 5. </s>
  <s xml:id="echoid-s538" xml:space="preserve">MB x MG =
<pb o="(12)" file="0024" n="24"/>
MA x MH: </s>
  <s xml:id="echoid-s539" xml:space="preserve">but MA x MH = MD x MN by Conſtruction; </s>
  <s xml:id="echoid-s540" xml:space="preserve">therefore MB x <lb/>MG = MD x MN, and the points B, G, N, D, are in a circle. </s>
  <s xml:id="echoid-s541" xml:space="preserve">But the point <lb/>G is alſo in the circle L; </s>
  <s xml:id="echoid-s542" xml:space="preserve">therefore theſe circles either touch or cut each other in <lb/>the point G. </s>
  <s xml:id="echoid-s543" xml:space="preserve">Now the circles BND and BCI touch one another in B by con-<lb/>ſtruction; </s>
  <s xml:id="echoid-s544" xml:space="preserve">therefore the ſegment BC is ſimilar to the ſegment BG; </s>
  <s xml:id="echoid-s545" xml:space="preserve">and alſo by <lb/>conſtruction the ſegment BC is ſimilar to the ſegment FG; </s>
  <s xml:id="echoid-s546" xml:space="preserve">and therefore the <lb/>ſegment FG is ſimilar to the ſegment BG; </s>
  <s xml:id="echoid-s547" xml:space="preserve">and hence the circles FGE and BGD <lb/>touch one another in the point G.</s>
  <s xml:id="echoid-s548" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s549" xml:space="preserve"><emph style="sc">The</emph> Caſes are three.</s>
  <s xml:id="echoid-s550" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s551" xml:space="preserve"><emph style="sc">Case</emph> Iſt. </s>
  <s xml:id="echoid-s552" xml:space="preserve">Iſ the circle be required to touch and include both the given ones; <lb/></s>
  <s xml:id="echoid-s553" xml:space="preserve">then M muſt be taken in KL produced; </s>
  <s xml:id="echoid-s554" xml:space="preserve">and MN muſt be taken a fourth pro-<lb/>portional to MD, MA, MH, A being the moſt diſtant point of interſection in <lb/>the circle K, and H the neareſt point of interſection in the circle L.</s>
  <s xml:id="echoid-s555" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s556" xml:space="preserve"><emph style="sc">Case</emph> 2d. </s>
  <s xml:id="echoid-s557" xml:space="preserve">If the circle be required to touch both the given ones externally; <lb/></s>
  <s xml:id="echoid-s558" xml:space="preserve">then alſo M muſt be taken in KL produced; </s>
  <s xml:id="echoid-s559" xml:space="preserve">and MN taken a fourth proportional <lb/>to MD, MA, MH, A being the neareſt point of interſection in the circle K, <lb/>and H the moſt diſtant in the circle L.</s>
  <s xml:id="echoid-s560" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s561" xml:space="preserve"><emph style="sc">Case</emph> 3d. </s>
  <s xml:id="echoid-s562" xml:space="preserve">If the circle be required to touch and include the circle K, and to <lb/>touch L externally; </s>
  <s xml:id="echoid-s563" xml:space="preserve">then M muſt be taken in KL itſelf; </s>
  <s xml:id="echoid-s564" xml:space="preserve">and MN a fourth pro-<lb/>portional to MD, MA, MH, A being the moſt diſtant point in K, and H the <lb/>neareſt in L.</s>
  <s xml:id="echoid-s565" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div29" type="section" level="1" n="29">
<head xml:id="echoid-head34" xml:space="preserve">PROBLEM XIV.</head>
<p>
  <s xml:id="echoid-s566" xml:space="preserve"><emph style="sc">Having</emph> three circles given whoſe centers are A, B, and D; </s>
  <s xml:id="echoid-s567" xml:space="preserve">to draw a fourth <lb/>which ſhall touch all three.</s>
  <s xml:id="echoid-s568" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s569" xml:space="preserve"><emph style="sc">Let</emph> that whoſe center is A be called the 1ſt, that whoſe center is B the 2d, <lb/>and that whoſe center is D the 3d. </s>
  <s xml:id="echoid-s570" xml:space="preserve">Then with center B, and radius equal to <lb/>the differcnce, or ſum, as the caſe requires, of the ſemidiameters of the 1ſt and <lb/>2d circles, let an auxiliary circle be deſcribed; </s>
  <s xml:id="echoid-s571" xml:space="preserve">and likewiſe with D center, and <lb/>radius equal to the difference, or ſum, as the caſe requires, of the ſemidiameters <lb/>of the 1ſt and 3d circles, let another auxiliary circle be deſcribed; </s>
  <s xml:id="echoid-s572" xml:space="preserve">and laſtly by <lb/>the preceding Problem draw a circle which ſhall touch the two auxiliary ones, <lb/>and likewiſe paſs through the point A which is the center of the firſt circle. <lb/></s>
  <s xml:id="echoid-s573" xml:space="preserve">Let the center of this laſt deſcribed circle be E, and the ſame point E will like-<lb/>wiſe be the center of the circle required; </s>
  <s xml:id="echoid-s574" xml:space="preserve">as will appear by adding equals to <lb/>equals, or taking equals from equals, as the caſe requires.</s>
  <s xml:id="echoid-s575" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s576" xml:space="preserve"><emph style="sc">The</emph> Caſes are theſe.</s>
  <s xml:id="echoid-s577" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s578" xml:space="preserve"><emph style="sc">Case</emph> 1ſt. </s>
  <s xml:id="echoid-s579" xml:space="preserve">Iſ it be required that the circle ſhould touch and include all the other <lb/>three; </s>
  <s xml:id="echoid-s580" xml:space="preserve">then let A be the center of the greateſt given circle, B of the next, and D <lb/>of the leaſt: </s>
  <s xml:id="echoid-s581" xml:space="preserve">and let BG = the difference of the ſemidiameters of the 1ſt and 2d,
<pb o="(13)" file="0025" n="25"/>
and DF = the Difference of the ſemidiameters of the 1ſt and 3d, and through A <lb/>deſcribe a circle which ſhall touch the two auxiliary ones externally.</s>
  <s xml:id="echoid-s582" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s583" xml:space="preserve"><emph style="sc">Case</emph> 2d. </s>
  <s xml:id="echoid-s584" xml:space="preserve">If it be required that the circle ſhould touch all the other three <lb/>externally; </s>
  <s xml:id="echoid-s585" xml:space="preserve">then the circles being the ſame as before in reſpect to their magnitude, <lb/>let BG and DF be alſo the ſame as in the 1ſt Caſe, but through A deſcribe a circle <lb/>which ſhall touch and include the two auxiliary ones.</s>
  <s xml:id="echoid-s586" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s587" xml:space="preserve"><emph style="sc">Case</emph> 3d. </s>
  <s xml:id="echoid-s588" xml:space="preserve">If it be required that the circle ſhould touch and include one of the <lb/>given circles A, and touch the other two externally; </s>
  <s xml:id="echoid-s589" xml:space="preserve">then let BG and DF = the <lb/>the ſum of the ſemidiameters of the 1ſt and 2d, and ſum of the ſemidiameters of <lb/>of the 1ſt and 3d reſpectively; </s>
  <s xml:id="echoid-s590" xml:space="preserve">and through A deſcribe a circle which ſhall touch <lb/>the two auxiliary ones externally.</s>
  <s xml:id="echoid-s591" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s592" xml:space="preserve"><emph style="sc">Case</emph> 4th. </s>
  <s xml:id="echoid-s593" xml:space="preserve">If it be required that the circle ſhould touch externally one of the <lb/>given circles A, and ſhould touch and include the other two; </s>
  <s xml:id="echoid-s594" xml:space="preserve">then let BG and <lb/>DF = the ſums as in Caſe 3d, but through A deſcribe a circle which ſhall touch <lb/>and include the two auxiliary ones.</s>
  <s xml:id="echoid-s595" xml:space="preserve"/>
</p>
<pb o="(14)" file="0026" n="26"/>
</div>
<div xml:id="echoid-div30" type="section" level="1" n="30">
<head xml:id="echoid-head35" xml:space="preserve">SUPPLEMENT.</head>
<head xml:id="echoid-head36" xml:space="preserve">PROBLEM I.</head>
<p>
  <s xml:id="echoid-s596" xml:space="preserve">HAVING two points given A and B, to determine the Locus of the cen-<lb/>ters of all ſuch circles as may be drawn through A and B.</s>
  <s xml:id="echoid-s597" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s598" xml:space="preserve"><emph style="sc">Join</emph> AB and biſect it in the point C, and through C, draw a perpendicular <lb/>to it CE, and continue it both ways in infinitum, and it is evident that this line <lb/>CE will be the Locus required.</s>
  <s xml:id="echoid-s599" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div31" type="section" level="1" n="31">
<head xml:id="echoid-head37" xml:space="preserve">PROBLEM II.</head>
<p>
  <s xml:id="echoid-s600" xml:space="preserve"><emph style="sc">Having</emph> two right lines given AB and CD, to determine the Locus of the <lb/>centers of all ſuch circles as may be drawn touching both the ſaid lines.</s>
  <s xml:id="echoid-s601" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s602" xml:space="preserve"><emph style="sc">Case</emph> 1ft. </s>
  <s xml:id="echoid-s603" xml:space="preserve">Suppoſe AB and CD to be parallel; </s>
  <s xml:id="echoid-s604" xml:space="preserve">then drawing GI perpendicu-<lb/>lar to them both; </s>
  <s xml:id="echoid-s605" xml:space="preserve">biſect it in H, and through H draw EHF parallel to the two <lb/>given lines, and it will be the Locus required.</s>
  <s xml:id="echoid-s606" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s607" xml:space="preserve"><emph style="sc">Case</emph> 2d. </s>
  <s xml:id="echoid-s608" xml:space="preserve">Suppoſe the given lines being produced meet each other in E, <lb/>then biſecting the angle BED by the line EHF, this line EHF will be the Locus <lb/>required.</s>
  <s xml:id="echoid-s609" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div32" type="section" level="1" n="32">
<head xml:id="echoid-head38" xml:space="preserve">PROBLEM III.</head>
<p>
  <s xml:id="echoid-s610" xml:space="preserve"><emph style="sc">Having</emph> two circles given whoſe centers are A and B; </s>
  <s xml:id="echoid-s611" xml:space="preserve">to determine the Locus <lb/>of the centers of all ſuch circles as ſhall touch the two given ones.</s>
  <s xml:id="echoid-s612" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s613" xml:space="preserve"><emph style="sc">Cases</emph> 1ſt and 2d. </s>
  <s xml:id="echoid-s614" xml:space="preserve">Suppoſe it be required that the circles be touched outwardly <lb/>by both the given ones; </s>
  <s xml:id="echoid-s615" xml:space="preserve">or that they be touched inwardly by both the given <lb/>ones.</s>
  <s xml:id="echoid-s616" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s617" xml:space="preserve"><emph style="sc">Then</emph> joining the centers A and B, let AB cut the circumferences in C and <lb/>D and produced in P and O: </s>
  <s xml:id="echoid-s618" xml:space="preserve">let CD which is intercepted between the convex <lb/>circumſerences be biſected in E: </s>
  <s xml:id="echoid-s619" xml:space="preserve">ſet off from E towards B the center of the
<pb o="(15)" file="0027" n="27"/>
greater of the given circles the line EH = the difference of the Radii of the given <lb/>circles, and with A and B as Foci and EH Tranverſe Axis, let two oppoſite <lb/>Hyperbolas be deſcribed KEK and LHL: </s>
  <s xml:id="echoid-s620" xml:space="preserve">then I ſay that KEK will be the <lb/>locus of the centers of all the circles which can be drawn ſo as to be touched <lb/>outwardly by both the given circles; </s>
  <s xml:id="echoid-s621" xml:space="preserve">and LHL will be the locus of the centers <lb/>of all the circles which can be drawn ſo as to be touched inwardly by both the <lb/>given circles. </s>
  <s xml:id="echoid-s622" xml:space="preserve">For taking any point K in the Hyperbola KEK, and drawing KA <lb/>and KB, let theſe lines cut the given circumferences in F and G reſpectively: <lb/></s>
  <s xml:id="echoid-s623" xml:space="preserve">and make KQ = KA: </s>
  <s xml:id="echoid-s624" xml:space="preserve">then from the nature of the curve QB = EH, but by con-<lb/>ſtruction EH = BG - AF, therefore QB = BG - AF and hence QG = AF, and <lb/>then taking equals from equals KG = KF, which is a demonſtration of the 1ſt <lb/>Caſe.</s>
  <s xml:id="echoid-s625" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s626" xml:space="preserve"><emph style="sc">Then</emph> with regard to the 2d Caſe, taking any point L in the Hyperbola <lb/>LHL, and drawing LB and LA and producing them to meet the concave cir-<lb/>cumferences in M and N, let alſo LR be taken equal to LB; </s>
  <s xml:id="echoid-s627" xml:space="preserve">then from the pro-<lb/>perty of the curve AR = EH, but EH (by conſtruction) = BM - NA; </s>
  <s xml:id="echoid-s628" xml:space="preserve">there-<lb/>fore AR = BM - NA, and NR = BM, and then adding equals to equals LN = <lb/>LM, which is a demonſtration of the 2d Caſe.</s>
  <s xml:id="echoid-s629" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s630" xml:space="preserve"><emph style="sc">Case</emph> 3d. </s>
  <s xml:id="echoid-s631" xml:space="preserve">Suppoſe it be required that the circles to be deſcribed be touched <lb/>outwardly by one of the given circles, and inwardly by the other.</s>
  <s xml:id="echoid-s632" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s633" xml:space="preserve">Then drawing AB, let it cut the convex circumferences in C and D, and the <lb/>concave ones in P and O, biſect PD in E, and from E towards B ſet off EH = <lb/>the ſum of the given radii. </s>
  <s xml:id="echoid-s634" xml:space="preserve">Then with A and B foci and EH tranſverſe axis, <lb/>let two oppoſite hyperbolas be deſcribed KEK and LHL: </s>
  <s xml:id="echoid-s635" xml:space="preserve">and KEK will be the <lb/>locus of the centers of the circles which are touched inwardly by the circle A <lb/>and outwardly by the circle B; </s>
  <s xml:id="echoid-s636" xml:space="preserve">and LHL will be the locus of the centers of <lb/>thoſe circles which are touched inwardly by B and outwardly by A. </s>
  <s xml:id="echoid-s637" xml:space="preserve">The demon-<lb/>ſtration mutatis mutandis is the ſame as before.</s>
  <s xml:id="echoid-s638" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s639" xml:space="preserve"><emph style="sc">Case</emph> 4th. </s>
  <s xml:id="echoid-s640" xml:space="preserve">Suppoſe the given circle A to include B, and it be required that <lb/>the circles to be deſcribed be touched outwardly by them both.</s>
  <s xml:id="echoid-s641" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s642" xml:space="preserve">Let AB cut the circumferences in C and D, P and O: </s>
  <s xml:id="echoid-s643" xml:space="preserve">and biſecting CD in <lb/>I, and ſetting off from I towards P, IL = the ſum of the ſemidiameters of the <lb/>given circles, and with A and B foci, and IL tranſverſe axis, deſcribing an <lb/>ellipſe LKI, it will be the locus of the centers of the circles required. </s>
  <s xml:id="echoid-s644" xml:space="preserve">For <lb/>taking any point K in the ellipſe, and drawing AK and BK, let AK be con-<lb/>tinued to meet one of the given circumferences in G, and let BK meet the other <lb/>F. </s>
  <s xml:id="echoid-s645" xml:space="preserve">Then from the property of the curve AK + BK = IL = AG + BF (by con-
<pb o="(16)" file="0028" n="28"/>
ſtruction:) </s>
  <s xml:id="echoid-s646" xml:space="preserve">hence by ſubſtraction BK = KG + BF, and by ſubtraction again <lb/>FK = KG.</s>
  <s xml:id="echoid-s647" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s648" xml:space="preserve"><emph style="sc">Case</emph> 5th. </s>
  <s xml:id="echoid-s649" xml:space="preserve">Suppoſe the given circle A to include B, and it be required that <lb/>the circles to be deſcribed be touched outwardly by A and inwardly by B.</s>
  <s xml:id="echoid-s650" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s651" xml:space="preserve"><emph style="sc">Then</emph> let AB cut the circumferences in C and D, P and O: </s>
  <s xml:id="echoid-s652" xml:space="preserve">and biſecting <lb/>CO in I, and ſetting off from I towards P, IL = the difference of the ſemidia-<lb/>meters of the given circles, and with A and B foci and IL tranſverſe axis de-<lb/>ſcribing an ellipſe LKI, it will be the locus of the centers of the circles deſcribed, <lb/>and the demonſtration, mutatis mutandis, is the ſame as in the laſt caſe.</s>
  <s xml:id="echoid-s653" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s654" xml:space="preserve"><emph style="sc">Cases</emph> 6th and 7th. </s>
  <s xml:id="echoid-s655" xml:space="preserve">Suppoſe the two given circles cut each other, and it be <lb/>required that the circles to be deſcribed either be touched and included in them <lb/>both, or be touched by them both and at the ſame time include them both.</s>
  <s xml:id="echoid-s656" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s657" xml:space="preserve"><emph style="sc">These</emph> two caſes are ſimilar to caſes 1ſt and 2d, and as there, ſo alſo here, <lb/>the tranſverſe axis of the two oppoſite hyperbolas, which are the loci required, <lb/>muſt be taken = the difference of the ſemidiameters of the given circles. </s>
  <s xml:id="echoid-s658" xml:space="preserve">The <lb/>demonſtration is ſo alike, it need not be repeated.</s>
  <s xml:id="echoid-s659" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div33" type="section" level="1" n="33">
<head xml:id="echoid-head39" xml:space="preserve">PROBLEM IV.</head>
<p>
  <s xml:id="echoid-s660" xml:space="preserve"><emph style="sc">Having</emph> a given point A, and a given right line BC, to determine the locus <lb/>of the centers of thoſe circles which ſhall paſs through A and touch BC.</s>
  <s xml:id="echoid-s661" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s662" xml:space="preserve"><emph style="sc">From</emph> A draw AG perpendicular to BC, then with focus A and directrix BC <lb/>let a parabola be deſcribed, and it will be the locus required; </s>
  <s xml:id="echoid-s663" xml:space="preserve">for by the propert <lb/>of the curve FA always equals FG drawn perpendicular to the directrix.</s>
  <s xml:id="echoid-s664" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div34" type="section" level="1" n="34">
<head xml:id="echoid-head40" xml:space="preserve">PROBLEM V.</head>
<p>
  <s xml:id="echoid-s665" xml:space="preserve"><emph style="sc">Having</emph> a given point A, and a given circle whoſe center is B, to determine <lb/>the locus of the centers of all thoſe circles, which paſs through A, and at the <lb/>ſame time are touched by the given circle.</s>
  <s xml:id="echoid-s666" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s667" xml:space="preserve"><emph style="sc">Cases</emph> 1ſt and 2d. </s>
  <s xml:id="echoid-s668" xml:space="preserve">Suppoſe the point A to lie out of the given circle, and <lb/>it be required that the circles to be deſcribed be either touched outwardly by the <lb/>given circle, or inwardly by it.</s>
  <s xml:id="echoid-s669" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s670" xml:space="preserve"><emph style="sc">Let</emph> AB be drawn, and let it cut the given circumference where it is convex <lb/>towards A in the point C, and where it is concave in the point O: </s>
  <s xml:id="echoid-s671" xml:space="preserve">then biſecting <lb/>AC in E, and ſetting off from E towards B, EH = BC the given radius, and <lb/>with A and B foci and EH tranſverſe Axis deſcribing two oppoſite Hyperbolas <lb/>KEK and LHL, it is evident that KEK will be the locus of the centers of thoſe <lb/>circles which paſs thrugh A and are touched outwardly by the given circle, and <lb/>LHL will be the locus of the centers of thoſe circles which paſs through A and <lb/>are touched inwardly by the given circle.</s>
  <s xml:id="echoid-s672" xml:space="preserve"/>
</p>
<pb o="(17)" file="0029" n="29"/>
<p>
  <s xml:id="echoid-s673" xml:space="preserve"><emph style="sc">Case</emph> 3d. </s>
  <s xml:id="echoid-s674" xml:space="preserve">Suppoſe the given point A to lie in the given circle, whoſe center <lb/>is B.</s>
  <s xml:id="echoid-s675" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s676" xml:space="preserve"><emph style="sc">Then</emph> joining AB and continuing it to meet the given circumference in C and <lb/>O, biſect AC in E, and from E towards O ſetting off EH = BC the given <lb/>Radius, and with A and B Foci and EH tranſverſe Axis deſcribing an Ellipſe <lb/>EKH, it will evidently be the Locus required.</s>
  <s xml:id="echoid-s677" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div35" type="section" level="1" n="35">
<head xml:id="echoid-head41" xml:space="preserve">PROBLEM VI.</head>
<p>
  <s xml:id="echoid-s678" xml:space="preserve"><emph style="sc">Having</emph> a right line BC given, and alſo a circle whoſe center is A, to deter-<lb/>mine the Locus of the centers of the circles which ſhall be touched both by the <lb/>given right line and alſo by the given circle.</s>
  <s xml:id="echoid-s679" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s680" xml:space="preserve"><emph style="sc">There</emph> are three Caſes, but they are all comprchended under one general <lb/>ſolution.</s>
  <s xml:id="echoid-s681" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s682" xml:space="preserve"><emph style="sc">Case</emph> 1ſt. </s>
  <s xml:id="echoid-s683" xml:space="preserve">Let the given right line be without the given circle, and let it be <lb/>required that the circles to be deſcribed be touched outwardly by the given circle.</s>
  <s xml:id="echoid-s684" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s685" xml:space="preserve"><emph style="sc">Case</emph> 2d. </s>
  <s xml:id="echoid-s686" xml:space="preserve">Let the given right line be without the given circle, and let it be <lb/>required that the circles to be deſcribed, be touched inwardly by the given <lb/>circle.</s>
  <s xml:id="echoid-s687" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s688" xml:space="preserve"><emph style="sc">Case</emph> 3d. </s>
  <s xml:id="echoid-s689" xml:space="preserve">Let the given right line be within the given circle, and then the <lb/>circles to be deſcribed muſt be touched outwardly by the given circle.</s>
  <s xml:id="echoid-s690" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div36" type="section" level="1" n="36">
<head xml:id="echoid-head42" xml:space="preserve"><emph style="sc">General</emph> <emph style="sc">Solution.</emph></head>
<p>
  <s xml:id="echoid-s691" xml:space="preserve"><emph style="sc">From</emph> the given center A let fall a perpendicular AG to the given line BC, <lb/>which meets the given circumference in D [or in Caſes 2d and 3d is produced <lb/>to meet it in D] and biſecting DG in F, and ſetting off FM = FA (which is the <lb/>ſame thing as making GM = AD the given Radius) and through M drawing <lb/>MLK parallel to the given line BC, with A Focus and LK Directrix deſcribe a <lb/>Parabola, and it will be the Locus of the centers of the circles required; </s>
  <s xml:id="echoid-s692" xml:space="preserve">for <lb/>from the property of the Curve FA = FM, and adding equals to equals, or <lb/>ſubtracting equals from equals, as the Caſe requires, FD = FG.</s>
  <s xml:id="echoid-s693" xml:space="preserve"/>
</p>
<pb file="0030" n="30"/>
<pb o="(19)" file="0031" n="31"/>
</div>
<div xml:id="echoid-div37" type="section" level="1" n="37">
<head xml:id="echoid-head43" xml:space="preserve">A SECOND <lb/>SUPPLEMENT, <lb/>BEING <lb/>Monſ. DE FERMAT’S Treatiſe on <lb/>Spherical Tangencies.</head>
<head xml:id="echoid-head44" xml:space="preserve">PROBLEM I.</head>
<p>
  <s xml:id="echoid-s694" xml:space="preserve">HAVING four points N, O, M, F, given, to deſcribe a ſphere which <lb/>ſhall paſs through them all.</s>
  <s xml:id="echoid-s695" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s696" xml:space="preserve"><emph style="sc">Taking</emph> any three of them N, O, M, ad libitum, they will form a triangle, <lb/>about which a circle ANOM may be circumſcribed, which will be given in <lb/>magnitude and poſition. </s>
  <s xml:id="echoid-s697" xml:space="preserve">That this circle is in the ſurface of the ſphere <lb/>ſought appears from hence; </s>
  <s xml:id="echoid-s698" xml:space="preserve">becauſe if a ſphere be cut by any plane, the <lb/>ſection will be a circle; </s>
  <s xml:id="echoid-s699" xml:space="preserve">but only one circle can be drawn to paſs through the <lb/>three given points N, O, M; </s>
  <s xml:id="echoid-s700" xml:space="preserve">therefore this circle muſt be in the ſurface of <lb/>the ſphere. </s>
  <s xml:id="echoid-s701" xml:space="preserve">Let the center of this circle be C, from whence let CB be <lb/>erected perpendicular to it’s plane; </s>
  <s xml:id="echoid-s702" xml:space="preserve">it is evident that the center of the ſphere <lb/>ſought will be in this line CB. </s>
  <s xml:id="echoid-s703" xml:space="preserve">From the fourth given point F let FB be <lb/>drawn perpendicular to CB, which FB will be alſo given in magnitude and <lb/>poſition. </s>
  <s xml:id="echoid-s704" xml:space="preserve">Through C draw ACD parallel to FB, and this line will be a
<pb o="(20)" file="0032" n="32"/>
diameter of the circle given in poſition, and therefore the points A and D <lb/>will alſo be given.</s>
  <s xml:id="echoid-s705" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s706" xml:space="preserve"><emph style="sc">Let</emph>us now ſuppoſe the thing done, and that the center of the ſphere <lb/>ſought is E, which, as obſerved before, muſt be in the line CB. </s>
  <s xml:id="echoid-s707" xml:space="preserve">Drawing <lb/>EF, EA, ED, theſe lines muſt be equal, ſince the points F, A, D, have <lb/>been ſhewn to be in the ſurface of the ſphere. </s>
  <s xml:id="echoid-s708" xml:space="preserve">But theſe lines EF, EA, ED, <lb/>are in the ſame plane, ſince FB and AD are parallel, and BC perpendicular <lb/>to each of them. </s>
  <s xml:id="echoid-s709" xml:space="preserve">If therefore a circle be deſcribed to paſs through the three <lb/>points F, A, D, whoſe center is E, it will be in the line CB, and will be <lb/>the center of the ſphere required.</s>
  <s xml:id="echoid-s710" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div38" type="section" level="1" n="38">
<head xml:id="echoid-head45" xml:space="preserve">PROBLEM II.</head>
<p>
  <s xml:id="echoid-s711" xml:space="preserve"><emph style="sc">Having</emph> three points N, O, M, given, and a plane AD, to deſcribe a-<lb/>ſphere which ſhall paſs through the three given points; </s>
  <s xml:id="echoid-s712" xml:space="preserve">and alſo touch the <lb/>given plane.</s>
  <s xml:id="echoid-s713" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s714" xml:space="preserve"><emph style="sc">Let</emph> a circle ENOM be deſcribed to paſs through the three given points, <lb/>it will be in the ſurface of the ſphere ſought, from what was ſaid under the <lb/>former Problem. </s>
  <s xml:id="echoid-s715" xml:space="preserve">From it’s center I let a perpendicular to it’s plane IA be <lb/>erected; </s>
  <s xml:id="echoid-s716" xml:space="preserve">the center of the ſphere ſought will be in this line IA; </s>
  <s xml:id="echoid-s717" xml:space="preserve">let the line <lb/>IA meet the given plane in the point A, which point will be therefore given. <lb/></s>
  <s xml:id="echoid-s718" xml:space="preserve">From the center of the given circle ENOM, let a perpendicular to the given <lb/>plane ID be drawn, the point D will then be given, and therefore the line <lb/>AD both in poſition and magnitude, as likewiſe the lines ID, IA, and the <lb/>plane of the triangle ADI. </s>
  <s xml:id="echoid-s719" xml:space="preserve">But the plane of the circle NOM is alſo given <lb/>in poſition, and therefore alſo their common ſection EIF, and hence the <lb/>points E and F.</s>
  <s xml:id="echoid-s720" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s721" xml:space="preserve">Suppoſe now the thing done, and that the center of the ſphere ſought is B. <lb/></s>
  <s xml:id="echoid-s722" xml:space="preserve">Draw BE, BF, and BC parallel to ID. </s>
  <s xml:id="echoid-s723" xml:space="preserve">Since the triangle ADI, and the <lb/>line EIF are in the ſame plane, therefore BE, BF, BC, will be in the ſame <lb/>plane. </s>
  <s xml:id="echoid-s724" xml:space="preserve">But the line ID is perpendicular to the given plane, therefore the <lb/>line BC parallel to it, will alſo be perpendicular to the given plane. </s>
  <s xml:id="echoid-s725" xml:space="preserve">Since <lb/>then a ſphere is to be deſcribed to touch the plane AD, a perpendicular BC <lb/>from it’s center B will give the point of contact C; </s>
  <s xml:id="echoid-s726" xml:space="preserve">and BC, BE, BF will be <lb/>equal, and it has been proved that they are in a plane given in poſition, in <lb/>which plane is alſo the right line AD. </s>
  <s xml:id="echoid-s727" xml:space="preserve">The queſtion is then reduced to this, <lb/>Having two points E and F given, as alſo a right line AD in the ſame plane,
<pb o="(21)" file="0033" n="33"/>
to find the center of a circle which will paſs through the two points, and like-<lb/>wiſe touch the right line, which is the VIIth of the preceeding Problems.</s>
  <s xml:id="echoid-s728" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div39" type="section" level="1" n="39">
<head xml:id="echoid-head46" xml:space="preserve">PROBLEM III.</head>
<p>
  <s xml:id="echoid-s729" xml:space="preserve"><emph style="sc">Having</emph> three points N, O, M, given, as likewiſe a ſphere IG, to de-<lb/>ſcribe a ſphere which will paſs through the three given points, and likewiſe <lb/>touch the given ſphere.</s>
  <s xml:id="echoid-s730" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s731" xml:space="preserve">The circle NOM in the ſurface of the ſphere ſought is given, and a per-<lb/>pendicular to its plane from it’s center FA being drawn, the center of the <lb/>ſphere required will be in this line. </s>
  <s xml:id="echoid-s732" xml:space="preserve">From I the center of the given ſphere <lb/>let IB be drawn perpendicular to FA, and through F, ED parallel to IB, <lb/>which, from what has been before proved, will be in the plane of the circle <lb/>NOM, and the points E and D will be given.</s>
  <s xml:id="echoid-s733" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s734" xml:space="preserve">Suppoſe now the thing done, and that the center of the ſphere required is <lb/>C. </s>
  <s xml:id="echoid-s735" xml:space="preserve">Then the lines CI, CE, CD, will be in the ſame plane, which is given, as <lb/>the points I, E, and D are given. </s>
  <s xml:id="echoid-s736" xml:space="preserve">But the point of contact of two ſpheres is <lb/>in the line joining their centers; </s>
  <s xml:id="echoid-s737" xml:space="preserve">therefore the ſphere ſought will touch the <lb/>ſphere given in the point G, and the line IC will exceed the lines EC, ED, by <lb/>IG the radius of the given ſphere: </s>
  <s xml:id="echoid-s738" xml:space="preserve">with center I therefore and this diſtance <lb/>IG let a circle be deſcribed in the plane of the lines CI, CE, CD, and it <lb/>will paſs through the point G and be given in magnitude and poſition; </s>
  <s xml:id="echoid-s739" xml:space="preserve">but <lb/>the points D and E are alſo in the ſame plane; </s>
  <s xml:id="echoid-s740" xml:space="preserve">and therefore the queſtion is <lb/>reduced to this, Having two points E and D given, as likewiſe a circle <lb/>IGH, to find the center of a circle which will paſs through the two points <lb/>and likewiſe touch the circle, which is the XIIth of the preceeding Problems.</s>
  <s xml:id="echoid-s741" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div40" type="section" level="1" n="40">
<head xml:id="echoid-head47" xml:space="preserve">PROBLEM IV.</head>
<p>
  <s xml:id="echoid-s742" xml:space="preserve"><emph style="sc">Having</emph> four planes AH, AB, BC, HG, given; </s>
  <s xml:id="echoid-s743" xml:space="preserve">it is required to de-<lb/>ſcribe a ſphere which ſhall touch them all four.</s>
  <s xml:id="echoid-s744" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s745" xml:space="preserve"><emph style="sc">If</emph> two planes touch a ſphere, the center of that ſphere will be in a plane <lb/>beſecting the inclination of the other two. </s>
  <s xml:id="echoid-s746" xml:space="preserve">And if the planes be parallel, it <lb/>will be in a parallel plane beſecting their interval. </s>
  <s xml:id="echoid-s747" xml:space="preserve">This being allowed, <lb/>which is too evident to need further proof; </s>
  <s xml:id="echoid-s748" xml:space="preserve">the center of the ſphere ſought <lb/>will be in a plane biſecting the inclination of two planes CB and BA; </s>
  <s xml:id="echoid-s749" xml:space="preserve">it will <lb/>likewiſe be in another plane biſecting the inclination of the two planes BA and <lb/>AH; </s>
  <s xml:id="echoid-s750" xml:space="preserve">and therefore in a right line, which is the common ſection of theſe two
<pb o="(22)" file="0034" n="34"/>
biſecting planes; </s>
  <s xml:id="echoid-s751" xml:space="preserve">let this right line be EF. </s>
  <s xml:id="echoid-s752" xml:space="preserve">Moreover, the center of the <lb/>ſphere ſought will alſo be in a plane biſecting the inclination of the two planes <lb/>AH and GH, and the interſection of this laſt biſecting plane with the right <lb/>line EF will give a point D, which will be the center of the ſphere required.</s>
  <s xml:id="echoid-s753" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div41" type="section" level="1" n="41">
<head xml:id="echoid-head48" xml:space="preserve">PROBLEM V.</head>
<p>
  <s xml:id="echoid-s754" xml:space="preserve"><emph style="sc">Having</emph> three planes AB, BC, CD, given, and alſo a point H; </s>
  <s xml:id="echoid-s755" xml:space="preserve">to find <lb/>a ſphere which ſhall paſs through the given point, and likewiſe touch the <lb/>three given planes.</s>
  <s xml:id="echoid-s756" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s757" xml:space="preserve"><emph style="sc">Suppose</emph> it done. </s>
  <s xml:id="echoid-s758" xml:space="preserve">The three planes, by what was ſaid under the laſt pro-<lb/>poſition, will give a right line in poſition, in which will be the center of the <lb/>ſphere required. </s>
  <s xml:id="echoid-s759" xml:space="preserve">Let this right line be GE, perpendicular to which from H <lb/>the given point let HI be drawn, which therefore will be given in magnitude <lb/>and poſition. </s>
  <s xml:id="echoid-s760" xml:space="preserve">Let HI be produced, and FI taken equal to HI; </s>
  <s xml:id="echoid-s761" xml:space="preserve">the point <lb/>F will then be given. </s>
  <s xml:id="echoid-s762" xml:space="preserve">Now ſince the center of the ſphere required is in the <lb/>line GE, and FH is perpendicular thereto and biſected thereby, and one <lb/>extreme H is by hypotheſis in the ſurface of the ſaid ſphere, the other extreme <lb/>F will be ſo too. </s>
  <s xml:id="echoid-s763" xml:space="preserve">Nay even a circle deſcribed with I center and IH radius <lb/>in a plane perpendicular to GE will be in the ſaid ſpherical ſurface. </s>
  <s xml:id="echoid-s764" xml:space="preserve">Here <lb/>then we have a circle given in magnitude and poſition, and taking any one <lb/>of the given planes AB, by an evident corollary from Problem II. </s>
  <s xml:id="echoid-s765" xml:space="preserve">of this <lb/>Supplement, a ſphere may be deſcribed which will touch the given plane, <lb/>and likewiſe have the given circle in it’s ſurface; </s>
  <s xml:id="echoid-s766" xml:space="preserve">and ſuch a ſphere will <lb/>anſwer every thing here required.</s>
  <s xml:id="echoid-s767" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div42" type="section" level="1" n="42">
<head xml:id="echoid-head49" xml:space="preserve">PROBLEM VI.</head>
<p>
  <s xml:id="echoid-s768" xml:space="preserve"><emph style="sc">Having</emph> three planes ED, DB, BC, given, and alſo a ſphere RM, to <lb/>conſtruct a ſphere which ſhall touch the given one, and likewiſe the three <lb/>given planes.</s>
  <s xml:id="echoid-s769" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s770" xml:space="preserve"><emph style="sc">Suppose</emph> it done, and that the ſphere ERCA is the required one, viz. <lb/></s>
  <s xml:id="echoid-s771" xml:space="preserve">touches the ſphere in R, and the planes in E, A, C. </s>
  <s xml:id="echoid-s772" xml:space="preserve">Let the center of this <lb/>ſphere be O; </s>
  <s xml:id="echoid-s773" xml:space="preserve">then drawing RO, EO, AO, CO, they will all be equal, and <lb/>RO will paſs through M the center of the given ſphere; </s>
  <s xml:id="echoid-s774" xml:space="preserve">and EO, AO, CO, <lb/>will be perpendicular to the planes ED, DB, BC. </s>
  <s xml:id="echoid-s775" xml:space="preserve">Let OU, OG, OI, be <lb/>made each equal to OM; </s>
  <s xml:id="echoid-s776" xml:space="preserve">and through the points U, G and I, let the planes <lb/>UP, GH, IN, be ſuppoſed drawn parallel to the given ones ED, DB, BC, <lb/>reſpectively. </s>
  <s xml:id="echoid-s777" xml:space="preserve">Since OR is equal to OE, and OM equal to OU, RM will be
<pb o="(23)" file="0035" n="35"/>
equal to UE: </s>
  <s xml:id="echoid-s778" xml:space="preserve">but RM is given in magnitude, being the radius of the given <lb/>ſphere, therefore UE is alſo given in magnitude. </s>
  <s xml:id="echoid-s779" xml:space="preserve">And ſince OE is perpen-<lb/>dicular to the plane DE, it will be alſo to the plane PU which is parallel <lb/>thereto. </s>
  <s xml:id="echoid-s780" xml:space="preserve">UE then being given in magnitude, and being the interval be-<lb/>tween two parallel planes DE, PU, whereof DE is given in poſition by hypo-<lb/>theſis, the other PU will alſo be given in poſition. </s>
  <s xml:id="echoid-s781" xml:space="preserve">In the ſame manner it <lb/>may be proved that the planes GH, IN, are given in poſition, and that the <lb/>lines OG, OI, are perpendicular thereto reſpectively, and each alſo equal to <lb/>OM. </s>
  <s xml:id="echoid-s782" xml:space="preserve">A ſphere therefore deſcribed with center O and OM radius will touch <lb/>the three planes PU, GH, IN, given in poſition: </s>
  <s xml:id="echoid-s783" xml:space="preserve">but the point M is given, <lb/>being the center of the given ſphere. </s>
  <s xml:id="echoid-s784" xml:space="preserve">The queſtion is then reduced to this, <lb/>Having three planes given PU, GH, IN, and a point M, to find the radius <lb/>of a ſphere which ſhall touch the given planes, and paſs through the given <lb/>point; </s>
  <s xml:id="echoid-s785" xml:space="preserve">which is the ſame as the preceeding Problem. </s>
  <s xml:id="echoid-s786" xml:space="preserve">[And this radius be-<lb/>ing increaſed or diminiſhed by MR, according as R is taken in the further or <lb/>nearer ſurface of the given ſphere, will give the radius of a ſphere which will <lb/>touch the three given planes DE, DB, BC, and likewiſe the given <lb/>ſphere.</s>
  <s xml:id="echoid-s787" xml:space="preserve">]</s>
</p>
<p>
  <s xml:id="echoid-s788" xml:space="preserve"><emph style="sc">By</emph> a like method, when among the Data there are no points, but only <lb/>planes and ſpheres, we ſhall always be able to ſubſtitute a given point in the <lb/>place of a given ſphere.</s>
  <s xml:id="echoid-s789" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div43" type="section" level="1" n="43">
<head xml:id="echoid-head50" xml:space="preserve">PROBLEM VII.</head>
<p>
  <s xml:id="echoid-s790" xml:space="preserve"><emph style="sc">Having</emph> two points H, M, as alſo two planes AB, BC, given, to find a <lb/>ſphere which ſhall paſs through the given points, and touch the given <lb/>planes.</s>
  <s xml:id="echoid-s791" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s792" xml:space="preserve"><emph style="sc">Draw</emph> HM and biſect it in I, the point I will be given, through the <lb/>point I let a plane be erected perpendicular to the right line HM, this plane <lb/>will be given in poſition, and the center of the ſphere required will be in this <lb/>plane. </s>
  <s xml:id="echoid-s793" xml:space="preserve">But becauſe it is alſo to touch the planes AB, BC, its center will be <lb/>alſo in another plane given in poſition (by what has been proved, Prob. </s>
  <s xml:id="echoid-s794" xml:space="preserve">IV.) <lb/></s>
  <s xml:id="echoid-s795" xml:space="preserve">and therefore in a right line which is their interſection, given in poſition, <lb/>which let be GE; </s>
  <s xml:id="echoid-s796" xml:space="preserve">to which line GE from one of the given points M demit-<lb/>ting a perpendicular MF, it will be given in magnitude and poſition, and <lb/>being continued to D ſo that FD equals MF, the point D will be given; </s>
  <s xml:id="echoid-s797" xml:space="preserve"><lb/>and, from what has been proved before, will be in the ſpherical ſurface.</s>
  <s xml:id="echoid-s798" xml:space="preserve">
<pb o="(24)" file="0036" n="36"/>
Thercfore there are given three points H, M, D, as likewiſe a plane AB, <lb/>or AC, through which points the ſphere is to paſs, and alſo touch the given <lb/>plane. </s>
  <s xml:id="echoid-s799" xml:space="preserve">Hence it appears that this Problem is reduced to the IId of this <lb/>Supplement.</s>
  <s xml:id="echoid-s800" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s801" xml:space="preserve"><emph style="sc">Before</emph> we proceed, the following eaſy Lemmas muſt be premiſed.</s>
  <s xml:id="echoid-s802" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div44" type="section" level="1" n="44">
<head xml:id="echoid-head51" xml:space="preserve">LEMMA I.</head>
<p>
  <s xml:id="echoid-s803" xml:space="preserve"><emph style="sc">Let</emph> there be a circle BCD, and a point E taken without it, and iſ from <lb/>E a line EDOB be drawn to paſs through the center, and another line ECA <lb/>to cut it any ways; </s>
  <s xml:id="echoid-s804" xml:space="preserve">we know from the Elements that the rectangle AEC is <lb/>equal to the rectangle BED. </s>
  <s xml:id="echoid-s805" xml:space="preserve">Let us now ſuppoſe a ſphere whoſe center is O, <lb/>and one of whoſe great circles is ACDB; </s>
  <s xml:id="echoid-s806" xml:space="preserve">if from the ſame point E a line <lb/>ECA be any-how drawn to meet the ſpherical ſurface in the points C and A, <lb/>I ſay the rectangle AEC will ſtill be equal to the rectangle BED. </s>
  <s xml:id="echoid-s807" xml:space="preserve">For if we <lb/>ſuppoſe the circle and right line ECA to revolve upon EDB as an immove-<lb/>able axis, the lines EC and EA will not be changed, becauſe the points C <lb/>and A deſcribe circles whoſe planes are perpendicular to that axis; </s>
  <s xml:id="echoid-s808" xml:space="preserve">and <lb/>therefore the rectangle AEC will in any plane be ſtill equal to the rectangle <lb/>BED.</s>
  <s xml:id="echoid-s809" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div45" type="section" level="1" n="45">
<head xml:id="echoid-head52" xml:space="preserve">LEMMA II.</head>
<p>
  <s xml:id="echoid-s810" xml:space="preserve"><emph style="sc">By</emph> the ſame method of reaſoning, the Vth Lemma immediately preceed-<lb/>ing Problem XIII, in the Treatiſe of Circular Tangencies, may be extended <lb/>alſo to ſpheres, viz. </s>
  <s xml:id="echoid-s811" xml:space="preserve">that in any plane (ſee the Figures belonging to that <lb/>Lemma) MG X MB = MH X MA. </s>
  <s xml:id="echoid-s812" xml:space="preserve">And alſo that MF X MC = ME X MI.</s>
  <s xml:id="echoid-s813" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div46" type="section" level="1" n="46">
<head xml:id="echoid-head53" xml:space="preserve">LEMMA III.</head>
<p>
  <s xml:id="echoid-s814" xml:space="preserve"><emph style="sc">Let</emph> there be two ſpheres YN, XM, through whoſe centers let the right <lb/>line RYNXMU paſs, and let it be as the radius YN to the radius XM, ſo <lb/>YU to XU; </s>
  <s xml:id="echoid-s815" xml:space="preserve">and from the point U let a line UTS be drawn in any plane, <lb/>and let the rectangle S U T be equal to the rectangle RUM; </s>
  <s xml:id="echoid-s816" xml:space="preserve">I fay that if <lb/>any ſphere OTS be deſcribed to paſs through the points T and S, and to <lb/>touch one of the given ſpheres XM as in O, it will alſo touch the other <lb/>given ſphere YN. </s>
  <s xml:id="echoid-s817" xml:space="preserve">For joining UO, and producing it to meet the ſurface of <lb/>the ſphere OTS in Q; </s>
  <s xml:id="echoid-s818" xml:space="preserve">the rectangle QUO = the rectangle SUT, by <lb/>Lemma I. </s>
  <s xml:id="echoid-s819" xml:space="preserve">but the rectangle SUT = the rectangle RUM. </s>
  <s xml:id="echoid-s820" xml:space="preserve">by conſtruction,
<pb o="(25)" file="0037" n="37"/>
which RUM by Lemma II. </s>
  <s xml:id="echoid-s821" xml:space="preserve">is equal to a rectangle under UO and a line <lb/>drawn through the points U and O to the further ſurface of the ſphere YN. <lb/></s>
  <s xml:id="echoid-s822" xml:space="preserve">Therefore the point Q is in the ſurface of the ſphere YN; </s>
  <s xml:id="echoid-s823" xml:space="preserve">it is therefore <lb/>common to the ſpheres YN and OTS; </s>
  <s xml:id="echoid-s824" xml:space="preserve">and I ſay that theſe ſpheres touch in <lb/>the ſaid point Q. </s>
  <s xml:id="echoid-s825" xml:space="preserve">For from the point U let a line UZ be drawn in any <lb/>plane of the ſphere OTS, and being produced let it cut the three <lb/>ſpheres in the points Z, D, H, K, P, B. </s>
  <s xml:id="echoid-s826" xml:space="preserve">The rectangle ZUB in the <lb/>ſphere OTS is by Lemma I. </s>
  <s xml:id="echoid-s827" xml:space="preserve">and II. </s>
  <s xml:id="echoid-s828" xml:space="preserve">equal to the rectangle DUP terminated <lb/>by the ſpheres XM and YN. </s>
  <s xml:id="echoid-s829" xml:space="preserve">But DU is greater than ZU, becauſe the <lb/>ſpheres XM and OTS touch in the point O, and therefore any other line from <lb/>U but UO muſt meet the ſurface of OTS before it meets the ſurface XM. </s>
  <s xml:id="echoid-s830" xml:space="preserve"><lb/>Since then ZUB = DUP, and DU is greater than ZU, UP muſt be leſs <lb/>than UB, and the point B will fall without the ſphere YN; </s>
  <s xml:id="echoid-s831" xml:space="preserve">and by the <lb/>fame reaſon, all other points in the ſurface of the ſphere OTS, except the <lb/>point Q.</s>
  <s xml:id="echoid-s832" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s833" xml:space="preserve"><emph style="sc">The</emph> Demonſtration is ſimilar and equally eaſy in all caſes, whether the <lb/>ſpheres touch exlernally or internally.</s>
  <s xml:id="echoid-s834" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div47" type="section" level="1" n="47">
<head xml:id="echoid-head54" xml:space="preserve">LEMMA IV.</head>
<p>
  <s xml:id="echoid-s835" xml:space="preserve"><emph style="sc">Let</emph> there be a plane AC, and a ſphere FGD through whoſe center O let <lb/>FODB be drawn perpendicular to the plane, and from F any right line <lb/>FGA cutting the ſphere in G and the plane in A; </s>
  <s xml:id="echoid-s836" xml:space="preserve">I ſay that the rectangle <lb/>AFG = the rectangle BFD. </s>
  <s xml:id="echoid-s837" xml:space="preserve">For let the given ſphere and plane be cut by <lb/>the plane of the triangle ABF, the ſection of the one will be the circle GDF, <lb/>and of the other the right line ABC. </s>
  <s xml:id="echoid-s838" xml:space="preserve">Since the line FB is perpendicular <lb/>to the plane AC, it will be alſo to the right line AC. </s>
  <s xml:id="echoid-s839" xml:space="preserve">Having then a circle <lb/>FDB, and a right line AC in the ſame plane; </s>
  <s xml:id="echoid-s840" xml:space="preserve">and a line FDB paſſing thro' <lb/>the center perpendicular to AC, join D and G, and in the quadrilateral <lb/>figure ABDG the angles at B and G being both right ones, it will be in a <lb/>circle, and the rectangle AFG = the rectangle BFD; </s>
  <s xml:id="echoid-s841" xml:space="preserve">and the ſame may be <lb/>proved in any other ſection of the ſphere.</s>
  <s xml:id="echoid-s842" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div48" type="section" level="1" n="48">
<head xml:id="echoid-head55" xml:space="preserve">LEMMA V.</head>
<p>
  <s xml:id="echoid-s843" xml:space="preserve"><emph style="sc">Let</emph> there be a plane ABD and a ſphere EGF, through whoſe center O <lb/>let FOEC be drawn perpendicular to the plane, and in any other plane let <lb/>FHI be drawn ſo that the rectangle IFH = the rectangle CFE: </s>
  <s xml:id="echoid-s844" xml:space="preserve">if through
<pb o="(26)" file="0038" n="38"/>
the points I and H a ſphere be deſcribed which touches the plane AC, I ſay <lb/>it will alſo touch the ſphere EGF. </s>
  <s xml:id="echoid-s845" xml:space="preserve">From F draw FB to the point of contact <lb/>of the ſphere and plane, and make the rectangle BFN = the rectangle CFE, <lb/>and the point N will be in the ſurface of the ſphere EGF, by Lemma IV. <lb/></s>
  <s xml:id="echoid-s846" xml:space="preserve">But the rectangle CFE, by conſtruction, = the rectangle IFH; </s>
  <s xml:id="echoid-s847" xml:space="preserve">therefore <lb/>IFH = BFN, and the point N will be alſo in the ſurſace of the ſphere IHB. </s>
  <s xml:id="echoid-s848" xml:space="preserve"><lb/>It remains then to be proved that theſe ſpheres touch in N, which is very eaſy <lb/>to be done. </s>
  <s xml:id="echoid-s849" xml:space="preserve">For from the point F through any point R in the ſpherical ſur-<lb/>face EGF let the line FR be drawn, which may cut the ſpherical ſurface <lb/>IBH in L and P, and the plane AC in K. </s>
  <s xml:id="echoid-s850" xml:space="preserve">The rectangle KFR = the rect-<lb/>angle CFE, by Lemma IV. </s>
  <s xml:id="echoid-s851" xml:space="preserve">= the rectangle IFH, by conſtruction, = the <lb/>rectangle PFL. </s>
  <s xml:id="echoid-s852" xml:space="preserve">Since then KFR = PFL, and KF is greater than PF, be-<lb/>cauſe the ſphere IHB touches the plane AC in B, therefore FR is leſs than <lb/>FL, and the point R is without the ſphere IHB, and the ſame may be ſhewn <lb/>of every other point in the ſpherical ſurface EGF, except the point N.</s>
  <s xml:id="echoid-s853" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s854" xml:space="preserve">Theſe Lemmas, though they be very eaſy, are very elegant and valuable, <lb/>eſpecially the IIId and Vth. </s>
  <s xml:id="echoid-s855" xml:space="preserve">In the IIId. </s>
  <s xml:id="echoid-s856" xml:space="preserve">though there be an inſinite num-<lb/>ber of ſpheres which, paſſing through the points T and S, may touch the <lb/>ſphere XM, yet they will all alſo touch the ſphere YN, by what is there <lb/>proved. </s>
  <s xml:id="echoid-s857" xml:space="preserve">In the Vth, though there be an infinite number of ſpheres which, <lb/>paſſing through the points I and H, may touch the plane AC, yet they will <lb/>all alſo touch the ſphere EGF, by what is there proved.</s>
  <s xml:id="echoid-s858" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s859" xml:space="preserve">We ſhall now be able to go through the remaining Problems with eaſe.</s>
  <s xml:id="echoid-s860" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div49" type="section" level="1" n="49">
<head xml:id="echoid-head56" xml:space="preserve">PROBLEM VIII.</head>
<p>
  <s xml:id="echoid-s861" xml:space="preserve">Let there be given a plane ABC, and two points H and M, and alſo a <lb/>ſphere DFE; </s>
  <s xml:id="echoid-s862" xml:space="preserve">to find a ſphere which ſhall paſs through the given points, and <lb/>touch the given plane, and likewiſe the given ſphere.</s>
  <s xml:id="echoid-s863" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s864" xml:space="preserve">Through the center O of the given ſphere let EODB be demitted perpen-<lb/>dicular to the given plane ABC, and let HE be drawn, and make the rect-<lb/>angle HEG equal to the rectangle BED, and G will then be given. </s>
  <s xml:id="echoid-s865" xml:space="preserve">Find <lb/>then a ſphere, by Problem II. </s>
  <s xml:id="echoid-s866" xml:space="preserve">which ſhall paſs through the three points M, <lb/>H, G, and touch the plane ABC, and it will be the ſphere here required. <lb/></s>
  <s xml:id="echoid-s867" xml:space="preserve">For it paſſes through the points M and H, and touches the plane ABC, <lb/>by conſtruction; </s>
  <s xml:id="echoid-s868" xml:space="preserve">it likewiſe touches the ſphere DFE, by Lemma V. </s>
  <s xml:id="echoid-s869" xml:space="preserve">For <lb/>ſince the rectangle HEG = the rectangle BED, every ſphere which paſſes
<pb o="(27)" file="0039" n="39"/>
through the points H and G, and touches the plane ABC, touches likewiſe <lb/>the ſphere DFE.</s>
  <s xml:id="echoid-s870" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div50" type="section" level="1" n="50">
<head xml:id="echoid-head57" xml:space="preserve">PROBLEM IX.</head>
<p>
  <s xml:id="echoid-s871" xml:space="preserve"><emph style="sc">Let</emph> there be given two ſpheres AB, DE, as alſo two points H and M; <lb/></s>
  <s xml:id="echoid-s872" xml:space="preserve">to find a ſphere which ſhall paſs through the two given points, and likewiſe <lb/>touch the two given ſpheres.</s>
  <s xml:id="echoid-s873" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s874" xml:space="preserve"><emph style="sc">Let</emph> the right line AF be drawn paſſing through the centers of the <lb/>ſpheres, and as the radius AB is to the radius DE, ſo make BF to EF, and <lb/>the point F will be given. </s>
  <s xml:id="echoid-s875" xml:space="preserve">Make the rectangle HFG = the rectangle NFA, <lb/>and the point G will be given. </s>
  <s xml:id="echoid-s876" xml:space="preserve">Now having given three points M, H, G, <lb/>as alſo a ſphere DE; </s>
  <s xml:id="echoid-s877" xml:space="preserve">find a ſphere by Problem III, which ſhall paſs through <lb/>the given points, and touch the given ſphere; </s>
  <s xml:id="echoid-s878" xml:space="preserve">and, by Lemma III, it will be <lb/>the ſphere here required.</s>
  <s xml:id="echoid-s879" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div51" type="section" level="1" n="51">
<head xml:id="echoid-head58" xml:space="preserve">PROBLEM X.</head>
<p>
  <s xml:id="echoid-s880" xml:space="preserve"><emph style="sc">Let</emph> there be given two planes AB, BD, a point H, and a ſphere <lb/>EGF; </s>
  <s xml:id="echoid-s881" xml:space="preserve">to find a ſphere which ſhall paſs through the given point, and touch <lb/>the given ſphere, as alſo the two given planes.</s>
  <s xml:id="echoid-s882" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s883" xml:space="preserve"><emph style="sc">Through</emph> the center O of the given ſphere let a perpendicular to either of <lb/>the given planes CEOF be demitted, and make the rectangle HFI = the <lb/>rectangle CFE. </s>
  <s xml:id="echoid-s884" xml:space="preserve">Then having given the two points H and I, as alſo the <lb/>two planes AB, BD; </s>
  <s xml:id="echoid-s885" xml:space="preserve">find a ſphere, by Problem VII, which ſhall paſs <lb/>through the two given points, and likewiſe touch the two given planes; </s>
  <s xml:id="echoid-s886" xml:space="preserve">and, <lb/>by Lemma V, it will be the ſphere required.</s>
  <s xml:id="echoid-s887" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div52" type="section" level="1" n="52">
<head xml:id="echoid-head59" xml:space="preserve">PROBLEM XI.</head>
<p>
  <s xml:id="echoid-s888" xml:space="preserve"><emph style="sc">Let</emph> there be given a point, a plane, and two ſpheres; </s>
  <s xml:id="echoid-s889" xml:space="preserve">to find a ſphere <lb/>which ſhall paſs through the point, touch the plane, and alſo the two <lb/>ſpheres.</s>
  <s xml:id="echoid-s890" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s891" xml:space="preserve">This Problem, by a like method of reaſoning, is immediately reduced to <lb/>the VIIIth, where two points, a plane, and a ſphere are given, and that by <lb/>means of the Vth Lemma. </s>
  <s xml:id="echoid-s892" xml:space="preserve">But if you chuſe to uſe the IIId Lemma, it will <lb/>be reduced to the ſame Problem by a different method, and a different <lb/>conſtruction.</s>
  <s xml:id="echoid-s893" xml:space="preserve"/>
</p>
<pb o="(28)" file="0040" n="40"/>
</div>
<div xml:id="echoid-div53" type="section" level="1" n="53">
<head xml:id="echoid-head60" xml:space="preserve">PROBLEM XII.</head>
<p>
  <s xml:id="echoid-s894" xml:space="preserve"><emph style="sc">Let</emph> there be given a point and three ſpheres, to ſind a ſphere which ſhall <lb/>paſs through the point, and touch all the three ſpheres.</s>
  <s xml:id="echoid-s895" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s896" xml:space="preserve"><emph style="sc">We</emph> aſſign no figure to this Problem alſo, becauſe, by help of Lemma III, <lb/>it may immediately be reduced to Problem IX, where two points and two <lb/>ſpheres are given.</s>
  <s xml:id="echoid-s897" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div54" type="section" level="1" n="54">
<head xml:id="echoid-head61" xml:space="preserve">PROBLEM XIII.</head>
<p>
  <s xml:id="echoid-s898" xml:space="preserve"><emph style="sc">Let</emph> there be two planes, and alſo two ſpheres given; </s>
  <s xml:id="echoid-s899" xml:space="preserve">to find a ſphere <lb/>which ſhall touch the planes, as alſo the ſpheres.</s>
  <s xml:id="echoid-s900" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s901" xml:space="preserve">Suppoſe the thing done. </s>
  <s xml:id="echoid-s902" xml:space="preserve">If therefore we imagine another ſpherical ſurface <lb/>parallel to that which is required, and which we now ſuppoſe found, and <lb/>whoſe radius is leſs than it's by the radius of the leſſer of the two given <lb/>ſpheres; </s>
  <s xml:id="echoid-s903" xml:space="preserve">this new ſpherical ſurface will touch two planes parallel to the two <lb/>given ones, and whoſe diſtance therefrom will be equal to the radius of the <lb/>leſſer of the given ſpheres; </s>
  <s xml:id="echoid-s904" xml:space="preserve">it will alſo touch a ſphere concentric to the <lb/>greater given one whoſe radius is leſs than it's by the radius of the leſſer given <lb/>one; </s>
  <s xml:id="echoid-s905" xml:space="preserve">and it will likewife paſs through the center of the leſſer given one. <lb/></s>
  <s xml:id="echoid-s906" xml:space="preserve">The Queſtion is then reduced to Problem X, where a point, two planes and <lb/>a ſphere are given.</s>
  <s xml:id="echoid-s907" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div55" type="section" level="1" n="55">
<head xml:id="echoid-head62" xml:space="preserve">PROBLEM XIV.</head>
<p>
  <s xml:id="echoid-s908" xml:space="preserve"><emph style="sc">Having</emph> three ſpheres and a plane given; </s>
  <s xml:id="echoid-s909" xml:space="preserve">to find a ſphere which ſhall <lb/>touch them all.</s>
  <s xml:id="echoid-s910" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s911" xml:space="preserve"><emph style="sc">By</emph> a like method to what is uſed in the preceeding, and in the VIth Pro-<lb/>blem, this is reduced to Problem XI, where a point, a plane, and two <lb/>ſpheres are given.</s>
  <s xml:id="echoid-s912" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div56" type="section" level="1" n="56">
<head xml:id="echoid-head63" xml:space="preserve">PROBLEM XV.</head>
<p>
  <s xml:id="echoid-s913" xml:space="preserve"><emph style="sc">Having</emph> four ſpheres given; </s>
  <s xml:id="echoid-s914" xml:space="preserve">to ſind a ſphere which ſhall touch them all.</s>
  <s xml:id="echoid-s915" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s916" xml:space="preserve"><emph style="sc">Suppose</emph> the thing done. </s>
  <s xml:id="echoid-s917" xml:space="preserve">As, in the treatiſe of Circular Tangencies, the <lb/>laſt Problem, where it is required, having three circles given, to find a fourth <lb/>which ſhall touch them all, is reduced to another, where a point and two <lb/>circles are given; </s>
  <s xml:id="echoid-s918" xml:space="preserve">ſo alſo this, by a like method, and ſimilar to what has been <lb/>uſed in the preceding Problems, is reduced to Problem XII, where three <lb/>ſpheres and a point are given.</s>
  <s xml:id="echoid-s919" xml:space="preserve"/>
</p>
<pb o="(29)" file="0041" n="41"/>
<p>
  <s xml:id="echoid-s920" xml:space="preserve"><emph style="sc">The</emph> various Caſes, Determinations and other Minuliæ we have taken no <lb/>notice of: </s>
  <s xml:id="echoid-s921" xml:space="preserve">for if we had, this Treatiſe would have very much exceeded that <lb/>to which it was intended as a Supplement.</s>
  <s xml:id="echoid-s922" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div57" type="section" level="1" n="57">
<head xml:id="echoid-head64" xml:space="preserve">Synopſis of the PROBLEMS.</head>
<p style="it">
  <s xml:id="echoid-s923" xml:space="preserve">N. </s>
  <s xml:id="echoid-s924" xml:space="preserve">B. </s>
  <s xml:id="echoid-s925" xml:space="preserve">A point is repreſented by.</s>
  <s xml:id="echoid-s926" xml:space="preserve">, a plane by 1, and a ſphere by 0.</s>
  <s xml:id="echoid-s927" xml:space="preserve"/>
</p>
<note position="right" xml:space="preserve"> <lb/>1. .... # 4. 1111 # 15. 0000 <lb/>2. ...1 # 5. 111. # 12. 000. <lb/>3. ...0 # 6. 1110 # 14. 0001 <lb/>7. ..11 # 13. 1100 # 9. 00.. <lb/>8. ..10 # 10. 11.0 # 11. 00.1 <lb/></note>
<pb file="0042" n="42"/>
<pb file="0043" n="43"/>
  <figure>
    <image file="0043-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0043-01"/>
  </figure>
<pb file="0043a" n="44"/>
  <figure>
    <image file="0043a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0043a-01"/>
  </figure>
<pb file="0044" n="45"/>
<pb file="0045" n="46"/>
  <figure>
    <image file="0045-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0045-01"/>
  </figure>
<pb file="0045a" n="47"/>
  <figure>
    <image file="0045a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0045a-01"/>
  </figure>
<pb file="0046" n="48"/>
<pb file="0047" n="49"/>
  <figure>
    <image file="0047-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0047-01"/>
  </figure>
<pb file="0047a" n="50"/>
  <figure>
    <image file="0047a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0047a-01"/>
  </figure>
<pb file="0048" n="51"/>
<pb file="0049" n="52"/>
  <figure>
    <image file="0049-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0049-01"/>
  </figure>
<pb file="0049a" n="53"/>
  <figure>
    <image file="0049a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0049a-01"/>
  </figure>
<pb file="0050" n="54"/>
<pb file="0051" n="55"/>
  <figure>
    <image file="0051-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0051-01"/>
  </figure>
<pb file="0051a" n="56"/>
  <figure>
    <image file="0051a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0051a-01"/>
  </figure>
<pb file="0052" n="57"/>
<pb file="0053" n="58"/>
  <figure>
    <image file="0053-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0053-01"/>
  </figure>
<pb file="0053a" n="59"/>
  <figure>
    <image file="0053a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0053a-01"/>
  </figure>
<pb file="0054" n="60"/>
<pb file="0055" n="61"/>
  <figure>
    <image file="0055-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0055-01"/>
  </figure>
<pb file="0055a" n="62"/>
  <figure>
    <image file="0055a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0055a-01"/>
  </figure>
<pb file="0056" n="63"/>
<pb file="0057" n="64"/>
</div>
<div xml:id="echoid-div58" type="section" level="1" n="58">
<head xml:id="echoid-head65" xml:space="preserve">THE <lb/>TWO BOOKS <lb/>OF <lb/>APOLLONIUS PERGÆUS, <lb/>CONCERNING <lb/>DETERMINATE SECTION, <lb/>As they have been Reſtored by <lb/>WILLEBRORDUS SNELLIUS.</head>
<head xml:id="echoid-head66" xml:space="preserve">By JOHN LAWSON, B. D. Rector of Swanſcombe, Kent.</head>
<head xml:id="echoid-head67" xml:space="preserve">TO WHICH ARE ADDED, <lb/>THE SAME TWO BOOKS, BY WILLIAM WALES, <lb/>BEING <lb/>AN ENTIRE NEW WORK.</head>
<head xml:id="echoid-head68" xml:space="preserve">LONDON: <lb/>Printed by G. BIGG, Succeſſor to D. LEACH. <lb/>And ſold by B. <emph style="sc">White</emph>, in Fleet-Street; L. <emph style="sc">Davis</emph>, in Holborne; J. <emph style="sc">Nourse</emph>, in the <lb/>Strand; and T. <emph style="sc">Payne</emph>, near the Mews-Gate<unsure/>.</head>
<head xml:id="echoid-head69" xml:space="preserve">MDCC LXXII.</head>
<pb file="0058" n="65"/>
<pb o="[i]" file="0059" n="66"/>
</div>
<div xml:id="echoid-div59" type="section" level="1" n="59">
<head xml:id="echoid-head70" xml:space="preserve">ADVERTISEMENT.</head>
<p>
  <s xml:id="echoid-s928" xml:space="preserve">SINCE the publication of the preceding Tract on <lb/><emph style="sc">Tangencies</emph>, the Tranſlator thereof has obſerved, <lb/>that thoſe pieces of <emph style="sc">Willebrordus</emph> <emph style="sc">Snellius</emph>, which he <lb/>mentioned in his Preface thereto, are exceeding ſcarce <lb/>in England. </s>
  <s xml:id="echoid-s929" xml:space="preserve">His Reſuſcitata Geometria de ſectione rationis <lb/>&amp; </s>
  <s xml:id="echoid-s930" xml:space="preserve">ſpatii, 1607, he has never once had an opportunity <lb/>of ſeeing; </s>
  <s xml:id="echoid-s931" xml:space="preserve">but ſuppoſing this ſhould in a ſhort time be <lb/>loſt, more than ample amends is made by what <lb/>Dr. </s>
  <s xml:id="echoid-s932" xml:space="preserve"><emph style="sc">Halley</emph> has done on the ſame ſubject. </s>
  <s xml:id="echoid-s933" xml:space="preserve">Leſt the <lb/>other Tract, De Sectione Determinatâ, ſhould undergo <lb/>the ſame fate with the original <emph style="sc">Apollonius</emph>, he was <lb/>determined to reſcue it therefrom, or reſpite it at leaſt <lb/>for ſome time, by putting it into an Engliſh dreſs. <lb/></s>
  <s xml:id="echoid-s934" xml:space="preserve">While he was doing this, he happened to communicate <lb/>the piece to ſome friends; </s>
  <s xml:id="echoid-s935" xml:space="preserve">one of whom has ventured, <lb/>after <emph style="sc">Snellius</emph>, on this ſubject, and he preſumes with ſome <lb/>ſucceſs, as every Reader will allow, when he peruſes the <lb/>Propoſitions here printed after thoſe of <emph style="sc">Snellius</emph>. </s>
  <s xml:id="echoid-s936" xml:space="preserve">Yet, <lb/>notwithſtanding this, the Editor perſiſted in his reſolu-<lb/>tion of printing his tranſlation of <emph style="sc">Snellius</emph>, as the work
<pb o="[ii]" file="0060" n="67"/>
has much merit, and was in danger of being loſt, and as he <lb/>was the firſt that conſtructed Quadratic Equations after <lb/>this particular manner, as Dr. </s>
  <s xml:id="echoid-s937" xml:space="preserve"><emph style="sc">Simson</emph> obſerves in his <lb/>Note on <emph style="sc">Euc</emph>. </s>
  <s xml:id="echoid-s938" xml:space="preserve">VI. </s>
  <s xml:id="echoid-s939" xml:space="preserve">28 and 29.</s>
  <s xml:id="echoid-s940" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s941" xml:space="preserve">The Editor leaves his Friend to ſpeak for himſelf <lb/>in relation to what he has done, and truſts that the <lb/>candid Reader will not think more meanly of his <lb/>performance from the modeſt manner in which he <lb/>ſpeaks of it himſelf.</s>
  <s xml:id="echoid-s942" xml:space="preserve"/>
</p>
<pb o="[iii]" file="0061" n="68"/>
</div>
<div xml:id="echoid-div60" type="section" level="1" n="60">
<head xml:id="echoid-head71" xml:space="preserve">EXTRACT from PAPPUS's Preface to his Seventh Book <lb/>in Dr. HALLEY's Tranſlation.</head>
<head xml:id="echoid-head72" xml:space="preserve">DE SECTIONE DETERMINATA II.</head>
<p>
  <s xml:id="echoid-s943" xml:space="preserve">HIS ſubjiciuntur libri duo de Sectione Determinatâ, <lb/>quas etiam ad modum præcedentium unam pro-<lb/>poſitionem dicere liceat, ſed disjunctam: </s>
  <s xml:id="echoid-s944" xml:space="preserve">quæ hujuſ-<lb/>modi eſt. </s>
  <s xml:id="echoid-s945" xml:space="preserve">“ Datam rectam infinitam in uno puncto ſe-<lb/>care, ita, ut è rectis interceptis inter illud &amp; </s>
  <s xml:id="echoid-s946" xml:space="preserve">puncta <lb/>in illâ data, vel quadratum ex unâ, vel rectangulum <lb/>ex duabus interceptis, datam habeat rationem, vel ad <lb/>contentum ſub aliâ unâ interceptâ &amp; </s>
  <s xml:id="echoid-s947" xml:space="preserve">datá quâdum; <lb/></s>
  <s xml:id="echoid-s948" xml:space="preserve">vel etiam ad contentum ſub duabus aliis interceptis: </s>
  <s xml:id="echoid-s949" xml:space="preserve"><lb/>idque ad quam partem velis punctorum datorum.</s>
  <s xml:id="echoid-s950" xml:space="preserve">” <lb/>Hujus autem, quaſi bis disjunctæ, &amp; </s>
  <s xml:id="echoid-s951" xml:space="preserve">intricatos Dio-<lb/>riſmos habentis, per plura neceſſario facta eſt demon-<lb/>ſtratio. </s>
  <s xml:id="echoid-s952" xml:space="preserve">Hanc autem dedit Apollonius communi methodo <lb/>tentamen faciens, ac ſolis rectis lineis uſus, ad exemplum <lb/>ſecundi libri Elementorum <anchor type="note" xlink:href="" symbol="*"/> primorum Euclidis: </s>
  <s xml:id="echoid-s953" xml:space="preserve">ac rur-
ſus idem demonſtravit ingenioſe quidem, &amp; </s>
  <s xml:id="echoid-s954" xml:space="preserve">magis ad <lb/>inſtitutionem accomodate, per ſemicirculos. </s>
  <s xml:id="echoid-s955" xml:space="preserve">Habet <lb/>
<anchor type="note" xlink:label="note-0061-01a" xlink:href="note-0061-01"/>
<pb o="[iv]" file="0062" n="69"/>
autem primus liber Problemata ſex, Epitagmata, ſive <lb/>Diſpoſitiones punctorum, ſedecim; </s>
  <s xml:id="echoid-s956" xml:space="preserve">Dioriſmos quinque: <lb/></s>
  <s xml:id="echoid-s957" xml:space="preserve">quorum quatuor quidem Maximi ſunt, Minimus vero <lb/>unus. </s>
  <s xml:id="echoid-s958" xml:space="preserve">Sunt autem maximi, ad ſecundum Epitagma ſe-<lb/>cundi problematis; </s>
  <s xml:id="echoid-s959" xml:space="preserve">item ad tertium quarti problematis; </s>
  <s xml:id="echoid-s960" xml:space="preserve"><lb/>ad tertium quinti &amp; </s>
  <s xml:id="echoid-s961" xml:space="preserve">ad tertium ſexti. </s>
  <s xml:id="echoid-s962" xml:space="preserve">Minimus vero <lb/>eſt ad tertium Epitagma tertii problematis.</s>
  <s xml:id="echoid-s963" xml:space="preserve">-Secundus <lb/>liber de Sectione Determinatâ tria habet Problemata, <lb/>Diſpoſitiones novem, Determinationes tres; </s>
  <s xml:id="echoid-s964" xml:space="preserve">e quibus <lb/>Minima ſunt ad tertium primi, ut &amp; </s>
  <s xml:id="echoid-s965" xml:space="preserve">ad tertium ſecun-<lb/>di; </s>
  <s xml:id="echoid-s966" xml:space="preserve">Maximum autem eſt ad tertium tertii problematis. </s>
  <s xml:id="echoid-s967" xml:space="preserve"><lb/>-Lemmata habit liber primus XXVII, ſecundus vero <lb/>XXIV. </s>
  <s xml:id="echoid-s968" xml:space="preserve">Inſunt autem in utroque libro de Sectione <lb/>determinatâ Theorenata octoginta tria.</s>
  <s xml:id="echoid-s969" xml:space="preserve"/>
</p>
<div xml:id="echoid-div60" type="float" level="2" n="1">
<note symbol="*" position="foot" xlink:label="note-0061-01" xlink:href="note-0061-01a" xml:space="preserve">From hence it appears that <emph style="sc">Euclid's</emph> were called the firſt Elements, <lb/>and that the other Analytical Tracts, recited by <emph style="sc">Pappus,</emph> were called the <lb/>ſecond Elements.</note>
</div>
<pb o="[v]" file="0063" n="70"/>
</div>
<div xml:id="echoid-div62" type="section" level="1" n="61">
<head xml:id="echoid-head73" xml:space="preserve">THE <lb/>PREFACE.</head>
<p>
  <s xml:id="echoid-s970" xml:space="preserve">HAD not a motive more prevalent than Cuſtom induced me to <lb/>ſay ſomething by way of Preface to the Performance which <lb/>I herewith offer to the Public, the difficulty I find in doing it with <lb/>propriety, would have determined me to remain entirely ſilent.</s>
  <s xml:id="echoid-s971" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s972" xml:space="preserve">The ſubject has employed the Pen of one of the ableſt Geometers <lb/>of the laſt Century; </s>
  <s xml:id="echoid-s973" xml:space="preserve">it may therefore ſeem very preſumptuous, in <lb/>me at leaſt, who am but young in theſe matters, to attempt it after <lb/>him. </s>
  <s xml:id="echoid-s974" xml:space="preserve">To obviate, if poſſible, this Cenſure is my only intention <lb/>here; </s>
  <s xml:id="echoid-s975" xml:space="preserve">and I hope I ſhall not be deemed impertinent, if I attempt to <lb/>ſhew wherein I apprehend I have come nearer to the great original <lb/>than he hath done.</s>
  <s xml:id="echoid-s976" xml:space="preserve"/>
</p>
<pb o="[vi]" file="0064" n="71"/>
<p>
  <s xml:id="echoid-s977" xml:space="preserve"><emph style="sc">Pappus</emph>, in his preface to the ſeventh Book of Matbematical <lb/>Collections, tells us that this Tract of <emph style="sc">Apollonius</emph> was divided into <lb/>two Books; </s>
  <s xml:id="echoid-s978" xml:space="preserve">that the firſt Book contained ſix Problems, and the <lb/>ſecond three: </s>
  <s xml:id="echoid-s979" xml:space="preserve">now <emph style="sc">Snellius'</emph> whole work contains but four; </s>
  <s xml:id="echoid-s980" xml:space="preserve">and <lb/>it ſeemed to me difficult to ſhew how thoſe could contain the ſub-<lb/>ſtance of nine, and yet the ſix firſt have ſixteen Epitagmas, or ge-<lb/>neral Caſes, and the three laſt nine. </s>
  <s xml:id="echoid-s981" xml:space="preserve">I firſt, therefore began with <lb/>inquiring whether, or no, other Problems could not be found, <lb/>wherein the ſection of an indefinite ſtraight line is propoſed to be <lb/>effected, “So, that of the ſegments contained between the point of <lb/>ſection ſought, and given points in the ſaid line, either the ſquare <lb/>on one of them, or the rectangle contained by two of them, may <lb/>have a given ratio to the rectangle contained by one of them and a <lb/>given external line, or to the rectangle contained by two of them;</s>
  <s xml:id="echoid-s982" xml:space="preserve">“ <lb/>as is deſcribed by P<emph style="sc">APPUS.</emph></s>
</p>
<p>
  <s xml:id="echoid-s983" xml:space="preserve">In this inquiry it ſoon occurred to me, that the three problems <lb/>which make my firſt, ſecond and fourth, come within the account <lb/>given by P<emph style="sc">APPUS</emph>;</s>
  <s xml:id="echoid-s984" xml:space="preserve">and therefore are properly Problems in Determi-<lb/>ate Section, to be added to the four given by <emph style="sc">Snellius</emph>: </s>
  <s xml:id="echoid-s985" xml:space="preserve">and it does <lb/>not appear to me that more can be found which ſhould. </s>
  <s xml:id="echoid-s986" xml:space="preserve">Hence I <lb/>concluded, that there were in theſe, ſome, more general than thoſe <lb/>of <emph style="sc">Apollonius</emph>, which ought therefore to be divided.</s>
  <s xml:id="echoid-s987" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s988" xml:space="preserve">My next buſineſs was, if poſſible, to find out the order in which <lb/><emph style="sc">Apollonius</emph> had arranged them: </s>
  <s xml:id="echoid-s989" xml:space="preserve">and here, with reſpect to the <lb/>firſt Book, I had no other information to guide me, but what is to
<pb o="[vii]" file="0065" n="72"/>
be met with in the above mentioned Preface of <emph style="sc">Pappus</emph>; </s>
  <s xml:id="echoid-s990" xml:space="preserve">where he <lb/>tells us that in the ſix Problems of Book I. </s>
  <s xml:id="echoid-s991" xml:space="preserve">there were “Sixteen <lb/>Epitagmas, or general Caſes, five Determinations; </s>
  <s xml:id="echoid-s992" xml:space="preserve">and of theſe, <lb/>four were Maxima, and one a minimum: </s>
  <s xml:id="echoid-s993" xml:space="preserve">That the maxima are at the <lb/>ſecond Epitagma of the ſecond Problem, at the third of the fourth, <lb/>the third of the fifth, and the third of the ſixth; </s>
  <s xml:id="echoid-s994" xml:space="preserve">but that the minimum <lb/>was at the third Epitagma of the third problem.</s>
  <s xml:id="echoid-s995" xml:space="preserve"><anchor type="note" xlink:href="" symbol="*"/>” It moreover ſeem-
ed reaſonable to me, that theſe Problems wherein the feweſt points <lb/>are given, would be antecedent to thoſe wherein there were more; <lb/></s>
  <s xml:id="echoid-s996" xml:space="preserve">and of theſe wherein the number of given points are the ſame, that <lb/>thoſe would be prior to the others, wherein there was a given ex-<lb/>ternal line concerned: </s>
  <s xml:id="echoid-s997" xml:space="preserve">and laſtly, that when the number of given <lb/>points were two, the ſecond Caſe, or Epitagma, would naturally <lb/>be when the required point O is ſought between the two given <lb/>ones.</s>
  <s xml:id="echoid-s998" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s999" xml:space="preserve">Now the three new Problems, together with the three firſt of <lb/><emph style="sc">Snellius</emph>, making exactly ſixteen Epitagmas, viz. </s>
  <s xml:id="echoid-s1000" xml:space="preserve">one in the firſt, <lb/>and three in each of the others; </s>
  <s xml:id="echoid-s1001" xml:space="preserve">it ſeemed highly probable, that <lb/>theſe compoſed the firſt book. </s>
  <s xml:id="echoid-s1002" xml:space="preserve">Alſo that the Problem, wherein <lb/>only one point was given, would be the firſt; </s>
  <s xml:id="echoid-s1003" xml:space="preserve">and it ſeemed eaſy <lb/>to aſſign the ſecond, becauſe it is the only one wherein the limita-<lb/>
<anchor type="note" xlink:label="note-0065-01a" xlink:href="note-0065-01"/>
<pb o="[viii]" file="0066" n="73"/>
tion is at the ſecond Epitagma; </s>
  <s xml:id="echoid-s1004" xml:space="preserve">and farther, the Limiting Ratio is <lb/>therein a maximum, as it ought. </s>
  <s xml:id="echoid-s1005" xml:space="preserve">Again, the Problem, wherein <lb/>it is propoſed to make the ſquare on AO in a given ratio to the rect-<lb/>angle contained by EO and P, has its limiting ratio a minimum <lb/>when the required point is ſought beyond (E) that of the given ones <lb/>which bounds the ſegment concerned in the conſequent term of <lb/>the ratio; </s>
  <s xml:id="echoid-s1006" xml:space="preserve">which, therefore, I apprehend muſt have been the third <lb/>Epitagma, and if ſo, this of courſe muſt have been the third Pro-<lb/>blem: </s>
  <s xml:id="echoid-s1007" xml:space="preserve">and as there remains only one wherein the number of given <lb/>pointsare two, I make that the fourth. </s>
  <s xml:id="echoid-s1008" xml:space="preserve">With reſpect to the fifth and <lb/>ſixth Problems, in which three points are given, it ſhould ſeem <lb/>that that would be the firſt in order, wherein there is a given ex-<lb/>ternal line concerned.</s>
  <s xml:id="echoid-s1009" xml:space="preserve"/>
</p>
<div xml:id="echoid-div62" type="float" level="2" n="1">
<note symbol="*" position="foot" xlink:label="note-0065-01" xlink:href="note-0065-01a" xml:space="preserve">The words which are in Italics were entirely omitted in <emph style="sc">Snellius's</emph> Extract <lb/>from Pappus, both in the Greek and Latin, and are read with ſome variation in <lb/><emph style="sc">Commandine's</emph> tranſlation; but are here printed according to Dr. <emph style="sc">Halley</emph>: and <lb/>though I know not whether in this particular place he had the Authority of either <lb/>of the Savilian MSS, yet I hope I run no great riſk in ſubſcribing to the opinion <lb/>of ſo excellent a Geometer.</note>
</div>
<p>
  <s xml:id="echoid-s1010" xml:space="preserve">But it ſhould, by no means, be diſſembled that objections may be <lb/>brought againſt the identity, and arrangment of ſome of theſe Pro-<lb/>blems. </s>
  <s xml:id="echoid-s1011" xml:space="preserve">For firſt, <emph style="sc">Pappus</emph> no where expreſsly ſays that <emph style="sc">Apollo</emph>-<lb/><emph style="sc">NIUS</emph> compared together two ſquares, wherefore, if this cannot be <lb/>implied, the identity of the fourth Problem is deeply ſtruck at: <lb/></s>
  <s xml:id="echoid-s1012" xml:space="preserve">and moreover, this fourth Problem perhaps cannot with propriety, <lb/>be ſaid to have its limiting ratio either maximum or minimum, un-<lb/>leſs the ratio of equality, can be admitted as ſuch. </s>
  <s xml:id="echoid-s1013" xml:space="preserve">Laſtly, in the <lb/>fifth Problem, the ſaid limiting ratio is a minimum, and not a maxi-<lb/>mum as it is ſaid to have been by <emph style="sc">Pappus</emph>: </s>
  <s xml:id="echoid-s1014" xml:space="preserve">either, therefore, a
<pb o="[ix]" file="0067" n="74"/>
miſtake muſt be admitted in this Author, or the fifth Problem is <lb/>wrong placed. </s>
  <s xml:id="echoid-s1015" xml:space="preserve">I am not prepared farther to obviate theſe objec-<lb/>tions, and only mention them to ſhew that although I ſaw them <lb/>in their full force, I could by no means agree, that they are pow-<lb/>erful enough to overturn thoſe already advanced for what I have <lb/>
<anchor type="handwritten" xlink:label="hd-0067-01a" xlink:href="hd-0067-01"/>
done.</s>
  <s xml:id="echoid-s1016" xml:space="preserve"/>
</p>
<div xml:id="echoid-div63" type="float" level="2" n="2">
  <handwritten xlink:label="hd-0067-01" xlink:href="hd-0067-01a"/>
</div>
<p>
  <s xml:id="echoid-s1017" xml:space="preserve">I come now to Book II, which if I am not much miſtaken, was <lb/>entirely employed about what <emph style="sc">Snellius</emph> makes his fourth Problem. <lb/></s>
  <s xml:id="echoid-s1018" xml:space="preserve">In this I am confirmed not only by the account which <emph style="sc">Pappus</emph> gives <lb/>in his Preface, but much more by the Lemmas of <emph style="sc">Apollonius</emph> <lb/>
<anchor type="handwritten" xlink:label="hd-0067-01a" xlink:href="hd-0067-01"/>
which he hath left us. </s>
  <s xml:id="echoid-s1019" xml:space="preserve">For we there find that <emph style="sc">Lemma</emph> 21, where-<lb/>in is aſſigned the leaſt ratio which the rectangle contained by AO <lb/>and UO can bear to that contained by EO and IO, when O is ſought <lb/>between the two mean points of the four given ones, is ſaid to be <lb/>concerned in determining the μοναχὴ, or ſingle Caſe <anchor type="note" xlink:href="" symbol="*"/>, of Problem 1.</s>
  <s xml:id="echoid-s1020" xml:space="preserve">
This Problem therefore of <emph style="sc">Apollonius</emph> contained only thoſe <lb/>Caſes of the general one, where O is ſought between the two mean <lb/>points. </s>
  <s xml:id="echoid-s1021" xml:space="preserve">In like manner, we gather from Lemma 22, that his ſe-<lb/>cond Problem was concerned in determining the point O when ſought <lb/>between a mean point, and an extreme one. </s>
  <s xml:id="echoid-s1022" xml:space="preserve">And laſtly, from <lb/>Lemma 24, that the third Problem of Book II. </s>
  <s xml:id="echoid-s1023" xml:space="preserve">determined <lb/>the point O when required without all the given ones. <lb/></s>
  <s xml:id="echoid-s1024" xml:space="preserve">
<anchor type="note" xlink:label="note-0067-01a" xlink:href="note-0067-01"/>
<pb o="[x]" file="0068" n="75"/>
The Limitations of the two former are ſaid by <emph style="sc">Pappus</emph> to have <lb/>been minimums, and that of the third a maximum, in conformity, <lb/>to which, I have here made them ſo; </s>
  <s xml:id="echoid-s1025" xml:space="preserve">although I cannot ſee with <lb/>what propriety: </s>
  <s xml:id="echoid-s1026" xml:space="preserve">each of them admitting, in ſome Caſes, of a maxi-<lb/>mum and in others of minimum, as I have intimated in a ſcbolium at <lb/>the end of each Problem. </s>
  <s xml:id="echoid-s1027" xml:space="preserve">But notwithſtanding I have conformed to <lb/>the manner of <emph style="sc">Apollonius</emph> in dividing this Problem into three, <lb/>which it muſt be confeſſed contributes much to order in enumera-<lb/>ting ſuch a multitude of Caſes, yet have I previouſly ſhewn how <lb/>the whole may be generally conſtructed at once; </s>
  <s xml:id="echoid-s1028" xml:space="preserve">and that by a me-<lb/>thod, which I flatter myſelf will not be found inferior to any that <lb/>hath heretofore been given of this very intricate and general <lb/>Problem.</s>
  <s xml:id="echoid-s1029" xml:space="preserve"/>
</p>
<div xml:id="echoid-div64" type="float" level="2" n="3">
  <handwritten xlink:label="hd-0067-01" xlink:href="hd-0067-01a"/>
<note symbol="*" position="foot" xlink:label="note-0067-01" xlink:href="note-0067-01a" xml:space="preserve">So called, I conceive, becauſe in every other Caſe of the third Epitagma, <lb/>except this extreme, or limiting one, there are two points which will ſatisſy <lb/>the Problem.</note>
</div>
<p>
  <s xml:id="echoid-s1030" xml:space="preserve">Such are the things that I have attempted, and ſuch the reaſons <lb/>for what I have done in the following little Tract. </s>
  <s xml:id="echoid-s1031" xml:space="preserve">The merit due <lb/>to each of them I chearfully ſubmit (where every one ought) to <lb/>the deciſion of the impartial Reader. </s>
  <s xml:id="echoid-s1032" xml:space="preserve">In the Conſtructions, my <lb/>chief Aim was novelty and uniformity: </s>
  <s xml:id="echoid-s1033" xml:space="preserve">I could have given more <lb/>ſimple conſtructions to one or two of them; </s>
  <s xml:id="echoid-s1034" xml:space="preserve">in particular the ſixth <lb/>of Book I: </s>
  <s xml:id="echoid-s1035" xml:space="preserve">but it was not my intention to give any thing that I <lb/>knew had been done before. </s>
  <s xml:id="echoid-s1036" xml:space="preserve">I know of many imperfections, but <lb/>no falſe reaſonings, and hope none will be found; </s>
  <s xml:id="echoid-s1037" xml:space="preserve">but if there <lb/>ſhould, I hope the candid Geometer will be more inclined to ex-<lb/>cuſe than exult, when I aſſure him the greateſt part of the work <lb/>has been executed at different times, amidſt the hurry and perplexi-<lb/>ties which it may eaſily be conceived attend the fitting out for a <lb/>three years Voyage to the ſouth ſeas.</s>
  <s xml:id="echoid-s1038" xml:space="preserve"/>
</p>
<pb o="[xi]" file="0069" n="76"/>
<p>
  <s xml:id="echoid-s1039" xml:space="preserve">I cannot conclude without acknowledging, in the warmeſt man-<lb/>ner, the obligations I am under to my truly worthy and ingenious <lb/>friend, the Tranſlator of <emph style="sc">Snellius</emph>; </s>
  <s xml:id="echoid-s1040" xml:space="preserve">for the great pains and trou-<lb/>ble he hath taken to furniſh me with tranſlations from various Au-<lb/>thors, which my utter want of the Greek, and little acquaintance <lb/>with the Latin Language made abſolutely neceſſary to me: </s>
  <s xml:id="echoid-s1041" xml:space="preserve">And <lb/>after all, had it not been for his kindneſs, this attempt might ſtill <lb/>have remained in as great obſcurity as its Author.</s>
  <s xml:id="echoid-s1042" xml:space="preserve"/>
</p>
<pb file="0070" n="77"/>
<pb file="0071" n="78"/>
</div>
<div xml:id="echoid-div66" type="section" level="1" n="62">
<head xml:id="echoid-head74" xml:space="preserve">PROBLEMS <lb/>CONCERNING <lb/>DETERMINATE SECTION.</head>
<head xml:id="echoid-head75" xml:space="preserve">PROBLEM I.</head>
<p>
  <s xml:id="echoid-s1043" xml:space="preserve">TO cut a given indefinite right line in one point, ſo that of the ſegments <lb/>intercepted between that point and two other points given in the inde-<lb/>finite right line, the ſquare of one of them may be to the rectangle under the <lb/>other and a given external right line, in a given ratio.</s>
  <s xml:id="echoid-s1044" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1045" xml:space="preserve">In the given indefinite right line let be aſſigned the points A and E, it is then <lb/>required to cut it in the point O, ſo that <emph style="ol">AO</emph><emph style="sub">2</emph> may be to OE into a given <lb/>line AU in the ratio of R to S; </s>
  <s xml:id="echoid-s1046" xml:space="preserve">which ratio let be expreſſed by AI to AU, <lb/>ſetting off AI from A either way, either towards E or the contrary; </s>
  <s xml:id="echoid-s1047" xml:space="preserve">and <lb/>then from A and I erect two perpendiculars AY equal to AE, and IR <lb/>equal to AI, and theſe on the ſame ſide of the given indefinite line, if AI <lb/>was ſet off towards E; </s>
  <s xml:id="echoid-s1048" xml:space="preserve">but on oppoſite ſides, if AI was ſet off the other way. <lb/></s>
  <s xml:id="echoid-s1049" xml:space="preserve">The former conſtruction I will beg leave to call Homotactical, and the latter <lb/>Antitactical. </s>
  <s xml:id="echoid-s1050" xml:space="preserve">Let now the extremities of theſe perpendiculars Y and R be <lb/>joined, and upon YR as a diameter let a circle be deſcribed, I ſay that the <lb/>interſection of this circle with the given indefinite line ſolves the Problem. </s>
  <s xml:id="echoid-s1051" xml:space="preserve"><lb/>If it interſects the line in two places, the Problem admits of two Solutions;</s>
  <s xml:id="echoid-s1052" xml:space="preserve">
<pb o="[2]" file="0072" n="79"/>
but if it only touches, then only of one; </s>
  <s xml:id="echoid-s1053" xml:space="preserve">if it neither touches nor cuts, it is <lb/>then impoſſible.</s>
  <s xml:id="echoid-s1054" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1055" xml:space="preserve"><emph style="sc">Demonstration</emph>. </s>
  <s xml:id="echoid-s1056" xml:space="preserve">Let a point of interſection then be O, and join O Y <lb/>and OR. </s>
  <s xml:id="echoid-s1057" xml:space="preserve">The angles AYO and IOR are equal, the angle AOY being the <lb/>complement of each of them to a right one, and hence the triangles AOY and <lb/>IOR are ſimilar.</s>
  <s xml:id="echoid-s1058" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1059" xml:space="preserve">Hence AY = AE: </s>
  <s xml:id="echoid-s1060" xml:space="preserve">AO:</s>
  <s xml:id="echoid-s1061" xml:space="preserve">: OI: </s>
  <s xml:id="echoid-s1062" xml:space="preserve">IR = AI <lb/>And by div . </s>
  <s xml:id="echoid-s1063" xml:space="preserve">or comp . </s>
  <s xml:id="echoid-s1064" xml:space="preserve">EO: </s>
  <s xml:id="echoid-s1065" xml:space="preserve">AO:</s>
  <s xml:id="echoid-s1066" xml:space="preserve">: AO: </s>
  <s xml:id="echoid-s1067" xml:space="preserve">AI <lb/>And <emph style="ol">AO</emph><emph style="sub">2</emph> = EO x AI <lb/>Therefore <emph style="ol">AO</emph><emph style="sub">2</emph> (= EO x AI): </s>
  <s xml:id="echoid-s1068" xml:space="preserve">EO x AU:</s>
  <s xml:id="echoid-s1069" xml:space="preserve">: AI: </s>
  <s xml:id="echoid-s1070" xml:space="preserve">AU:</s>
  <s xml:id="echoid-s1071" xml:space="preserve">: R: </s>
  <s xml:id="echoid-s1072" xml:space="preserve">S <lb/>Q. </s>
  <s xml:id="echoid-s1073" xml:space="preserve">E. </s>
  <s xml:id="echoid-s1074" xml:space="preserve">D.</s>
  <s xml:id="echoid-s1075" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1076" xml:space="preserve">This Problem admits of two Caſes. </s>
  <s xml:id="echoid-s1077" xml:space="preserve">The 1ſt determinate or limited, the 2d <lb/>unlimited.</s>
  <s xml:id="echoid-s1078" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1079" xml:space="preserve"><emph style="sc">Case</emph> I. </s>
  <s xml:id="echoid-s1080" xml:space="preserve">Is when OE the co-efficient of the given external line AU is part of <lb/>AO the ſide of the required ſquare [fig. </s>
  <s xml:id="echoid-s1081" xml:space="preserve">1. </s>
  <s xml:id="echoid-s1082" xml:space="preserve">2.</s>
  <s xml:id="echoid-s1083" xml:space="preserve">] and here the <emph style="sc">Llmitation</emph> is, <lb/>that AI muſt not be given leſs than four times AE, as appears from fig. </s>
  <s xml:id="echoid-s1084" xml:space="preserve">2. <lb/></s>
  <s xml:id="echoid-s1085" xml:space="preserve">for AE: </s>
  <s xml:id="echoid-s1086" xml:space="preserve">AO:</s>
  <s xml:id="echoid-s1087" xml:space="preserve">: OI: </s>
  <s xml:id="echoid-s1088" xml:space="preserve">AI; </s>
  <s xml:id="echoid-s1089" xml:space="preserve">and here OI being the half of AI, AE will be <lb/>the half of AO, or the fourth part of AI. </s>
  <s xml:id="echoid-s1090" xml:space="preserve">In this Caſe the Homotactical Con-<lb/>ſtruction is uſed.</s>
  <s xml:id="echoid-s1091" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1092" xml:space="preserve"><emph style="sc">Case</emph> II. </s>
  <s xml:id="echoid-s1093" xml:space="preserve">Is when AO the ſide of the required ſquare is part of OE the co-<lb/>efficient of the given external line AU, [fig. </s>
  <s xml:id="echoid-s1094" xml:space="preserve">3.</s>
  <s xml:id="echoid-s1095" xml:space="preserve">] and this is unlimited, for here <lb/>the Anlitactical Conſtruction is uſed. </s>
  <s xml:id="echoid-s1096" xml:space="preserve">Or if O be required between A and E, <lb/>this is effected by the ſame Conſtruction.</s>
  <s xml:id="echoid-s1097" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div67" type="section" level="1" n="63">
<head xml:id="echoid-head76" xml:space="preserve">LEMMA I.</head>
<p>
  <s xml:id="echoid-s1098" xml:space="preserve">If from the extremes of any diameter perpendiculars be let fall upon any <lb/>Chord, I ſay that the ſegments of theſe perpendiculars intercepted by like Arcs <lb/>are equal, and moreover alſo the ſegments of the Chords themſelves.</s>
  <s xml:id="echoid-s1099" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1100" xml:space="preserve">That YO is equal to IU may be thus ſhewn. </s>
  <s xml:id="echoid-s1101" xml:space="preserve">Having joined YI, the <lb/>angle IYE is a right one, being in a ſemicircle, and the angle at O is right by <lb/>hypotbeſis; </s>
  <s xml:id="echoid-s1102" xml:space="preserve">hence YI is parallel to the Chord, and YOUI is a parallelogram, <lb/>and the oppoſite ſides YO and IU will be equal. </s>
  <s xml:id="echoid-s1103" xml:space="preserve">In the ſame manner OE is <lb/>proved equal to UL. </s>
  <s xml:id="echoid-s1104" xml:space="preserve">And as to the ſegments of the Chord, it is thus ſhewn. <lb/></s>
  <s xml:id="echoid-s1105" xml:space="preserve">By Euc. </s>
  <s xml:id="echoid-s1106" xml:space="preserve">III. </s>
  <s xml:id="echoid-s1107" xml:space="preserve">35. </s>
  <s xml:id="echoid-s1108" xml:space="preserve">and 36, the rect. </s>
  <s xml:id="echoid-s1109" xml:space="preserve">EOY = rect. </s>
  <s xml:id="echoid-s1110" xml:space="preserve">SOR, and rect. </s>
  <s xml:id="echoid-s1111" xml:space="preserve">LUI = rect. </s>
  <s xml:id="echoid-s1112" xml:space="preserve"><lb/>SUR. </s>
  <s xml:id="echoid-s1113" xml:space="preserve">But, by what has been juſt proved, rect. </s>
  <s xml:id="echoid-s1114" xml:space="preserve">EOY = rect. </s>
  <s xml:id="echoid-s1115" xml:space="preserve">LUI; </s>
  <s xml:id="echoid-s1116" xml:space="preserve">hence <lb/>rect. </s>
  <s xml:id="echoid-s1117" xml:space="preserve">SOR = rect. </s>
  <s xml:id="echoid-s1118" xml:space="preserve">SUR, and the ſegments SO and OR are reſpectively equal <lb/>to the ſegments UR and SU.</s>
  <s xml:id="echoid-s1119" xml:space="preserve"/>
</p>
<pb o="[3]" file="0073" n="80"/>
</div>
<div xml:id="echoid-div68" type="section" level="1" n="64">
<head xml:id="echoid-head77" xml:space="preserve">LEMMA II.</head>
<p>
  <s xml:id="echoid-s1120" xml:space="preserve">If of four proportionals the ſum of two, being either extremes or means, be <lb/>greater than the ſum of the other two; </s>
  <s xml:id="echoid-s1121" xml:space="preserve">then I ſay theſe will be greateſt and <lb/>leaſt of all.</s>
  <s xml:id="echoid-s1122" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1123" xml:space="preserve">This is the converſe of Euc. </s>
  <s xml:id="echoid-s1124" xml:space="preserve">V. </s>
  <s xml:id="echoid-s1125" xml:space="preserve">25. </s>
  <s xml:id="echoid-s1126" xml:space="preserve">and may be thus demonſtrated. </s>
  <s xml:id="echoid-s1127" xml:space="preserve">Draw <lb/>a circle whoſe diameter may be equal to the greater ſum; </s>
  <s xml:id="echoid-s1128" xml:space="preserve">and in it inſcribe the <lb/>leſſer ſum IO, which will therefore not paſs through the center, and let the <lb/>parts be IU and UO; </s>
  <s xml:id="echoid-s1129" xml:space="preserve">then through U draw a diameter AUE, and the other <lb/>two terms will be AU and EU, of which AU is greateſt of all and EU leaſt of <lb/>all, and IU and UO of intermediate magnitude, by Euc. </s>
  <s xml:id="echoid-s1130" xml:space="preserve">III. </s>
  <s xml:id="echoid-s1131" xml:space="preserve">7.</s>
  <s xml:id="echoid-s1132" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div69" type="section" level="1" n="65">
<head xml:id="echoid-head78" xml:space="preserve">LEMMA III.</head>
<p>
  <s xml:id="echoid-s1133" xml:space="preserve">If of four proportionals the difference of two, being either extremes or <lb/>means, be greater than the difference of the other two, then I ſay theſe will be <lb/>the greateſt and leaſt of all.</s>
  <s xml:id="echoid-s1134" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1135" xml:space="preserve">This is demonſtrated in the ſame manner as the preceding by Euc. </s>
  <s xml:id="echoid-s1136" xml:space="preserve">III. </s>
  <s xml:id="echoid-s1137" xml:space="preserve">8.</s>
  <s xml:id="echoid-s1138" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div70" type="section" level="1" n="66">
<head xml:id="echoid-head79" xml:space="preserve">PROBLEM II.</head>
<p>
  <s xml:id="echoid-s1139" xml:space="preserve">To cut a given indefinite right line in one point, ſo that, of the three ſeg-<lb/>ments intercepted between the ſaid point and three points given in the ſame in-<lb/>definite right line, the rectangle under one of them and a given external right <lb/>line may be to the rectangle under the other two in a given ratio.</s>
  <s xml:id="echoid-s1140" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1141" xml:space="preserve">In the given indefinite line let the aſſigned points be A, E, I. </s>
  <s xml:id="echoid-s1142" xml:space="preserve">It is then <lb/>required to cut it again in the point O, ſo that AO into a given external line <lb/>R may be to EO x IO as R to S. </s>
  <s xml:id="echoid-s1143" xml:space="preserve">If A be an extreme point and E the <lb/>middle one, then ſet off IU = AE the contrary way from A; </s>
  <s xml:id="echoid-s1144" xml:space="preserve">but if A be the <lb/>middle point, then ſet it off towards A. </s>
  <s xml:id="echoid-s1145" xml:space="preserve">Then from U ſet off UN = S the <lb/>conſequent of the given ratio, either towards A, or the contrary way; </s>
  <s xml:id="echoid-s1146" xml:space="preserve">for as <lb/>the Caſes vary, it’s poſition will vary. </s>
  <s xml:id="echoid-s1147" xml:space="preserve">From A and N erect perpendiculars <lb/>AY and NM to the given indefinite right line equal to AE and AI re-<lb/>ſpectively, and theſe bomotactical if A be an extreme point, but antitactical if <lb/>A be the middle point of the three given ones. </s>
  <s xml:id="echoid-s1148" xml:space="preserve">Join the extremes of theſe <lb/>perpendiculars Y and M, and upon YM as a Diameter deſcribe a circle. </s>
  <s xml:id="echoid-s1149" xml:space="preserve">I ſay <lb/>that the interſection of this circle with the given indefinite line ſolves the <lb/>Problem. </s>
  <s xml:id="echoid-s1150" xml:space="preserve">If it interſects the line in two points, then the Problem admits of
<pb o="[4]" file="0074" n="81"/>
two ſolutions; </s>
  <s xml:id="echoid-s1151" xml:space="preserve">if it only touches, then but of one; </s>
  <s xml:id="echoid-s1152" xml:space="preserve">if it neither cuts nor <lb/>touches, it is then impoſſible.</s>
  <s xml:id="echoid-s1153" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1154" xml:space="preserve"><emph style="sc">Demonstration</emph>. </s>
  <s xml:id="echoid-s1155" xml:space="preserve">Let the point of interſection then be O or o. </s>
  <s xml:id="echoid-s1156" xml:space="preserve">We ſhall <lb/>have, by Lemma I. </s>
  <s xml:id="echoid-s1157" xml:space="preserve">AO x ON = MN x NK = MN x AY = AI x AE, by <lb/>conſtruction. </s>
  <s xml:id="echoid-s1158" xml:space="preserve">Let now from N be ſet off NL = AI in the ſame direction as A <lb/>is from I; </s>
  <s xml:id="echoid-s1159" xml:space="preserve">then by what has been demonſtrated NL: </s>
  <s xml:id="echoid-s1160" xml:space="preserve">ON:</s>
  <s xml:id="echoid-s1161" xml:space="preserve">: AO: </s>
  <s xml:id="echoid-s1162" xml:space="preserve">AE.</s>
  <s xml:id="echoid-s1163" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1164" xml:space="preserve">And by Diviſion or Compoſition OL: </s>
  <s xml:id="echoid-s1165" xml:space="preserve">ON:</s>
  <s xml:id="echoid-s1166" xml:space="preserve">: OE: </s>
  <s xml:id="echoid-s1167" xml:space="preserve">AE <lb/>And by Permutation OL: </s>
  <s xml:id="echoid-s1168" xml:space="preserve">OE:</s>
  <s xml:id="echoid-s1169" xml:space="preserve">: ON: </s>
  <s xml:id="echoid-s1170" xml:space="preserve">AE <lb/>But by what has been proved ON: </s>
  <s xml:id="echoid-s1171" xml:space="preserve">AE:</s>
  <s xml:id="echoid-s1172" xml:space="preserve">: AI: </s>
  <s xml:id="echoid-s1173" xml:space="preserve">AO <lb/>Therefore by Equality OL: </s>
  <s xml:id="echoid-s1174" xml:space="preserve">OE:</s>
  <s xml:id="echoid-s1175" xml:space="preserve">: AI: </s>
  <s xml:id="echoid-s1176" xml:space="preserve">AO <lb/>And by Diviſion or Compoſition LE: </s>
  <s xml:id="echoid-s1177" xml:space="preserve">OE:</s>
  <s xml:id="echoid-s1178" xml:space="preserve">: OI: </s>
  <s xml:id="echoid-s1179" xml:space="preserve">AO <lb/>And LE x AO = OE x OI <lb/>But LE = NU. </s>
  <s xml:id="echoid-s1180" xml:space="preserve">for NL was put = AI, and IU = AE <lb/>Hence NU x AO = S x AO = OE x OI <lb/>And R: </s>
  <s xml:id="echoid-s1181" xml:space="preserve">S:</s>
  <s xml:id="echoid-s1182" xml:space="preserve">: R x AO: </s>
  <s xml:id="echoid-s1183" xml:space="preserve">S x AO or OE x OI <lb/>Q. </s>
  <s xml:id="echoid-s1184" xml:space="preserve">E. </s>
  <s xml:id="echoid-s1185" xml:space="preserve">D.</s>
  <s xml:id="echoid-s1186" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1187" xml:space="preserve">This Problem may be conſidered as having two <emph style="sc">Epitacmas</emph>, the firſt, <lb/>when the ſegment aſſigned for the coefficient of the given external line R is <lb/>terminated by an extreme point of the three given ones and the point ſought; <lb/></s>
  <s xml:id="echoid-s1188" xml:space="preserve">and this again admits of three Caſes. </s>
  <s xml:id="echoid-s1189" xml:space="preserve">The other is when the aforeſaid ſeg-<lb/>ment is terminated by the middle point of the three given ones and the <lb/>point ſought.</s>
  <s xml:id="echoid-s1190" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1191" xml:space="preserve"><emph style="sc">Epitagma</emph> I. </s>
  <s xml:id="echoid-s1192" xml:space="preserve"><emph style="sc">Case</emph> I. </s>
  <s xml:id="echoid-s1193" xml:space="preserve">Let the aſſigned points be A, E, I. </s>
  <s xml:id="echoid-s1194" xml:space="preserve">A an extreme <lb/>and E the middle one. </s>
  <s xml:id="echoid-s1195" xml:space="preserve">And let the point O ſought (ſuch that AO x R: </s>
  <s xml:id="echoid-s1196" xml:space="preserve">OE <lb/>x OI:</s>
  <s xml:id="echoid-s1197" xml:space="preserve">: R: </s>
  <s xml:id="echoid-s1198" xml:space="preserve">S) be required to lie between A and E, or elſe beyond I, which <lb/>will ariſe from the ſame conſtruction.</s>
  <s xml:id="echoid-s1199" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1200" xml:space="preserve">Here the Homotactical conſtruction is uſed, and IU as likewiſe UN is ſet off <lb/>in the ſame direction as AI. </s>
  <s xml:id="echoid-s1201" xml:space="preserve">And ſince AO: </s>
  <s xml:id="echoid-s1202" xml:space="preserve">AE:</s>
  <s xml:id="echoid-s1203" xml:space="preserve">: AI: </s>
  <s xml:id="echoid-s1204" xml:space="preserve">ON, and AO + ON <lb/>is greater than AE + AI or AU, by <emph style="sc">Lemma</emph> II. </s>
  <s xml:id="echoid-s1205" xml:space="preserve">AO and ON will be the <lb/>leaſt and greateſt of all; </s>
  <s xml:id="echoid-s1206" xml:space="preserve">and AO will therefore be leſs than AE, as likewiſe <lb/>Ao (being equal ON by <emph style="sc">Lemma</emph> I.) </s>
  <s xml:id="echoid-s1207" xml:space="preserve">greater than AI. </s>
  <s xml:id="echoid-s1208" xml:space="preserve">This Caſe is <lb/>unlimited.</s>
  <s xml:id="echoid-s1209" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1210" xml:space="preserve"><emph style="sc">Case</emph> II. </s>
  <s xml:id="echoid-s1211" xml:space="preserve">Let the aſſigned points be in the ſame poſition as before, and let <lb/>the point O ſought be required between E and I.</s>
  <s xml:id="echoid-s1212" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1213" xml:space="preserve">Here the conſtruction is Homotactical, and UN is ſet off the contraty way, viz. <lb/></s>
  <s xml:id="echoid-s1214" xml:space="preserve">in the direction IA. </s>
  <s xml:id="echoid-s1215" xml:space="preserve">And ſince AO: </s>
  <s xml:id="echoid-s1216" xml:space="preserve">AE:</s>
  <s xml:id="echoid-s1217" xml:space="preserve">: AI: </s>
  <s xml:id="echoid-s1218" xml:space="preserve">ON, and AO + ON is leſs <lb/>than AE + AI or AU, by <emph style="sc">Lemma</emph> II. </s>
  <s xml:id="echoid-s1219" xml:space="preserve">AE and AI will be the leaſt and
<pb o="[5]" file="0075" n="82"/>
greateſt of all, and AE will therefore be leſs than AO, and AI greater. </s>
  <s xml:id="echoid-s1220" xml:space="preserve">And <lb/>the ſame will hold with regard to Ao.</s>
  <s xml:id="echoid-s1221" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1222" xml:space="preserve">Here is a <emph style="sc">Limitation</emph>, which is this; </s>
  <s xml:id="echoid-s1223" xml:space="preserve">that UN or S the conſequent of the <lb/>given ratio, ſet off from R, muſt not be given greater than the difference of <lb/>the ſum of AE and AI and of a line whoſe ſquare is equal to four times their <lb/>rectangle [i. </s>
  <s xml:id="echoid-s1224" xml:space="preserve">e. </s>
  <s xml:id="echoid-s1225" xml:space="preserve">to expreſs it in the modern manner, UN muſt not exceed AI + <lb/>AE - √4 AI x AE*.</s>
  <s xml:id="echoid-s1226" xml:space="preserve">] This appears by Fig. </s>
  <s xml:id="echoid-s1227" xml:space="preserve">2. </s>
  <s xml:id="echoid-s1228" xml:space="preserve">to this Caſe, the circle there <lb/>touching the given indefinite line, and pointing out the Limit.</s>
  <s xml:id="echoid-s1229" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1230" xml:space="preserve"><emph style="sc">Case</emph> III. </s>
  <s xml:id="echoid-s1231" xml:space="preserve">Let the aſſigned points be ſtill in the ſame poſition, and let the <lb/>point ſought be now required on the contrary ſide of A.</s>
  <s xml:id="echoid-s1232" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1233" xml:space="preserve">Here the conſtruction is ſtill Homotactical, and UN is ſet off the ſame way as <lb/>in the laſt Caſe; </s>
  <s xml:id="echoid-s1234" xml:space="preserve">and the <emph style="sc">Limitation</emph> is, that UN muſt not be given leſs than <lb/>the ſum of AI, AE, and a line whoſe ſquare is equal to four times their rect-<lb/>angle [or expreſſing it Algebraically, UN muſt not be leſs than AI + AE + <lb/>√4 AI x AE*.</s>
  <s xml:id="echoid-s1235" xml:space="preserve">]</s>
</p>
<p>
  <s xml:id="echoid-s1236" xml:space="preserve"><emph style="sc">Epitagma</emph> II. </s>
  <s xml:id="echoid-s1237" xml:space="preserve"><emph style="sc">Case</emph> IV. </s>
  <s xml:id="echoid-s1238" xml:space="preserve">Let now A be the middle point of the given <lb/>ones, and let O the point ſought be required either between A and one of the <lb/>extremes, or beyond either of the extremes.</s>
  <s xml:id="echoid-s1239" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1240" xml:space="preserve">Here having ſet off IU = AE toward A, you may ſet off UN either way, <lb/>and uſing the Antitactical conſtruction, the ſolution will be unlimited. </s>
  <s xml:id="echoid-s1241" xml:space="preserve">The <lb/>only difference is, that if UN be in the direction UI, two ſolutions will ariſe, <lb/>whereof in one the point O will fall between A and E, and in the other be-<lb/>yond I; </s>
  <s xml:id="echoid-s1242" xml:space="preserve">but if UN be in the direction IU, two ſolutions will ariſe, whereof <lb/>in one the point will fall between A and I, and in the other beyond E. </s>
  <s xml:id="echoid-s1243" xml:space="preserve">In <lb/>proof of which <emph style="sc">Lemma</emph> III. </s>
  <s xml:id="echoid-s1244" xml:space="preserve">is to be uſed, as <emph style="sc">Lemma</emph> II. </s>
  <s xml:id="echoid-s1245" xml:space="preserve">was in Caſe I. </s>
  <s xml:id="echoid-s1246" xml:space="preserve">II.</s>
  <s xml:id="echoid-s1247" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1248" xml:space="preserve"><emph style="sc">Corollary</emph> I. </s>
  <s xml:id="echoid-s1249" xml:space="preserve">If then the given ratio be that of AT to TI, or of AE to <lb/>EP ſet off from A the other way, ſo that EP be leſs than AE, I ſay then <lb/>that O will fall between E and P, as likewiſe ο between T and I, provided o <lb/>falls beyond I.</s>
  <s xml:id="echoid-s1250" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1251" xml:space="preserve">For by conſtruction IU = AE, and UN = PE. </s>
  <s xml:id="echoid-s1252" xml:space="preserve">therefore IN = AP. </s>
  <s xml:id="echoid-s1253" xml:space="preserve">But by <lb/><emph style="sc">Lemma</emph> I. </s>
  <s xml:id="echoid-s1254" xml:space="preserve">oN = AO. </s>
  <s xml:id="echoid-s1255" xml:space="preserve">therefore (o falling beyond I by hypotbeſis) O will fall <lb/>beyond P; </s>
  <s xml:id="echoid-s1256" xml:space="preserve">but by hypotbeſis it falls ſhort of E; </s>
  <s xml:id="echoid-s1257" xml:space="preserve">therefore O falls between <lb/>P and E.</s>
  <s xml:id="echoid-s1258" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1259" xml:space="preserve">Next to ſhew that ο will fall between T and I, we have AT: </s>
  <s xml:id="echoid-s1260" xml:space="preserve">TI:</s>
  <s xml:id="echoid-s1261" xml:space="preserve">: AE: </s>
  <s xml:id="echoid-s1262" xml:space="preserve">EP</s>
</p>
<p>
  <s xml:id="echoid-s1263" xml:space="preserve">And by Diviſion AT: </s>
  <s xml:id="echoid-s1264" xml:space="preserve">AI:</s>
  <s xml:id="echoid-s1265" xml:space="preserve">: AE: </s>
  <s xml:id="echoid-s1266" xml:space="preserve">AP</s>
</p>
<p>
  <s xml:id="echoid-s1267" xml:space="preserve">Hence AT x AP = IAE or o AO</s>
</p>
<pb o="[6]" file="0076" n="83"/>
<p>
  <s xml:id="echoid-s1268" xml:space="preserve">Therefore AT x AO is greater than o AO</s>
</p>
<p>
  <s xml:id="echoid-s1269" xml:space="preserve">Or AT greater than Ao.</s>
  <s xml:id="echoid-s1270" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1271" xml:space="preserve"><emph style="sc">Corollary</emph> II. </s>
  <s xml:id="echoid-s1272" xml:space="preserve">If the three given points be I, A, E; </s>
  <s xml:id="echoid-s1273" xml:space="preserve">and O falls between <lb/>A and I, ſo as to make AO x PE: </s>
  <s xml:id="echoid-s1274" xml:space="preserve">IOE:</s>
  <s xml:id="echoid-s1275" xml:space="preserve">: AL: </s>
  <s xml:id="echoid-s1276" xml:space="preserve">LI, I ſay then O will fall <lb/>beyond L.</s>
  <s xml:id="echoid-s1277" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1278" xml:space="preserve">For let us ſuppoſe that O and L coincide; </s>
  <s xml:id="echoid-s1279" xml:space="preserve">then by hypotbeſis AL: </s>
  <s xml:id="echoid-s1280" xml:space="preserve">LI:</s>
  <s xml:id="echoid-s1281" xml:space="preserve">: <lb/>AL x PE: </s>
  <s xml:id="echoid-s1282" xml:space="preserve">IL x LE</s>
</p>
<p>
  <s xml:id="echoid-s1283" xml:space="preserve">And by the next following <emph style="sc">Lemma</emph> IV. </s>
  <s xml:id="echoid-s1284" xml:space="preserve">AL x IL: </s>
  <s xml:id="echoid-s1285" xml:space="preserve">IL x PE:</s>
  <s xml:id="echoid-s1286" xml:space="preserve">: AL: </s>
  <s xml:id="echoid-s1287" xml:space="preserve">LE <lb/>i. </s>
  <s xml:id="echoid-s1288" xml:space="preserve">e. </s>
  <s xml:id="echoid-s1289" xml:space="preserve">AL: </s>
  <s xml:id="echoid-s1290" xml:space="preserve">PE:</s>
  <s xml:id="echoid-s1291" xml:space="preserve">: AL: </s>
  <s xml:id="echoid-s1292" xml:space="preserve">LE</s>
</p>
<p>
  <s xml:id="echoid-s1293" xml:space="preserve">Hence PE is equal to LE, a part to the whole, which is manifeſtly abſurd.</s>
  <s xml:id="echoid-s1294" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div71" type="section" level="1" n="67">
<head xml:id="echoid-head80" xml:space="preserve">LEMMA IV.</head>
<p>
  <s xml:id="echoid-s1295" xml:space="preserve">If it be as a line to a line ſo a rectangle to a rectangle; </s>
  <s xml:id="echoid-s1296" xml:space="preserve">then I ſay it will be <lb/>as the flrſt line into the breadth of the ſecond rectangle to the ſecond line into <lb/>the breadth of the firſt rectangle, ſo the length of the firſt rectangle to the <lb/>length of the ſecond.</s>
  <s xml:id="echoid-s1297" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1298" xml:space="preserve">Suppoſition. </s>
  <s xml:id="echoid-s1299" xml:space="preserve">AE: </s>
  <s xml:id="echoid-s1300" xml:space="preserve">IO:</s>
  <s xml:id="echoid-s1301" xml:space="preserve">: UYN: </s>
  <s xml:id="echoid-s1302" xml:space="preserve">SRL.</s>
  <s xml:id="echoid-s1303" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1304" xml:space="preserve">Concluſion. </s>
  <s xml:id="echoid-s1305" xml:space="preserve">AE x RL: </s>
  <s xml:id="echoid-s1306" xml:space="preserve">IO x YN:</s>
  <s xml:id="echoid-s1307" xml:space="preserve">: UY: </s>
  <s xml:id="echoid-s1308" xml:space="preserve">SR.</s>
  <s xml:id="echoid-s1309" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1310" xml:space="preserve"><emph style="sc">Dem</emph>. </s>
  <s xml:id="echoid-s1311" xml:space="preserve">AE: </s>
  <s xml:id="echoid-s1312" xml:space="preserve">IO:</s>
  <s xml:id="echoid-s1313" xml:space="preserve">: AE x YN: </s>
  <s xml:id="echoid-s1314" xml:space="preserve">IO x YN:</s>
  <s xml:id="echoid-s1315" xml:space="preserve">: UYN: </s>
  <s xml:id="echoid-s1316" xml:space="preserve">SRL</s>
</p>
<p>
  <s xml:id="echoid-s1317" xml:space="preserve">And by Permutation AE x YN: </s>
  <s xml:id="echoid-s1318" xml:space="preserve">UYN:</s>
  <s xml:id="echoid-s1319" xml:space="preserve">: AE: </s>
  <s xml:id="echoid-s1320" xml:space="preserve">UY:</s>
  <s xml:id="echoid-s1321" xml:space="preserve">: IO x YN: </s>
  <s xml:id="echoid-s1322" xml:space="preserve">SRL</s>
</p>
<p>
  <s xml:id="echoid-s1323" xml:space="preserve">But SR: </s>
  <s xml:id="echoid-s1324" xml:space="preserve">AE:</s>
  <s xml:id="echoid-s1325" xml:space="preserve">: SRL:</s>
  <s xml:id="echoid-s1326" xml:space="preserve">: AE x RL</s>
</p>
<p>
  <s xml:id="echoid-s1327" xml:space="preserve">Therefore ex æquo perturbatè SR: </s>
  <s xml:id="echoid-s1328" xml:space="preserve">UY:</s>
  <s xml:id="echoid-s1329" xml:space="preserve">: IO x YN: </s>
  <s xml:id="echoid-s1330" xml:space="preserve">AE x RL</s>
</p>
<p>
  <s xml:id="echoid-s1331" xml:space="preserve">Q. </s>
  <s xml:id="echoid-s1332" xml:space="preserve">E. </s>
  <s xml:id="echoid-s1333" xml:space="preserve">D.</s>
  <s xml:id="echoid-s1334" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div72" type="section" level="1" n="68">
<head xml:id="echoid-head81" xml:space="preserve">LEMMA V.</head>
<p>
  <s xml:id="echoid-s1335" xml:space="preserve">If a right line be cut in two points, I fay the rectangle under the alternate <lb/>ſegments is equal to that under the whole and the middle ſegment, together <lb/>with the rectangle under the extremes.</s>
  <s xml:id="echoid-s1336" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1337" xml:space="preserve"><emph style="sc">Dem</emph>. </s>
  <s xml:id="echoid-s1338" xml:space="preserve">AI x IE + IO x IE = AO x IE.</s>
  <s xml:id="echoid-s1339" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1340" xml:space="preserve">Hence {AI x IE + IO x IE + AE x IO \\ i. </s>
  <s xml:id="echoid-s1341" xml:space="preserve">e. </s>
  <s xml:id="echoid-s1342" xml:space="preserve">AI x IE + AI x IO \\ i. </s>
  <s xml:id="echoid-s1343" xml:space="preserve">e. </s>
  <s xml:id="echoid-s1344" xml:space="preserve">AI x EO} = AO x IE + AE x IO.</s>
  <s xml:id="echoid-s1345" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1346" xml:space="preserve">Q. </s>
  <s xml:id="echoid-s1347" xml:space="preserve">E. </s>
  <s xml:id="echoid-s1348" xml:space="preserve">D.</s>
  <s xml:id="echoid-s1349" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1350" xml:space="preserve">N. </s>
  <s xml:id="echoid-s1351" xml:space="preserve">B. </s>
  <s xml:id="echoid-s1352" xml:space="preserve">Theſe two <emph style="sc">Lemmas</emph> ſave much Circumlocution and Tautology in <lb/>the two following Propoſitions, and indeed are highly uſeful in all caſes where <lb/>compound ratios are concerned.</s>
  <s xml:id="echoid-s1353" xml:space="preserve"/>
</p>
<pb o="[7]" file="0077" n="84"/>
</div>
<div xml:id="echoid-div73" type="section" level="1" n="69">
<head xml:id="echoid-head82" xml:space="preserve">PROBLEM III.</head>
<p>
  <s xml:id="echoid-s1354" xml:space="preserve">To cut a given indefinite right line in one point, ſo that of the three ſeg-<lb/>ments intercepted between the ſame, and three points given, the rectangle <lb/>under two of them may be to the ſquare of the remaining one in a given ratio.</s>
  <s xml:id="echoid-s1355" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1356" xml:space="preserve">In the indefinite line let the three points be A, E, I. </s>
  <s xml:id="echoid-s1357" xml:space="preserve">it is then required to be <lb/>cut again in O, ſo that OA x OE may be to <emph style="ol">OI</emph><emph style="sub">2</emph> (let the ſituation of I be <lb/>what it may) in a given ratio, which ratio let be expreſſed by EL to LI. <lb/></s>
  <s xml:id="echoid-s1358" xml:space="preserve">[And here I cannot but obſerve with <emph style="sc">Hugo</emph> D'<emph style="sc">Omerique</emph>, page 113. </s>
  <s xml:id="echoid-s1359" xml:space="preserve">that <lb/>this Problem, viz. </s>
  <s xml:id="echoid-s1360" xml:space="preserve">‘To exhibit two lines in a given ratio whoſe ſum, or whoſe <lb/>difference is given,’ ought to have had a place in the Elements as a Propoſition; </s>
  <s xml:id="echoid-s1361" xml:space="preserve"><lb/>or at leaſt to have been annext as a Scholium to the 9th or 10th of the VIth <lb/>Book.</s>
  <s xml:id="echoid-s1362" xml:space="preserve">] And be the ſituation of L alſo what it may, either between A and E, <lb/>or between A and I, or between E and I, or beyond either extreme. </s>
  <s xml:id="echoid-s1363" xml:space="preserve">To the three <lb/>points E, L, I, and the right line AI, let be found, by <emph style="sc">Problem</emph> II, a fourth <lb/>point O ſuch, that AI x OE: </s>
  <s xml:id="echoid-s1364" xml:space="preserve">OI x OL:</s>
  <s xml:id="echoid-s1365" xml:space="preserve">: EI: </s>
  <s xml:id="echoid-s1366" xml:space="preserve">IL. </s>
  <s xml:id="echoid-s1367" xml:space="preserve">And let ſuch a Caſe be <lb/>choſen of <emph style="sc">Problem</emph> II, that, according as AO is greater or leſs than AI, ſo of <lb/>the three rectangles, deſcribed in <emph style="sc">Lemma</emph> V, made by the four points E, O, <lb/>I, L, that of IO x EL may accordingly be greater or leſs than that of EI x OL.</s>
  <s xml:id="echoid-s1368" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1369" xml:space="preserve">D<emph style="sc">EMONSTRATION</emph> .</s>
  <s xml:id="echoid-s1370" xml:space="preserve">On ſuppoſition then that ſuch a Caſe of <emph style="sc">Problem</emph> II. </s>
  <s xml:id="echoid-s1371" xml:space="preserve">is <lb/>made uſe of, we have <lb/>AI x OE: </s>
  <s xml:id="echoid-s1372" xml:space="preserve">OI x OL:</s>
  <s xml:id="echoid-s1373" xml:space="preserve">: EI: </s>
  <s xml:id="echoid-s1374" xml:space="preserve">IL</s>
</p>
<p>
  <s xml:id="echoid-s1375" xml:space="preserve">And by <emph style="sc">Lemma</emph> IV, OL x EI: </s>
  <s xml:id="echoid-s1376" xml:space="preserve">OE x IL:</s>
  <s xml:id="echoid-s1377" xml:space="preserve">: AI: </s>
  <s xml:id="echoid-s1378" xml:space="preserve">OI</s>
</p>
<p>
  <s xml:id="echoid-s1379" xml:space="preserve">And by Diviſion or Compoſition EL x OI: </s>
  <s xml:id="echoid-s1380" xml:space="preserve">OE x IL:</s>
  <s xml:id="echoid-s1381" xml:space="preserve">: AO: </s>
  <s xml:id="echoid-s1382" xml:space="preserve">OI</s>
</p>
<p>
  <s xml:id="echoid-s1383" xml:space="preserve">This appears from <emph style="sc">Lemma</emph> V.</s>
  <s xml:id="echoid-s1384" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1385" xml:space="preserve">Then again by <emph style="sc">Lemma</emph> IV, AO x OE: </s>
  <s xml:id="echoid-s1386" xml:space="preserve"><emph style="ol">OI</emph><emph style="sub">2</emph>:</s>
  <s xml:id="echoid-s1387" xml:space="preserve">: EL: </s>
  <s xml:id="echoid-s1388" xml:space="preserve">IL.</s>
  <s xml:id="echoid-s1389" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1390" xml:space="preserve">Q. </s>
  <s xml:id="echoid-s1391" xml:space="preserve">E. </s>
  <s xml:id="echoid-s1392" xml:space="preserve">D.</s>
  <s xml:id="echoid-s1393" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1394" xml:space="preserve">This Problem has two <emph style="sc">Epitagmas</emph>. </s>
  <s xml:id="echoid-s1395" xml:space="preserve">The firſt wherein OI, whoſe ſquare is <lb/>ſought, is bounded by I an extreme point of the three given ones. </s>
  <s xml:id="echoid-s1396" xml:space="preserve">And this <lb/>again admits of three Caſes. </s>
  <s xml:id="echoid-s1397" xml:space="preserve">The ſecond is when the point I is the middle <lb/>point. </s>
  <s xml:id="echoid-s1398" xml:space="preserve">And this again has three caſes. </s>
  <s xml:id="echoid-s1399" xml:space="preserve">And there remain two Anomalous <lb/>Caſes, wherein Problem II. </s>
  <s xml:id="echoid-s1400" xml:space="preserve">is of no uſe, which muſt therefore be conſtructed <lb/>by themſelves.</s>
  <s xml:id="echoid-s1401" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1402" xml:space="preserve"><emph style="sc">Epitagma</emph> I. </s>
  <s xml:id="echoid-s1403" xml:space="preserve"><emph style="sc">Case</emph> I. </s>
  <s xml:id="echoid-s1404" xml:space="preserve">Let the ratio given, EL to LI, be inequalitatis <lb/>majoris, i. </s>
  <s xml:id="echoid-s1405" xml:space="preserve">e. </s>
  <s xml:id="echoid-s1406" xml:space="preserve">of a greater to a leſs; </s>
  <s xml:id="echoid-s1407" xml:space="preserve">and the point O ſought be required to lie <lb/>between I and the next point to it E, or elſe to lie beyond I the other way; <lb/></s>
  <s xml:id="echoid-s1408" xml:space="preserve">for the ſame conſtruction ſerves for both. </s>
  <s xml:id="echoid-s1409" xml:space="preserve">Here <emph style="sc">Case</emph> I. </s>
  <s xml:id="echoid-s1410" xml:space="preserve">of <emph style="sc">Problem</emph> II. </s>
  <s xml:id="echoid-s1411" xml:space="preserve">is to
<pb o="[8]" file="0078" n="85"/>
be uſed, and the point O will fall between E and I, and the point o beyond <lb/>L, much more beyond I.</s>
  <s xml:id="echoid-s1412" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1413" xml:space="preserve"><emph style="sc">Case</emph> II. </s>
  <s xml:id="echoid-s1414" xml:space="preserve">Let the given ratio, EL to LI, be inæqualitatis minoris, i. </s>
  <s xml:id="echoid-s1415" xml:space="preserve">e. </s>
  <s xml:id="echoid-s1416" xml:space="preserve">of a <lb/>leſs to a greater, and the point O ſought be required to lie between I and the <lb/>next point to it E; </s>
  <s xml:id="echoid-s1417" xml:space="preserve">or elſe to fall beyond A the other extreme. </s>
  <s xml:id="echoid-s1418" xml:space="preserve">For the ſame <lb/>conſtruction ſerves for both. </s>
  <s xml:id="echoid-s1419" xml:space="preserve">Here <emph style="sc">Case</emph> IV. </s>
  <s xml:id="echoid-s1420" xml:space="preserve">of <emph style="sc">Problem</emph> II. </s>
  <s xml:id="echoid-s1421" xml:space="preserve">is to be uſed, and <lb/>the point O will fall between E and I, and o beyond A, if we uſe one of the <lb/>conſtructions there recited: </s>
  <s xml:id="echoid-s1422" xml:space="preserve">but if we uſe the other, the points will ſhift places, <lb/>as was obſerved under that Caſe, viz. </s>
  <s xml:id="echoid-s1423" xml:space="preserve">O will fall beyond I the other way, and <lb/>o between L and E.</s>
  <s xml:id="echoid-s1424" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1425" xml:space="preserve"><emph style="sc">Case</emph> III. </s>
  <s xml:id="echoid-s1426" xml:space="preserve">Let now the point O be ſought between A and E. </s>
  <s xml:id="echoid-s1427" xml:space="preserve">Here ſet off <lb/>the given ratio in ſuch a manner that EI may be the ſum of the terms, and <lb/>make uſe of the IIId <emph style="sc">Case</emph> of <emph style="sc">Problem</emph> II. </s>
  <s xml:id="echoid-s1428" xml:space="preserve">and the <emph style="sc">Limitation</emph> here will <lb/>be evident from the <emph style="sc">Limitation</emph> there given, viz. </s>
  <s xml:id="echoid-s1429" xml:space="preserve">making EI: </s>
  <s xml:id="echoid-s1430" xml:space="preserve">IL:</s>
  <s xml:id="echoid-s1431" xml:space="preserve">: AI: </s>
  <s xml:id="echoid-s1432" xml:space="preserve">X, <lb/>the <emph style="sc">Limitation</emph> here is that X muſt not be leſs than IE + EL + √4 IEL*.</s>
  <s xml:id="echoid-s1433" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1434" xml:space="preserve"><emph style="sc">Epitagma</emph> II. </s>
  <s xml:id="echoid-s1435" xml:space="preserve"><emph style="sc">Case</emph> IV. </s>
  <s xml:id="echoid-s1436" xml:space="preserve">Here OI the line whoſe ſquare is concerned is <lb/>to be bounded by I the middle point of the three given ones, and O or o, its <lb/>other bound is to be ſought between I and either extreine A or E. </s>
  <s xml:id="echoid-s1437" xml:space="preserve">the ſame <lb/>conſtruction ſerving for both. </s>
  <s xml:id="echoid-s1438" xml:space="preserve">The given ratio muſt here be ſet off in ſuch a <lb/>manner that EI may be the ſum of the terms of it; </s>
  <s xml:id="echoid-s1439" xml:space="preserve">and make uſe <lb/>of Iſt <emph style="sc">Case</emph> of the IId <emph style="sc">Problem</emph>; </s>
  <s xml:id="echoid-s1440" xml:space="preserve">with this caution, that of the two ſegments <lb/>AI, IE, you choſe the leſſer IE whereon to exhibit the given ratio; </s>
  <s xml:id="echoid-s1441" xml:space="preserve">for then <lb/>it will appear by the work itſelf that O falling between E and L, o will alſo <lb/>fall between A and I: </s>
  <s xml:id="echoid-s1442" xml:space="preserve">otherwiſe, if AI was leſs than IE, there would want <lb/>ſome proof of this. </s>
  <s xml:id="echoid-s1443" xml:space="preserve">Therefore of the two extreme given points call that E <lb/>which bounds the leſſer ſegment, and then the general Demonſtration will fit <lb/>this Caſe as well as the reſt.</s>
  <s xml:id="echoid-s1444" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1445" xml:space="preserve"><emph style="sc">Case</emph> V. </s>
  <s xml:id="echoid-s1446" xml:space="preserve">Let the given ratio of EL to LI be inæqualitatis minoris; </s>
  <s xml:id="echoid-s1447" xml:space="preserve">and let <lb/>the point ſought be required to lie beyond either extreme. </s>
  <s xml:id="echoid-s1448" xml:space="preserve">The ſame con-<lb/>ſtruction ſerves for both. </s>
  <s xml:id="echoid-s1449" xml:space="preserve">Here we muſt uſe the IVth <emph style="sc">Case</emph> of the IId <emph style="sc">Pro-</emph> <lb/><emph style="sc">BLEM</emph>, and O being made to fall between E and L, o will fall always beyond <lb/>A, provided we call that point E which bounds the bigger ſegment. </s>
  <s xml:id="echoid-s1450" xml:space="preserve">I have <lb/>in the Figure made AI = IE on purpoſe to ſhew that in that caſe the point N <lb/>will coincide with A. </s>
  <s xml:id="echoid-s1451" xml:space="preserve">But if IE be greater than AI, the point N will <lb/>always fall beyond A, and conſequently the point o more ſo.</s>
  <s xml:id="echoid-s1452" xml:space="preserve"/>
</p>
<pb o="[9]" file="0079" n="86"/>
<p>
  <s xml:id="echoid-s1453" xml:space="preserve"><emph style="sc">Case</emph> VI. </s>
  <s xml:id="echoid-s1454" xml:space="preserve">Let the given ratio of EL to LI be inæqualitatis majoris, and <lb/>let the point ſought be required to lie beyond either extreme. </s>
  <s xml:id="echoid-s1455" xml:space="preserve">Here we muſt <lb/>uſe the IIId <emph style="sc">Case</emph> of the IId <emph style="sc">Problem</emph>; </s>
  <s xml:id="echoid-s1456" xml:space="preserve">and the <emph style="sc">Determination</emph> is that <lb/>UN (found in the ſame ratio to AI as IL is to IE) muſt not be leſs than <lb/>IE + EL + √4 IEL*.</s>
  <s xml:id="echoid-s1457" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1458" xml:space="preserve"><emph style="sc">Case</emph> VII. </s>
  <s xml:id="echoid-s1459" xml:space="preserve">Let the ſituation of O be required the ſame as in the two laſt <lb/>Caſes, but let the given ratio be that of equality, which was there ſuppoſed <lb/>of inequality. </s>
  <s xml:id="echoid-s1460" xml:space="preserve">Here the IId <emph style="sc">Problem</emph> will be of no uſe, and this Caſe <lb/>requires a particular conſtruction.</s>
  <s xml:id="echoid-s1461" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1462" xml:space="preserve">Let then the three Points be A, I, E, and I the middle one; </s>
  <s xml:id="echoid-s1463" xml:space="preserve">and let it <lb/>be required to find a fourth O beyond E, ſuch that AO x OE may equal <lb/><emph style="ol">OI</emph><emph style="sub">2</emph>.</s>
  <s xml:id="echoid-s1464" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1465" xml:space="preserve"><emph style="sc">Construction</emph>. </s>
  <s xml:id="echoid-s1466" xml:space="preserve">Upon AE diameter deſcribe a circle,, and let another YS <lb/>cut the former at right angles. </s>
  <s xml:id="echoid-s1467" xml:space="preserve">Join SI, and continue it to meet the circum-<lb/>ference in R. </s>
  <s xml:id="echoid-s1468" xml:space="preserve">From R draw a tangent to meet the given line in O, and I <lb/>ſay O is the point required.</s>
  <s xml:id="echoid-s1469" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1470" xml:space="preserve"><emph style="sc">Demonstration</emph>. </s>
  <s xml:id="echoid-s1471" xml:space="preserve">Joining YR, the triangles SUI and SYR will be ſimi-<lb/>lar, and the angle UIS or RIO = SYR. </s>
  <s xml:id="echoid-s1472" xml:space="preserve">But the angle IRO made by the <lb/>tangent and ſecant = SYR in the alternate ſegment. </s>
  <s xml:id="echoid-s1473" xml:space="preserve">Therefore RIO = IRO, <lb/>and OR = OI. </s>
  <s xml:id="echoid-s1474" xml:space="preserve">But by the property of the circle AOE = <emph style="ol">OR</emph><emph style="sub">2</emph>. </s>
  <s xml:id="echoid-s1475" xml:space="preserve">And <lb/>therefore AOE = <emph style="ol">OI</emph><emph style="sub">2</emph>.</s>
  <s xml:id="echoid-s1476" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1477" xml:space="preserve">Q. </s>
  <s xml:id="echoid-s1478" xml:space="preserve">E. </s>
  <s xml:id="echoid-s1479" xml:space="preserve">D.</s>
  <s xml:id="echoid-s1480" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1481" xml:space="preserve">The <emph style="sc">Determination</emph> is that AI muſt be greater than IE.</s>
  <s xml:id="echoid-s1482" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1483" xml:space="preserve"><emph style="sc">Case</emph> VIII. </s>
  <s xml:id="echoid-s1484" xml:space="preserve">Whereas in the Iſt and IId <emph style="sc">Cases</emph> the given ratio was that of <lb/>inequality, let us now ſuppoſe it that of equality; </s>
  <s xml:id="echoid-s1485" xml:space="preserve">and let the three points be <lb/>A, E, I, and E the middle one; </s>
  <s xml:id="echoid-s1486" xml:space="preserve">and let a fourth O be ſought between E and <lb/>I, ſuch that AOE may equal <emph style="ol">OI</emph><emph style="sub">2</emph>.</s>
  <s xml:id="echoid-s1487" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1488" xml:space="preserve">The <emph style="sc">Construction</emph> and <emph style="sc">Demonstration</emph> of this Caſe is in every reſpect <lb/>the ſame as that of the preceeding, as will appear by comparing the figures.</s>
  <s xml:id="echoid-s1489" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div74" type="section" level="1" n="70">
<head xml:id="echoid-head83" xml:space="preserve">PROBLEM IV.</head>
<p>
  <s xml:id="echoid-s1490" xml:space="preserve">To cut a given indefinite right line ſo in one point that, of the four ſeg-<lb/>ments intercepted between the ſame and four points given in the indefinite <lb/>line, the rectangle under any two aſſigned ones may be to the rectangle under <lb/>the two remaining ones in a given ratio.</s>
  <s xml:id="echoid-s1491" xml:space="preserve"/>
</p>
<pb o="[10]" file="0080" n="87"/>
<p>
  <s xml:id="echoid-s1492" xml:space="preserve">In the indefinite line let the four points be A, E, I, U. </s>
  <s xml:id="echoid-s1493" xml:space="preserve">It is then required <lb/>to be cut again in O ſo that OA x OU may be to OE x OI (be the Poſition <lb/>of the four given points what they may) in the ratio of AL to LE, let the <lb/>point L fall alſo as it may.</s>
  <s xml:id="echoid-s1494" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1495" xml:space="preserve"><emph style="sc">Construction</emph>. </s>
  <s xml:id="echoid-s1496" xml:space="preserve">To the three points E, L, U, and the right line UI, let <lb/>be found by the IId <emph style="sc">Problem</emph> a fourth point O, ſo that UI x OE may be to <lb/>OU x OL as AE to AL. </s>
  <s xml:id="echoid-s1497" xml:space="preserve">And let ſuch a <emph style="sc">Case</emph> be choſen of the IId <emph style="sc">Problem</emph> <lb/>that, according as UO is required greater or leſs than UI, or their ſum ſhall <lb/>conſtitute OI, ſo of the three rectangles deſcribed in the Vth <emph style="sc">Lemma</emph> made <lb/>by the four points E, O, A, L, that of OE x AL may accordingly be greater <lb/>or leſs than OL x AE, or their ſum conſtitute that of OA x EL.</s>
  <s xml:id="echoid-s1498" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1499" xml:space="preserve">N. </s>
  <s xml:id="echoid-s1500" xml:space="preserve">B. </s>
  <s xml:id="echoid-s1501" xml:space="preserve">U being now uſed to repreſent one of the given points, in all the <lb/>following Diagrams I have ſubſtituted V in the place where U was uſed <lb/>before.</s>
  <s xml:id="echoid-s1502" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1503" xml:space="preserve"><emph style="sc">Demonstration</emph>. </s>
  <s xml:id="echoid-s1504" xml:space="preserve">On ſuppoſition therefore that ſuch a <emph style="sc">Case</emph> of the <lb/>IId <emph style="sc">Problem</emph> is made uſe of, <lb/>We have UI x OE: </s>
  <s xml:id="echoid-s1505" xml:space="preserve">OU x OL:</s>
  <s xml:id="echoid-s1506" xml:space="preserve">: AE: </s>
  <s xml:id="echoid-s1507" xml:space="preserve">AL</s>
</p>
<p>
  <s xml:id="echoid-s1508" xml:space="preserve">And by Inverſion OU x OL: </s>
  <s xml:id="echoid-s1509" xml:space="preserve">UI x OE:</s>
  <s xml:id="echoid-s1510" xml:space="preserve">: AL: </s>
  <s xml:id="echoid-s1511" xml:space="preserve">AE</s>
</p>
<p>
  <s xml:id="echoid-s1512" xml:space="preserve">And by <emph style="sc">Lemma</emph> IV. </s>
  <s xml:id="echoid-s1513" xml:space="preserve">AL x OE: </s>
  <s xml:id="echoid-s1514" xml:space="preserve">AE x OL:</s>
  <s xml:id="echoid-s1515" xml:space="preserve">: OU: </s>
  <s xml:id="echoid-s1516" xml:space="preserve">UI</s>
</p>
<p>
  <s xml:id="echoid-s1517" xml:space="preserve">Hence by Compoſition or Diviſion, &amp;</s>
  <s xml:id="echoid-s1518" xml:space="preserve">c. </s>
  <s xml:id="echoid-s1519" xml:space="preserve">AL x OE: </s>
  <s xml:id="echoid-s1520" xml:space="preserve">OA x LE:</s>
  <s xml:id="echoid-s1521" xml:space="preserve">: OU <lb/>: </s>
  <s xml:id="echoid-s1522" xml:space="preserve">OI as appears by <emph style="sc">Lemma</emph> V.</s>
  <s xml:id="echoid-s1523" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1524" xml:space="preserve">Then again by <emph style="sc">Lemma</emph> IV. </s>
  <s xml:id="echoid-s1525" xml:space="preserve">OU x OA: </s>
  <s xml:id="echoid-s1526" xml:space="preserve">OI x OE:</s>
  <s xml:id="echoid-s1527" xml:space="preserve">: AL: </s>
  <s xml:id="echoid-s1528" xml:space="preserve">LE</s>
</p>
<p>
  <s xml:id="echoid-s1529" xml:space="preserve">Q. </s>
  <s xml:id="echoid-s1530" xml:space="preserve">E. </s>
  <s xml:id="echoid-s1531" xml:space="preserve">D.</s>
  <s xml:id="echoid-s1532" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1533" xml:space="preserve">This <emph style="sc">Problem</emph> has three <emph style="sc">Epitagmas</emph>. </s>
  <s xml:id="echoid-s1534" xml:space="preserve">The Iſt whereof is when of the <lb/>two aſſigned points A and U, the one of them is an extreme, and the other an <lb/>alternate mean; </s>
  <s xml:id="echoid-s1535" xml:space="preserve">and this admits of three <emph style="sc">Cases</emph>. </s>
  <s xml:id="echoid-s1536" xml:space="preserve">The IId is when A and U <lb/>are both of them extremes; </s>
  <s xml:id="echoid-s1537" xml:space="preserve">and this has four <emph style="sc">Cases</emph>. </s>
  <s xml:id="echoid-s1538" xml:space="preserve">The IIId is when of <lb/>A and U one of them is an extreme, and the other is the point next to it; </s>
  <s xml:id="echoid-s1539" xml:space="preserve">and <lb/>this has three Caſes. </s>
  <s xml:id="echoid-s1540" xml:space="preserve">And there remain three more Anomalous Caſes, wherein <lb/>the IId <emph style="sc">Problem</emph> is of no uſe, but which may be reduced to one, as ſhall be <lb/>ſhewn in it's proper place.</s>
  <s xml:id="echoid-s1541" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1542" xml:space="preserve"><emph style="sc">Epitagma</emph> I. </s>
  <s xml:id="echoid-s1543" xml:space="preserve"><emph style="sc">Case</emph> I. </s>
  <s xml:id="echoid-s1544" xml:space="preserve">Let A the firſt aſſigned point be an extreme, and <lb/>U the ſecond aſſigned point be an alternate mean; </s>
  <s xml:id="echoid-s1545" xml:space="preserve">and let the point O be <lb/>ſought between the firſt aſſigned A and the next point to it E; </s>
  <s xml:id="echoid-s1546" xml:space="preserve">or between <lb/>the ſecond aſſigned U and the laſt I. </s>
  <s xml:id="echoid-s1547" xml:space="preserve">For the ſame Conſtruction ſerves <lb/>for both.</s>
  <s xml:id="echoid-s1548" xml:space="preserve"/>
</p>
<pb o="[11]" file="0081" n="88"/>
<p>
  <s xml:id="echoid-s1549" xml:space="preserve">Here AE is to be made the ſum of the terms of the given ratio, and we are <lb/>to uſe the IVth <emph style="sc">Case</emph> of the IId <emph style="sc">Problem</emph>, whereby O falling between <lb/>L and E, o will fall beyond U; </s>
  <s xml:id="echoid-s1550" xml:space="preserve">and that it will fall ſhort of I appears from <lb/>the Iſt <emph style="sc">Corollary</emph> from the IVth <emph style="sc">Case</emph> of the IId <emph style="sc">Problem</emph>.</s>
  <s xml:id="echoid-s1551" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1552" xml:space="preserve"><emph style="sc">Case</emph> II. </s>
  <s xml:id="echoid-s1553" xml:space="preserve">The given ratio being inæqualitatis minoris, let the point ſought <lb/>be required between the ſecond aſſigned U and the ſecond in order E, or be-<lb/>yond the firſt A, which ariſes from the ſame Conſtruction. </s>
  <s xml:id="echoid-s1554" xml:space="preserve">Here AE is to be <lb/>made the difference of the terms of the given ratio, and we are to uſe the <lb/>IVth <emph style="sc">Case</emph> of the IId <emph style="sc">Problem</emph>, where O being made to fall between U and E, <lb/>o will fall beyond L, much more beyond A.</s>
  <s xml:id="echoid-s1555" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1556" xml:space="preserve"><emph style="sc">Case</emph> III. </s>
  <s xml:id="echoid-s1557" xml:space="preserve">The given ratio being inæqualitatis majoris, let the point ſought <lb/>be required between the ſecond aſſigned U and the ſecond in order E, or be-<lb/>yond the laſt I, which ariſes from the ſame Conſtruction. </s>
  <s xml:id="echoid-s1558" xml:space="preserve">Here AE is to be <lb/>made the difference of the terms of the given ratio, and L is to be ſet off the <lb/>contrary way to what it was in the laſt <emph style="sc">Case</emph>; </s>
  <s xml:id="echoid-s1559" xml:space="preserve">and we are to uſe the Iſt <emph style="sc">Case</emph> <lb/>of the IId <emph style="sc">Problem</emph>, whereby O being made to fall between E and L, or <lb/>between E and U, according as L or U is neareſt to the point E, o will fall <lb/>beyond I, as any one will ſee who conſiders the Conſtruction of that <emph style="sc">Case</emph> with <lb/>due attention.</s>
  <s xml:id="echoid-s1560" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1561" xml:space="preserve"><emph style="sc">Epitagma</emph> II. </s>
  <s xml:id="echoid-s1562" xml:space="preserve"><emph style="sc">Case</emph> IV. </s>
  <s xml:id="echoid-s1563" xml:space="preserve">Let the aſſigned points now be the extremes <lb/>A and U, and let O the point ſought be required now between the firſt <lb/>aſſigned A and the next to it E, or, which is effected by the ſame Conſtruction, <lb/>between the ſecond aſſigned U and the next to it I. </s>
  <s xml:id="echoid-s1564" xml:space="preserve">Here AE is to be made <lb/>the ſum of the terms of the given ratio, and the IVth <emph style="sc">Case</emph> of the IId <emph style="sc">Pro-</emph> <lb/><emph style="sc">BLEM</emph> is to be uſed, ſo that of the three points L, E, U, O being made to fall <lb/>beyond L one of the extremes, and o within U the other extreme, it will appear <lb/>from the Iſt <emph style="sc">Corollary</emph> from the IVth <emph style="sc">Case</emph> of the ſaid <emph style="sc">Problem</emph> that O <lb/>will fall between A and E, and o between U and I.</s>
  <s xml:id="echoid-s1565" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1566" xml:space="preserve"><emph style="sc">Case</emph> V. </s>
  <s xml:id="echoid-s1567" xml:space="preserve">The given ratio being inæqualitatis majoris, let the point ſought be <lb/>required between the ſecond and third in order, viz. </s>
  <s xml:id="echoid-s1568" xml:space="preserve">between E and I. </s>
  <s xml:id="echoid-s1569" xml:space="preserve">Here <lb/>AE muſt be the difference of the terms of the given ratio, and L ſet off to-<lb/>wards I, and the IId <emph style="sc">Case</emph> of the IId <emph style="sc">Problem</emph> uſed, and then O, as like-<lb/>wiſe o, will fall between E and I, if the <emph style="sc">Problem</emph> be poſſible.</s>
  <s xml:id="echoid-s1570" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1571" xml:space="preserve">As to the <emph style="sc">Determination</emph>, ſee <emph style="sc">Lemma</emph> VII. </s>
  <s xml:id="echoid-s1572" xml:space="preserve">following.</s>
  <s xml:id="echoid-s1573" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1574" xml:space="preserve"><emph style="sc">Case</emph> VI. </s>
  <s xml:id="echoid-s1575" xml:space="preserve">The given ratio being inæqualitatis majoris, let the point ſought <lb/>be required beyond the laſt aſſigned, that is the laſt in order, U. </s>
  <s xml:id="echoid-s1576" xml:space="preserve">Here AE <lb/>muſt be the difference of the terms of the given ratio, [and L muſt evidently
<pb o="[12]" file="0082" n="89"/>
fall beyond U, but for a more particular <emph style="sc">Determination</emph> ſee <emph style="sc">Lemma</emph> VII. <lb/></s>
  <s xml:id="echoid-s1577" xml:space="preserve">following] and the IId <emph style="sc">Case</emph> of the IId <emph style="sc">Problem</emph> is to be uſed, and then O, <lb/>as likewiſe o, will fall beyond U.</s>
  <s xml:id="echoid-s1578" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1579" xml:space="preserve"><emph style="sc">Case</emph> VII. </s>
  <s xml:id="echoid-s1580" xml:space="preserve">The given ratio being inæqualitatis minoris, let the point ſought <lb/>be required to lie beyond either of the aſſigned ones, i. </s>
  <s xml:id="echoid-s1581" xml:space="preserve">e. </s>
  <s xml:id="echoid-s1582" xml:space="preserve">beyond either ex-<lb/>treme, the ſame Conſtruction ſerving for both. </s>
  <s xml:id="echoid-s1583" xml:space="preserve">Here AE is to be the diſ-<lb/>ference of the terms of the given ratio, and L to be ſet off backwards beyond <lb/>A; </s>
  <s xml:id="echoid-s1584" xml:space="preserve">and the IVth <emph style="sc">Case</emph> of the IId <emph style="sc">Problem</emph> uſed, that ſo O being made to <lb/>fall beyond U, it will appear, by the IId <emph style="sc">Corollary</emph> from the IVth <emph style="sc">Case</emph> of <lb/>the ſaid <emph style="sc">Problem</emph>, that o will alſo fall beyond A.</s>
  <s xml:id="echoid-s1585" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1586" xml:space="preserve"><emph style="sc">Epitagma</emph> III. </s>
  <s xml:id="echoid-s1587" xml:space="preserve"><emph style="sc">Case</emph> VIII. </s>
  <s xml:id="echoid-s1588" xml:space="preserve">Let the aſſigned points A and U be now <lb/>one an extreme, and the other the point next it: </s>
  <s xml:id="echoid-s1589" xml:space="preserve">and let the point ſought be re-<lb/>quired to fall between the two aſſigned ones. </s>
  <s xml:id="echoid-s1590" xml:space="preserve">Here AE muſt be the ſum of <lb/>the terms of the given ratio, and the IId <emph style="sc">Case</emph> of the IId <emph style="sc">Problem</emph> uſed. <lb/></s>
  <s xml:id="echoid-s1591" xml:space="preserve">And ſo O, as likewiſe o, being made to fall between L and U, they will <lb/>much more fall between A and U.</s>
  <s xml:id="echoid-s1592" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1593" xml:space="preserve">The <emph style="sc">Limitation</emph> is, that VN (found in the ſame ratio to UI as AL to AE) <lb/>muſt not exceed LE + EU - √4 LEU*.</s>
  <s xml:id="echoid-s1594" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1595" xml:space="preserve"><emph style="sc">Case</emph> IX. </s>
  <s xml:id="echoid-s1596" xml:space="preserve">The given ratio being inæqualitatis minoris, let the point ſought <lb/>be required between the ſecond aſſigned U and the third in order E, or elſe <lb/>beyond the firſt aſſigned A, the ſame Conſtruction ſerving for both. </s>
  <s xml:id="echoid-s1597" xml:space="preserve">Here AE <lb/>is to be the difference of the terms of the given ratio, and L to be ſet off be-<lb/>yond A, and the Iſt <emph style="sc">Case</emph> of the IId <emph style="sc">Problem</emph> uſed: </s>
  <s xml:id="echoid-s1598" xml:space="preserve">and ſo O being made <lb/>to fall between E and U, o will fall beyond L, and much more be-<lb/>yond A.</s>
  <s xml:id="echoid-s1599" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1600" xml:space="preserve"><emph style="sc">Case</emph> X. </s>
  <s xml:id="echoid-s1601" xml:space="preserve">The given ratio being inæqualitatis majoris, let the point ſought <lb/>be required between the ſecond aſſigned U and the third in order E, or elſe <lb/>beyond the laſt in order I, the ſame conſtruction ſerving for both. </s>
  <s xml:id="echoid-s1602" xml:space="preserve">Here AE <lb/>is to be the difference of the terms of the given ratio, and L to be ſet off be-<lb/>yond E, and the IVth <emph style="sc">Case</emph> of the IId <emph style="sc">Problem</emph> uſed, ſo that O being <lb/>made to fall between U and E, o will fall beyond I, as any one will ſee who <lb/>conſiders the conſtruction of that <emph style="sc">Case</emph> with due attention.</s>
  <s xml:id="echoid-s1603" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1604" xml:space="preserve"><emph style="sc">Case</emph> XI. </s>
  <s xml:id="echoid-s1605" xml:space="preserve">As to the three Anomalous Caſes, in which the IId <emph style="sc">Problem</emph> is <lb/>of no uſe, and which I ſaid before might be reduced to one, they are theſe: <lb/></s>
  <s xml:id="echoid-s1606" xml:space="preserve">whereas in the IId and IIId <emph style="sc">Cases</emph>, as alſo in the VIth and VIIth, and like-<lb/>wiſe in the IXth and Xth the ratio given was that of inequality, let us now
<pb o="[13]" file="0083" n="90"/>
ſuppoſe it that of equality, and they may all be ſolved by one conſtruction, viz. <lb/></s>
  <s xml:id="echoid-s1607" xml:space="preserve">the rectangle AOU made equal to the rectangle EOI.</s>
  <s xml:id="echoid-s1608" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1609" xml:space="preserve">Let it be made as UI: </s>
  <s xml:id="echoid-s1610" xml:space="preserve">AE:</s>
  <s xml:id="echoid-s1611" xml:space="preserve">: UO: </s>
  <s xml:id="echoid-s1612" xml:space="preserve">EO</s>
</p>
<p>
  <s xml:id="echoid-s1613" xml:space="preserve">Then by permutation UO: </s>
  <s xml:id="echoid-s1614" xml:space="preserve">UI:</s>
  <s xml:id="echoid-s1615" xml:space="preserve">: EO: </s>
  <s xml:id="echoid-s1616" xml:space="preserve">AE</s>
</p>
<p>
  <s xml:id="echoid-s1617" xml:space="preserve">And by comp. </s>
  <s xml:id="echoid-s1618" xml:space="preserve">or diviſ. </s>
  <s xml:id="echoid-s1619" xml:space="preserve">UO: </s>
  <s xml:id="echoid-s1620" xml:space="preserve">OI:</s>
  <s xml:id="echoid-s1621" xml:space="preserve">: EO: </s>
  <s xml:id="echoid-s1622" xml:space="preserve">AO</s>
</p>
<p>
  <s xml:id="echoid-s1623" xml:space="preserve">Hence AOU = EOI.</s>
  <s xml:id="echoid-s1624" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1625" xml:space="preserve"><emph style="sc">Lemma</emph> VI. </s>
  <s xml:id="echoid-s1626" xml:space="preserve">Let there be two ſimilar triangles IAE, UAO, having their <lb/>baſes IE and UO parallel; </s>
  <s xml:id="echoid-s1627" xml:space="preserve">I ſay Iſt when they are right-angled, that the ex-<lb/>ceſs of the rectangle EAO, under the greater ſides of each, above the rect-<lb/>angle IAU, under the leſſer ſides of each, will be equal to the rectangle <lb/>IE x OU, under their baſes. </s>
  <s xml:id="echoid-s1628" xml:space="preserve">IIdly, When they are obtuſe-angled, that the <lb/>ſaid exceſs will be equal to the rectangle under the baſe of one and the ſum <lb/>of the diſtances of the angles at the baſe of the other from the perpendicular, <lb/>viz. </s>
  <s xml:id="echoid-s1629" xml:space="preserve">EI x <emph style="ol">OS + US</emph>. </s>
  <s xml:id="echoid-s1630" xml:space="preserve">IIIdly, When they are acute-angled, that then the ſaid <lb/>exceſs will be equal to the rectangle under the baſe of one and the difference <lb/>of the ſegments of the baſe of the other made by the perpendicular, viz. <lb/></s>
  <s xml:id="echoid-s1631" xml:space="preserve">OU x EL.</s>
  <s xml:id="echoid-s1632" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1633" xml:space="preserve"><emph style="sc">Demonstration</emph>. </s>
  <s xml:id="echoid-s1634" xml:space="preserve">Since EA: </s>
  <s xml:id="echoid-s1635" xml:space="preserve">AO:</s>
  <s xml:id="echoid-s1636" xml:space="preserve">: IA: </s>
  <s xml:id="echoid-s1637" xml:space="preserve">AU:</s>
  <s xml:id="echoid-s1638" xml:space="preserve">: EI: </s>
  <s xml:id="echoid-s1639" xml:space="preserve">OU, the rect-<lb/>angles EAO, IAU, and EI x OU will be ſimilar, and when Iſt the triangles <lb/>are right-angled EAO = IAU + EI x OU by Euc. </s>
  <s xml:id="echoid-s1640" xml:space="preserve">VI. </s>
  <s xml:id="echoid-s1641" xml:space="preserve">19. </s>
  <s xml:id="echoid-s1642" xml:space="preserve">and I. </s>
  <s xml:id="echoid-s1643" xml:space="preserve">47. </s>
  <s xml:id="echoid-s1644" xml:space="preserve">But if <lb/>they be oblique-angled, draw the perpendicular YAS. </s>
  <s xml:id="echoid-s1645" xml:space="preserve">Then IIdly, in caſe <lb/>they be obtuſe-angled, EAO = YAS + EY x OS by part Iſt; </s>
  <s xml:id="echoid-s1646" xml:space="preserve">and IAU = <lb/>YAS + IY x US by the ſame. </s>
  <s xml:id="echoid-s1647" xml:space="preserve">And therefore EAO - IAU = EY x OS -<lb/>IY x US = <emph style="ol">EY - IY</emph> or EI x <emph style="ol">OS + US</emph>. </s>
  <s xml:id="echoid-s1648" xml:space="preserve">But if IIIdly they be acute-angled, <lb/>and EY be greater than IY, then from Y ſet off YL = YI, and draw LAR <lb/>which will be equal and ſimilarly divided to IAU. </s>
  <s xml:id="echoid-s1649" xml:space="preserve">Then by part IId EAO <lb/>- LAR, i. </s>
  <s xml:id="echoid-s1650" xml:space="preserve">e. </s>
  <s xml:id="echoid-s1651" xml:space="preserve">EAO - IAU = EL x <emph style="ol">OS + RS</emph> = EL x OU.</s>
  <s xml:id="echoid-s1652" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1653" xml:space="preserve">Q. </s>
  <s xml:id="echoid-s1654" xml:space="preserve">E. </s>
  <s xml:id="echoid-s1655" xml:space="preserve">D.</s>
  <s xml:id="echoid-s1656" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1657" xml:space="preserve"><emph style="sc">Lemma</emph> VII. </s>
  <s xml:id="echoid-s1658" xml:space="preserve">If a right line VY, joining the tops of two perpendiculars <lb/>drawn from two points of the diameter of a circle E and I to the circum-<lb/>ference on oppoſite ſides of the diameter, cut the ſaid diameter in O, and <lb/>A and U be the extremes of the ſaid diameter, I ſay that the ratio of the <lb/>rectangle AOU to the rectangle EOI is a Minimum.</s>
  <s xml:id="echoid-s1659" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1660" xml:space="preserve">But if VY joins the tops of two perpendiculars from E and I drawn on <lb/>the ſame ſide of the diameter, and conſequently meets the diameter produced <lb/>in O, that then the ratio of AOU to EOI is a Maximum.</s>
  <s xml:id="echoid-s1661" xml:space="preserve"/>
</p>
<pb o="[14]" file="0084" n="91"/>
<p>
  <s xml:id="echoid-s1662" xml:space="preserve"><emph style="sc">Demonstration</emph>. </s>
  <s xml:id="echoid-s1663" xml:space="preserve">Through S, any other point taken at pleaſure, draw <lb/>LSM parallel to VOY, and join VS and produce it to meet the perpendi-<lb/>cular in N and the circumference in R. </s>
  <s xml:id="echoid-s1664" xml:space="preserve">Produce alfo the perpendicular YI <lb/>to meet the circumference again in F, and join RF; </s>
  <s xml:id="echoid-s1665" xml:space="preserve">then from ſrmilar tri-<lb/>angles it appears that the rectangle LSM: </s>
  <s xml:id="echoid-s1666" xml:space="preserve">ESI:</s>
  <s xml:id="echoid-s1667" xml:space="preserve">: VOY i. </s>
  <s xml:id="echoid-s1668" xml:space="preserve">e. </s>
  <s xml:id="echoid-s1669" xml:space="preserve">AOU: </s>
  <s xml:id="echoid-s1670" xml:space="preserve">EOI. <lb/></s>
  <s xml:id="echoid-s1671" xml:space="preserve">But the rectangle ASU i. </s>
  <s xml:id="echoid-s1672" xml:space="preserve">e. </s>
  <s xml:id="echoid-s1673" xml:space="preserve">VSR is greater than LSM. </s>
  <s xml:id="echoid-s1674" xml:space="preserve">(for LV i. </s>
  <s xml:id="echoid-s1675" xml:space="preserve">e. </s>
  <s xml:id="echoid-s1676" xml:space="preserve">MY x <lb/><emph style="ol">NI + MI</emph> together with VS x SN is by the preceeding <emph style="sc">Lemma</emph> equal to LSM. </s>
  <s xml:id="echoid-s1677" xml:space="preserve"><lb/>But ASU or VSR is equal to VS x SN, together with VS x NR; </s>
  <s xml:id="echoid-s1678" xml:space="preserve">for which <lb/>laſt rectangle we may ſubſtitute MY x NF: </s>
  <s xml:id="echoid-s1679" xml:space="preserve">for the triangles VLS and NRF <lb/>are ſimilar, being each of them ſimilar to VNY; </s>
  <s xml:id="echoid-s1680" xml:space="preserve">therefore VL or MY: </s>
  <s xml:id="echoid-s1681" xml:space="preserve">VS <lb/>:</s>
  <s xml:id="echoid-s1682" xml:space="preserve">: NR: </s>
  <s xml:id="echoid-s1683" xml:space="preserve">NF and MY x NF = VS x NR. </s>
  <s xml:id="echoid-s1684" xml:space="preserve">Now NF being always greater <lb/>than <emph style="ol">NI + MI</emph>, it appears from thence that ASU is greater than LSM.) </s>
  <s xml:id="echoid-s1685" xml:space="preserve"><lb/>Therefore the ratio of ASU to ESI is greater than that of AOU to EOI. </s>
  <s xml:id="echoid-s1686" xml:space="preserve"><lb/>And the ſame holds good with regard to any other point S taken between E <lb/>and I, ſo that the ratio of AOU to EOI is a Minimum, and ſingular, or <lb/>what the Antients called μοναχ@.</s>
  <s xml:id="echoid-s1687" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1688" xml:space="preserve">Let now VY join the tops of two perpendiculars drawn on the ſame ſide of <lb/>the diameter, and meet the diameter produced in O; </s>
  <s xml:id="echoid-s1689" xml:space="preserve">I ſay that the ratio of <lb/>AOU to EOI is a Maximum.</s>
  <s xml:id="echoid-s1690" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1691" xml:space="preserve">For uſing the ſame conſtruction as before, it will appear that the rectangle <lb/>LSM: </s>
  <s xml:id="echoid-s1692" xml:space="preserve">ESI:</s>
  <s xml:id="echoid-s1693" xml:space="preserve">: VOY or AOU: </s>
  <s xml:id="echoid-s1694" xml:space="preserve">EOI. </s>
  <s xml:id="echoid-s1695" xml:space="preserve">And it may be proved in the ſame <lb/>manner that VSR or ASU is leſs than LSM. </s>
  <s xml:id="echoid-s1696" xml:space="preserve">(LV i. </s>
  <s xml:id="echoid-s1697" xml:space="preserve">e. </s>
  <s xml:id="echoid-s1698" xml:space="preserve">MY x <emph style="ol">NI + MI</emph> or <lb/>(making IK = MI) MY x NK, together with LSM is by the preceding <emph style="sc">Lem</emph>-<lb/><emph style="sc">MA</emph> equal to VS x SN. </s>
  <s xml:id="echoid-s1699" xml:space="preserve">But VS x NR, together with VSR, is alſo equal to <lb/>VS x SN. </s>
  <s xml:id="echoid-s1700" xml:space="preserve">Now VS x NR is equal to MY x NF or LV x NF from the ſimi-<lb/>larity of the triangles LVS, NRF. </s>
  <s xml:id="echoid-s1701" xml:space="preserve">Therefore now alſo MY x NF together <lb/>with VSR is proved equal to VS x SN. </s>
  <s xml:id="echoid-s1702" xml:space="preserve">But as NK is leſs than NF, VSR <lb/>will be a leſs rectangle than LSM) Hence the ratio of LSM to ESI or it's <lb/>equal AOU to EOI is greater than the ratio of VSR or ASU to ESI. </s>
  <s xml:id="echoid-s1703" xml:space="preserve">And <lb/>the ſame holds with regard to any other point taken in the diameter pro-<lb/>duced. </s>
  <s xml:id="echoid-s1704" xml:space="preserve">Therefore the ratio of AOU to EOI is a Maximum.</s>
  <s xml:id="echoid-s1705" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1706" xml:space="preserve">Q. </s>
  <s xml:id="echoid-s1707" xml:space="preserve">E. </s>
  <s xml:id="echoid-s1708" xml:space="preserve">D.</s>
  <s xml:id="echoid-s1709" xml:space="preserve"/>
</p>
<pb o="[15]" file="0085" n="92"/>
</div>
<div xml:id="echoid-div75" type="section" level="1" n="71">
<head xml:id="echoid-head84" xml:space="preserve">DETERMINATE SECTION.</head>
<head xml:id="echoid-head85" xml:space="preserve">BOOK I.</head>
<head xml:id="echoid-head86" xml:space="preserve">PROBLEM I. (Fig. 1.)</head>
<p>
  <s xml:id="echoid-s1710" xml:space="preserve">In any indefinite ſtraight line, let the Point A be aſſigned; </s>
  <s xml:id="echoid-s1711" xml:space="preserve">it is required <lb/>to cut it in ſome other point O, ſo that the ſquare on the ſegment AO <lb/>may be to the ſquare on a given line, P, in the ratio of two given ſtraight <lb/>lines R and S.</s>
  <s xml:id="echoid-s1712" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1713" xml:space="preserve"><emph style="sc">Analysis</emph>. </s>
  <s xml:id="echoid-s1714" xml:space="preserve">Since, by Hypotheſis, the ſquare on AO muſt be to the <lb/>ſquare on P as R is to S, the ſquare on AO will be to the Square on P as <lb/>the ſquare on R is to the rectangle contained by R and S (<emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s1715" xml:space="preserve">V. </s>
  <s xml:id="echoid-s1716" xml:space="preserve">15.) <lb/></s>
  <s xml:id="echoid-s1717" xml:space="preserve">Let there be taken AD, a mean proportional between AB (R) and AC <lb/>(S); </s>
  <s xml:id="echoid-s1718" xml:space="preserve">then the Square on AO is to the ſquare on P as the ſquare on R is <lb/>to the ſquare on AD, or (<emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s1719" xml:space="preserve">VI. </s>
  <s xml:id="echoid-s1720" xml:space="preserve">22) AO is to P as R to AD; </s>
  <s xml:id="echoid-s1721" xml:space="preserve">conſe-<lb/>quently, AO is given by <emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s1722" xml:space="preserve">VI. </s>
  <s xml:id="echoid-s1723" xml:space="preserve">12.</s>
  <s xml:id="echoid-s1724" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1725" xml:space="preserve"><emph style="sc">Synthesis</emph>. </s>
  <s xml:id="echoid-s1726" xml:space="preserve">Make AB equal to R, AC equal to S, and deſcribe on <lb/>BC a ſemi-circle; </s>
  <s xml:id="echoid-s1727" xml:space="preserve">erect at A the indefinite perpendicular AF, meeting the <lb/>circle in D, and take AF equal to P; </s>
  <s xml:id="echoid-s1728" xml:space="preserve">draw DB, and parallel thereto FO, <lb/>meeting the indefinite line in O, the point required.</s>
  <s xml:id="echoid-s1729" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1730" xml:space="preserve">For, by reaſon of the ſimilar triangles ADB, AFO, AO is to AF (P) as <lb/>AB (R) is to AD; </s>
  <s xml:id="echoid-s1731" xml:space="preserve">therefore (<emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s1732" xml:space="preserve">VI. </s>
  <s xml:id="echoid-s1733" xml:space="preserve">22.) </s>
  <s xml:id="echoid-s1734" xml:space="preserve">the ſquare on AO is to the <lb/>ſquare on P as the ſquare on R is to the ſquare on AD; </s>
  <s xml:id="echoid-s1735" xml:space="preserve">but the ſquare on <lb/>AD is equal to the rectangle contained by AB (R) and AC (S) by <emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s1736" xml:space="preserve">VI. <lb/></s>
  <s xml:id="echoid-s1737" xml:space="preserve">13. </s>
  <s xml:id="echoid-s1738" xml:space="preserve">17; </s>
  <s xml:id="echoid-s1739" xml:space="preserve">and ſo the ſquare on AO is to the ſquare on P as the ſquare on R <lb/>is to the rectangle contained by R and S; </s>
  <s xml:id="echoid-s1740" xml:space="preserve">that is (<emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s1741" xml:space="preserve">V. </s>
  <s xml:id="echoid-s1742" xml:space="preserve">15.) </s>
  <s xml:id="echoid-s1743" xml:space="preserve">as R is to S.</s>
  <s xml:id="echoid-s1744" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1745" xml:space="preserve">Q. </s>
  <s xml:id="echoid-s1746" xml:space="preserve">E. </s>
  <s xml:id="echoid-s1747" xml:space="preserve">D.</s>
  <s xml:id="echoid-s1748" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1749" xml:space="preserve"><emph style="sc">Scholium</emph>. </s>
  <s xml:id="echoid-s1750" xml:space="preserve">Here are no limitations, nor any precautions whatever to be <lb/>obſerved, except that AB (R) muſt be ſet off from A that way which O <lb/>is required to fall.</s>
  <s xml:id="echoid-s1751" xml:space="preserve"/>
</p>
<pb o="[16]" file="0086" n="93"/>
</div>
<div xml:id="echoid-div76" type="section" level="1" n="72">
<head xml:id="echoid-head87" xml:space="preserve">PROBLEM II. (Fig. 2 and 3.)</head>
<p>
  <s xml:id="echoid-s1752" xml:space="preserve">In any indefinite ſtraight line, let there be aſſigned the points A and E; <lb/></s>
  <s xml:id="echoid-s1753" xml:space="preserve">it is required to cut it in another point O, ſo that the rectangle contained <lb/>by the ſegments AO, EO may be to the ſquare on a given line P, in the <lb/>ratio of two given ſtraight lines R and S.</s>
  <s xml:id="echoid-s1754" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1755" xml:space="preserve"><emph style="sc">Analysis</emph>. </s>
  <s xml:id="echoid-s1756" xml:space="preserve">Conceive the thing done, and O the point ſought: </s>
  <s xml:id="echoid-s1757" xml:space="preserve">then <lb/>would the rectangle AO, EO be to the ſquare on the given line P as R is <lb/>to S, by hypotheſis. </s>
  <s xml:id="echoid-s1758" xml:space="preserve">Make AQ to P as R is to S: </s>
  <s xml:id="echoid-s1759" xml:space="preserve">then the rectangle AO, <lb/>EO will be to the ſquare on P as AQ to P; </s>
  <s xml:id="echoid-s1760" xml:space="preserve">or (<emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s1761" xml:space="preserve">V. </s>
  <s xml:id="echoid-s1762" xml:space="preserve">15. </s>
  <s xml:id="echoid-s1763" xml:space="preserve">and 16) the <lb/>rectangle AO, EO will be to the rectangle AQ, EO as the ſquare on P is <lb/>to the rectangle EO, P; </s>
  <s xml:id="echoid-s1764" xml:space="preserve">and therefore AO is to AQ as P is to EO; </s>
  <s xml:id="echoid-s1765" xml:space="preserve">con-<lb/>ſequently (<emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s1766" xml:space="preserve">VI. </s>
  <s xml:id="echoid-s1767" xml:space="preserve">16.) </s>
  <s xml:id="echoid-s1768" xml:space="preserve">the rectangle AO, EO is equal to the rectangle <lb/>AQ, P; </s>
  <s xml:id="echoid-s1769" xml:space="preserve">and hence, as the ſum, or difference of AO and EO is alſo given, <lb/>theſe lines themſelves are given by the 85th or 86th of the Data.</s>
  <s xml:id="echoid-s1770" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1771" xml:space="preserve"><emph style="sc">Synthesis</emph>. </s>
  <s xml:id="echoid-s1772" xml:space="preserve">On AE deſcribe a circle, and erect at A the indefinite <lb/>perpendicular AK; </s>
  <s xml:id="echoid-s1773" xml:space="preserve">and, having taken AQ a fourth proportional to S, R <lb/>and P, take AD a mean proportional between AQ and P; </s>
  <s xml:id="echoid-s1774" xml:space="preserve">from D draw <lb/>DH, parallel to AE if O be required to fall between A and E; </s>
  <s xml:id="echoid-s1775" xml:space="preserve">but through <lb/>F, the center of the Circle on AE, if it be required beyond A or E, cutting <lb/>the circle in H; </s>
  <s xml:id="echoid-s1776" xml:space="preserve">laſtly, draw HO perpendicular to DH, meeting the inde-<lb/>finite line in O, the point required.</s>
  <s xml:id="echoid-s1777" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1778" xml:space="preserve">For it is manifeſt from the conſtruction that AD and HO are equal; <lb/></s>
  <s xml:id="echoid-s1779" xml:space="preserve">hence, and <emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s1780" xml:space="preserve">VI. </s>
  <s xml:id="echoid-s1781" xml:space="preserve">17, the rectangle AQ, P is equal to the ſquare on HO; </s>
  <s xml:id="echoid-s1782" xml:space="preserve"><lb/>conſequently equal to the rectangle AO, EO (<emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s1783" xml:space="preserve">III. </s>
  <s xml:id="echoid-s1784" xml:space="preserve">35. </s>
  <s xml:id="echoid-s1785" xml:space="preserve">36) and ſo <lb/>(<emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s1786" xml:space="preserve">VI. </s>
  <s xml:id="echoid-s1787" xml:space="preserve">16.) </s>
  <s xml:id="echoid-s1788" xml:space="preserve">AO is to P as AQ is to EO; </s>
  <s xml:id="echoid-s1789" xml:space="preserve">but by conſtruction AQ is to <lb/>P as R to S, therefore by compound ratio, the rectangle AO, AQ is to the <lb/>ſquare on P as the rectangle AQ, R is to the rectangle EO, S: </s>
  <s xml:id="echoid-s1790" xml:space="preserve">hence <lb/>(<emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s1791" xml:space="preserve">V. </s>
  <s xml:id="echoid-s1792" xml:space="preserve">15. </s>
  <s xml:id="echoid-s1793" xml:space="preserve">16.) </s>
  <s xml:id="echoid-s1794" xml:space="preserve">the rectangle AO, EO is to the ſquare on P as the rect-<lb/>angle EO, R is to the rectangle EO, S, that is, as R is to S. </s>
  <s xml:id="echoid-s1795" xml:space="preserve">Q. </s>
  <s xml:id="echoid-s1796" xml:space="preserve">E. </s>
  <s xml:id="echoid-s1797" xml:space="preserve">D.</s>
  <s xml:id="echoid-s1798" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1799" xml:space="preserve"><emph style="sc">Scholium</emph>. </s>
  <s xml:id="echoid-s1800" xml:space="preserve">This Problem has three Epitagmas. </s>
  <s xml:id="echoid-s1801" xml:space="preserve">Firſt, when O is ſought <lb/>beyond A; </s>
  <s xml:id="echoid-s1802" xml:space="preserve">ſecondly, when it is ſought between A and E, and laſtly, when <lb/>it is ſought beyond E. </s>
  <s xml:id="echoid-s1803" xml:space="preserve">The firſt and laſt of theſe are conſtructed by <lb/>Fig. </s>
  <s xml:id="echoid-s1804" xml:space="preserve">2, and have no limitations; </s>
  <s xml:id="echoid-s1805" xml:space="preserve">but in the ſecond, (Fig. </s>
  <s xml:id="echoid-s1806" xml:space="preserve">3.) </s>
  <s xml:id="echoid-s1807" xml:space="preserve">the given <lb/>ratio of R to S muſt not be greater than that which the ſquare on half <lb/>AE bears to the ſquare on P: </s>
  <s xml:id="echoid-s1808" xml:space="preserve">ſince if it be, a third proportional to S
<pb o="[17]" file="0087" n="94"/>
and P will be greater than one to R and half AE, and of courſe, AQ <lb/>(a fourth proportional to S, R and P) greater than a third proportional <lb/>to P and half AE; </s>
  <s xml:id="echoid-s1809" xml:space="preserve">in which caſe the rectangle AQ, P will be greater <lb/>than the ſquare on half AE, and ſo AD (a mean proportional between <lb/>AQ and P) greater than half AE; </s>
  <s xml:id="echoid-s1810" xml:space="preserve">but when this happens, it is plain that <lb/>DH can neither cut nor touch the circle on AE, and therefore, the <lb/>problem becomes impoſſible.</s>
  <s xml:id="echoid-s1811" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div77" type="section" level="1" n="73">
<head xml:id="echoid-head88" xml:space="preserve">PROBLEM III. (Fig. 4. and 5.)</head>
<p>
  <s xml:id="echoid-s1812" xml:space="preserve">In any indefinite ſtraight line let there be aſſigned the points A and E; <lb/></s>
  <s xml:id="echoid-s1813" xml:space="preserve">it is required to cut it in another point O, ſo that the ſquare on the ſegment <lb/>AO may be to the rectangle contained by the ſegment EO and a given line <lb/>P, in the ratio of two given ſtraight lines R and S.</s>
  <s xml:id="echoid-s1814" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1815" xml:space="preserve"><emph style="sc">Analysis</emph>. </s>
  <s xml:id="echoid-s1816" xml:space="preserve">Suppoſe the thing done, and that O is the point ſought: <lb/></s>
  <s xml:id="echoid-s1817" xml:space="preserve">then will the ſquare on AO be to the rectangle EO, P as R to S. </s>
  <s xml:id="echoid-s1818" xml:space="preserve">Make <lb/>AQ to P as R is to S; </s>
  <s xml:id="echoid-s1819" xml:space="preserve">then will the ſquare on AO be to the rectangle EO, <lb/>P as AQ is to P; </s>
  <s xml:id="echoid-s1820" xml:space="preserve">or (<emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s1821" xml:space="preserve">V. </s>
  <s xml:id="echoid-s1822" xml:space="preserve">15.) </s>
  <s xml:id="echoid-s1823" xml:space="preserve">the ſquare on AO is to the rectangle EO, <lb/>P as the rectangle AQ, AO is to the rectangle P, AO; </s>
  <s xml:id="echoid-s1824" xml:space="preserve">wherefore AO is to <lb/>EO as AQ to AO; </s>
  <s xml:id="echoid-s1825" xml:space="preserve">conſequently by compoſition, or diviſion, AO is to AE <lb/>as AQ is to OQ, and ſo (<emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s1826" xml:space="preserve">VI. </s>
  <s xml:id="echoid-s1827" xml:space="preserve">16.) </s>
  <s xml:id="echoid-s1828" xml:space="preserve">the rectangle AO, OQ is equal to <lb/>the rectangle AE, AQ; </s>
  <s xml:id="echoid-s1829" xml:space="preserve">and hence, as the ſum or difference of AO and OQ <lb/>is alſo given, theſe lines themſelves are given by the 85th or 86th of the <lb/>Data.</s>
  <s xml:id="echoid-s1830" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1831" xml:space="preserve"><emph style="sc">Synthesis</emph>. </s>
  <s xml:id="echoid-s1832" xml:space="preserve">Take AQ a fourth proportional to S, P and R, and <lb/>deſcribe thereon a circle; </s>
  <s xml:id="echoid-s1833" xml:space="preserve">erect at A, the indefinite perpendicular AK, and <lb/>take therein AD, a mean proportional between AE and AQ; </s>
  <s xml:id="echoid-s1834" xml:space="preserve">from D, <lb/>draw DH, parallel to AE, if O be required beyond E; </s>
  <s xml:id="echoid-s1835" xml:space="preserve">but through F the <lb/>center of the circle on AQ, if it be ſought beyond A, or between A and <lb/>E, cutting the ſaid circle in H: </s>
  <s xml:id="echoid-s1836" xml:space="preserve">Laſtly, from H draw HO perpendicular <lb/>to DH, which will cut the indefinite line in O, the point required.</s>
  <s xml:id="echoid-s1837" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1838" xml:space="preserve">For it is plain from the Conſtruction, that AD and HO are equal; </s>
  <s xml:id="echoid-s1839" xml:space="preserve">and <lb/>(<emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s1840" xml:space="preserve">VI. </s>
  <s xml:id="echoid-s1841" xml:space="preserve">17) the rectangle AE, AQ is equal to the ſquare on AD, and <lb/>therefore equal to the ſquare on HO; </s>
  <s xml:id="echoid-s1842" xml:space="preserve">but the ſquare on HO is equal to the <lb/>rectangle AO, OQ, (<emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s1843" xml:space="preserve">III. </s>
  <s xml:id="echoid-s1844" xml:space="preserve">35. </s>
  <s xml:id="echoid-s1845" xml:space="preserve">36) conſequently the rectangle AO, OQ
<pb o="[18]" file="0088" n="95"/>
is equal to the rectangle AE, AQ; </s>
  <s xml:id="echoid-s1846" xml:space="preserve">and hence (<emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s1847" xml:space="preserve">VI. </s>
  <s xml:id="echoid-s1848" xml:space="preserve">16.) </s>
  <s xml:id="echoid-s1849" xml:space="preserve">OQ is to AQ <lb/>as AE is to AO; </s>
  <s xml:id="echoid-s1850" xml:space="preserve">therefore, by compoſition or diviſion, AO is to AQ as <lb/>EO is to AO; </s>
  <s xml:id="echoid-s1851" xml:space="preserve">but by conſtruction, AQ is to R as P is to S, and ſo, by <lb/>compound ratio, the rectangle AO, AQ is to the rectangle R, AQ as the <lb/>rectangle EO, P is to the rectangle AO, S; </s>
  <s xml:id="echoid-s1852" xml:space="preserve">or (<emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s1853" xml:space="preserve">V. </s>
  <s xml:id="echoid-s1854" xml:space="preserve">15. </s>
  <s xml:id="echoid-s1855" xml:space="preserve">and 16.) </s>
  <s xml:id="echoid-s1856" xml:space="preserve">the ſquare <lb/>on AO is to the rectangle EO, P as the rectangle AO, R is to the rectangle <lb/>AO, S; </s>
  <s xml:id="echoid-s1857" xml:space="preserve">that is, as R to S.</s>
  <s xml:id="echoid-s1858" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1859" xml:space="preserve"><emph style="sc">Scholium</emph>. </s>
  <s xml:id="echoid-s1860" xml:space="preserve">This Problem hath three Epitagmas alſo, which I ſtill enu-<lb/>merate as before. </s>
  <s xml:id="echoid-s1861" xml:space="preserve">The firſt and ſecond are conſtructed by Fig. </s>
  <s xml:id="echoid-s1862" xml:space="preserve">4, where DH <lb/>is drawn through F, the center of the circle on AQ: </s>
  <s xml:id="echoid-s1863" xml:space="preserve">and theſe have no limi-<lb/>tations. </s>
  <s xml:id="echoid-s1864" xml:space="preserve">The third is conſtructed as in Fig. </s>
  <s xml:id="echoid-s1865" xml:space="preserve">5, where DH is drawn parallel <lb/>to AQ; </s>
  <s xml:id="echoid-s1866" xml:space="preserve">and here the given ratio of R to S muſt not be leſs than the ratio <lb/>which four times AE bears to P: </s>
  <s xml:id="echoid-s1867" xml:space="preserve">for if it be, AE will be greater than <lb/>one-fourth Part of AQ (a fourth proportional to S, R and P) in which <lb/>caſe the rectangle contained by AE and AQ will be greater than the ſquare <lb/>on half AQ, and conſequently AD (a mean proportional between AE and <lb/>AQ) greater than half AQ; </s>
  <s xml:id="echoid-s1868" xml:space="preserve">but it is plain when this is the caſe, that DH <lb/>will neither cut nor touch the circle on AQ, and therefore the problem is <lb/>impoſſible.</s>
  <s xml:id="echoid-s1869" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div78" type="section" level="1" n="74">
<head xml:id="echoid-head89" xml:space="preserve">PROBLEM IV. (Fig. 6. 7. and 8.)</head>
<p>
  <s xml:id="echoid-s1870" xml:space="preserve">In any indefinite ſtraight line, let there be aſſigned the points A and E; <lb/></s>
  <s xml:id="echoid-s1871" xml:space="preserve">it is required to cut it in another point O, ſo that the two ſquares on the <lb/>ſegments AO, EO, may obtain the Ratio of two given ſtraight lines, <lb/>R and S.</s>
  <s xml:id="echoid-s1872" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1873" xml:space="preserve"><emph style="sc">Analysis</emph>. </s>
  <s xml:id="echoid-s1874" xml:space="preserve">Imagine the thing to be effected, and that O is really the <lb/>point required: </s>
  <s xml:id="echoid-s1875" xml:space="preserve">then will the ſquare on AO be to the ſquare on EO as <lb/>R to S; </s>
  <s xml:id="echoid-s1876" xml:space="preserve">or (<emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s1877" xml:space="preserve">V. </s>
  <s xml:id="echoid-s1878" xml:space="preserve">15.) </s>
  <s xml:id="echoid-s1879" xml:space="preserve">the ſquare on AO will be to the ſquare on EO <lb/>as the ſquare on R is to the rectangle contained by R and S. </s>
  <s xml:id="echoid-s1880" xml:space="preserve">Let DE be <lb/>made a mean proportional between EB (R) and EC (S). </s>
  <s xml:id="echoid-s1881" xml:space="preserve">Then <lb/>(<emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s1882" xml:space="preserve">VI. </s>
  <s xml:id="echoid-s1883" xml:space="preserve">17.) </s>
  <s xml:id="echoid-s1884" xml:space="preserve">the ſquare on AO will be to the ſquare on EO as the <lb/>ſquare on R to the ſquare on DE; </s>
  <s xml:id="echoid-s1885" xml:space="preserve">and ſo (<emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s1886" xml:space="preserve">VI. </s>
  <s xml:id="echoid-s1887" xml:space="preserve">22.) </s>
  <s xml:id="echoid-s1888" xml:space="preserve">AO to EO as R to <lb/>DE; </s>
  <s xml:id="echoid-s1889" xml:space="preserve">and hence both AO and EO will be given by the converſe of Prop. </s>
  <s xml:id="echoid-s1890" xml:space="preserve">38. <lb/></s>
  <s xml:id="echoid-s1891" xml:space="preserve">of Eu. </s>
  <s xml:id="echoid-s1892" xml:space="preserve">Data.</s>
  <s xml:id="echoid-s1893" xml:space="preserve"/>
</p>
<pb o="[19]" file="0089" n="96"/>
<p>
  <s xml:id="echoid-s1894" xml:space="preserve"><emph style="sc">Synthesis</emph>. </s>
  <s xml:id="echoid-s1895" xml:space="preserve">Make EB equal to R, EC equal to S, and deſcribe on <lb/>BC a circle; </s>
  <s xml:id="echoid-s1896" xml:space="preserve">erect at E the perpendicular ED, meeting the periphery of the <lb/>circle in D; </s>
  <s xml:id="echoid-s1897" xml:space="preserve">alſo at A erect the perpendicular AF equal to R; </s>
  <s xml:id="echoid-s1898" xml:space="preserve">draw AD, <lb/>which produce, if neceſſary, to cut the indefinite line, as in O, which will <lb/>be the point required.</s>
  <s xml:id="echoid-s1899" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1900" xml:space="preserve">For becauſe of the ſimilar triangles AOF, EOD, AO is to EO as AF <lb/>(R) is to DE; </s>
  <s xml:id="echoid-s1901" xml:space="preserve">therefore the ſquare on AO is to the ſquare on EO as the <lb/>ſquare on R is to the ſquare on DE (<emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s1902" xml:space="preserve">VI. </s>
  <s xml:id="echoid-s1903" xml:space="preserve">22); </s>
  <s xml:id="echoid-s1904" xml:space="preserve">but the ſquare on DE <lb/>is equal to the rectangle contained by R and S; </s>
  <s xml:id="echoid-s1905" xml:space="preserve">therefore the ſquare on AO is <lb/>to the ſquare on EO as the ſquare on R is to the rectangle R, S; </s>
  <s xml:id="echoid-s1906" xml:space="preserve">that is as <lb/>R is to S, by <emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s1907" xml:space="preserve">V. </s>
  <s xml:id="echoid-s1908" xml:space="preserve">15.</s>
  <s xml:id="echoid-s1909" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1910" xml:space="preserve">Q. </s>
  <s xml:id="echoid-s1911" xml:space="preserve">E. </s>
  <s xml:id="echoid-s1912" xml:space="preserve">D.</s>
  <s xml:id="echoid-s1913" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1914" xml:space="preserve"><emph style="sc">Scholium</emph>. </s>
  <s xml:id="echoid-s1915" xml:space="preserve">This Problem alſo hath three Epitagmas, which I enumerate <lb/>as in the laſt. </s>
  <s xml:id="echoid-s1916" xml:space="preserve">The firſt is conſtructed by Fig. </s>
  <s xml:id="echoid-s1917" xml:space="preserve">6, wherein the perpendicu-<lb/>lars DE and AF are ſet off on the ſame ſide of the indefinite line; </s>
  <s xml:id="echoid-s1918" xml:space="preserve">the ſecond <lb/>by Fig. </s>
  <s xml:id="echoid-s1919" xml:space="preserve">7, where they are ſet off on contrary ſides, and the third by Fig. </s>
  <s xml:id="echoid-s1920" xml:space="preserve">8, <lb/>in which they are again ſet off on the ſame ſide. </s>
  <s xml:id="echoid-s1921" xml:space="preserve">The ſecond has no limits; <lb/></s>
  <s xml:id="echoid-s1922" xml:space="preserve">but in the firſt R muſt be leſs, and in the third greater than S, for reaſons <lb/>too obvious to be inſiſted on; </s>
  <s xml:id="echoid-s1923" xml:space="preserve">and hence, both theſe caſes are impoſſible <lb/>when the given ratio is that of equality.</s>
  <s xml:id="echoid-s1924" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div79" type="section" level="1" n="75">
<head xml:id="echoid-head90" xml:space="preserve">PROBLEM V. (Fig. 9. 10. 11. 12. 13. 14. 15. 16.)</head>
<p>
  <s xml:id="echoid-s1925" xml:space="preserve">In any indefinite ſtraight line let there be aſſigned the points A, E and I; <lb/></s>
  <s xml:id="echoid-s1926" xml:space="preserve">it is required to cut it in another point, O, ſo that the rectangle contained <lb/>by the ſegment AO and a given ſtraight line P may be to the rectangle <lb/>contained by the ſegments EO, IO in the ratio of two given ſtraight lines <lb/>R and S.</s>
  <s xml:id="echoid-s1927" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1928" xml:space="preserve"><emph style="sc">Analysis</emph>. </s>
  <s xml:id="echoid-s1929" xml:space="preserve">Conceive the thing done, and O the point ſought: </s>
  <s xml:id="echoid-s1930" xml:space="preserve">then <lb/>would the rectangle AO, P be to the rectangle EO, IO as R to S. </s>
  <s xml:id="echoid-s1931" xml:space="preserve">Make <lb/>IQ to P as S is to R; </s>
  <s xml:id="echoid-s1932" xml:space="preserve">then will the rectangle AO, P be to the rectangle <lb/>EO, IO as P is to IQ; </s>
  <s xml:id="echoid-s1933" xml:space="preserve">or (<emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s1934" xml:space="preserve">V. </s>
  <s xml:id="echoid-s1935" xml:space="preserve">15.) </s>
  <s xml:id="echoid-s1936" xml:space="preserve">the rectangle AO, P be to the <lb/>rectangle EO, IO as the rectangle IO, P is to the rectangle IO, IQ; </s>
  <s xml:id="echoid-s1937" xml:space="preserve">and <lb/>hence (<emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s1938" xml:space="preserve">V. </s>
  <s xml:id="echoid-s1939" xml:space="preserve">15. </s>
  <s xml:id="echoid-s1940" xml:space="preserve">16) AO is to EO as IO is to IQ; </s>
  <s xml:id="echoid-s1941" xml:space="preserve">whence, by compoſition <lb/>or diviſion, AE is to EO as OQ is to IQ: </s>
  <s xml:id="echoid-s1942" xml:space="preserve">therefore (<emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s1943" xml:space="preserve">VI. </s>
  <s xml:id="echoid-s1944" xml:space="preserve">16.) </s>
  <s xml:id="echoid-s1945" xml:space="preserve">the
<pb o="[20]" file="0090" n="97"/>
rectangle EO, OQ is equal to the rectangle AE, IQ; </s>
  <s xml:id="echoid-s1946" xml:space="preserve">conſequently, as the <lb/>ſum or difference of EO and OQ is alſo given, thoſe lines themſelves are <lb/>given by the 85th or 86th of the Data.</s>
  <s xml:id="echoid-s1947" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1948" xml:space="preserve"><emph style="sc">Synthesis</emph>. </s>
  <s xml:id="echoid-s1949" xml:space="preserve">Take IQ a fourth proportional to R, S and P, and <lb/>deſcribe on EQ a circle; </s>
  <s xml:id="echoid-s1950" xml:space="preserve">erect at E the indefinite perpendicular EK, and <lb/>take therein ED a mean proportional between AE and IQ; </s>
  <s xml:id="echoid-s1951" xml:space="preserve">from D draw <lb/>DH, parallel to EQ, if O muſt lie any where between the points E and Q; <lb/></s>
  <s xml:id="echoid-s1952" xml:space="preserve">but through F, the center of the circle on EQ if it muſt fall without them, <lb/>cutting the ſaid circle in H: </s>
  <s xml:id="echoid-s1953" xml:space="preserve">Laſtly, draw HO perpendicular to DH, which <lb/>will meet the indeſinite line in O, the point required.</s>
  <s xml:id="echoid-s1954" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1955" xml:space="preserve">For it is manifeſt from the conſtruction that ED and HO are equal; </s>
  <s xml:id="echoid-s1956" xml:space="preserve">and <lb/>(<emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s1957" xml:space="preserve">VI. </s>
  <s xml:id="echoid-s1958" xml:space="preserve">17.) </s>
  <s xml:id="echoid-s1959" xml:space="preserve">the rectangle AE, IQ is equal to the ſquare on ED, and <lb/>therefore equal to the ſquare on HO; </s>
  <s xml:id="echoid-s1960" xml:space="preserve">but the ſquare on HO is equal to the <lb/>rectangle EO, OQ (<emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s1961" xml:space="preserve">III. </s>
  <s xml:id="echoid-s1962" xml:space="preserve">35. </s>
  <s xml:id="echoid-s1963" xml:space="preserve">36.)</s>
  <s xml:id="echoid-s1964" xml:space="preserve">: </s>
  <s xml:id="echoid-s1965" xml:space="preserve">therefore the rectangle AE, IQ is <lb/>equal to the rectangle EO, OQ; </s>
  <s xml:id="echoid-s1966" xml:space="preserve">and hence (<emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s1967" xml:space="preserve">VI. </s>
  <s xml:id="echoid-s1968" xml:space="preserve">16.) </s>
  <s xml:id="echoid-s1969" xml:space="preserve">AE is to OE as OQ <lb/>to IQ, whence, by compoſition or diviſion, AO is to EO as OI to IQ; </s>
  <s xml:id="echoid-s1970" xml:space="preserve">but <lb/>IQ is to P as S to R, or inverſely, P is to IQ as R to S; </s>
  <s xml:id="echoid-s1971" xml:space="preserve">and ſo, by compound <lb/>ratio, the rectangle AO, P is to the rectangle EO, IQ as the rectangle IO, <lb/>R is to the rectangle IQ, S; </s>
  <s xml:id="echoid-s1972" xml:space="preserve">that is (<emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s1973" xml:space="preserve">V. </s>
  <s xml:id="echoid-s1974" xml:space="preserve">15 and 16.) </s>
  <s xml:id="echoid-s1975" xml:space="preserve">the rectangle AO, <lb/>P is to the rectangle IO, R as EO is to S; </s>
  <s xml:id="echoid-s1976" xml:space="preserve">or the rectangle AO, P is to the <lb/>rectangle IO, R as the rectangle IO, EO is to the rectangle IO, S <lb/>(<emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s1977" xml:space="preserve">V. </s>
  <s xml:id="echoid-s1978" xml:space="preserve">16.) </s>
  <s xml:id="echoid-s1979" xml:space="preserve">the rectangle AO, P is to the rectangle EO, IO as the rectangle <lb/>IO, R is to the rectangle IO, S; </s>
  <s xml:id="echoid-s1980" xml:space="preserve">that is (<emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s1981" xml:space="preserve">V. </s>
  <s xml:id="echoid-s1982" xml:space="preserve">15.) </s>
  <s xml:id="echoid-s1983" xml:space="preserve">as R is to S. </s>
  <s xml:id="echoid-s1984" xml:space="preserve">Q. </s>
  <s xml:id="echoid-s1985" xml:space="preserve">E. </s>
  <s xml:id="echoid-s1986" xml:space="preserve">D.</s>
  <s xml:id="echoid-s1987" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1988" xml:space="preserve"><emph style="sc">Scholium</emph>. </s>
  <s xml:id="echoid-s1989" xml:space="preserve">This Problem may be conſidered as having three Epitagmas, <lb/>or general Caſes, viz. </s>
  <s xml:id="echoid-s1990" xml:space="preserve">when A, the point which bounds the ſegment aſſigned <lb/>for the co efficient of the given line P being an extreme, O is ſought be-<lb/>tween it and the next thereto, or beyond all the points with reſpect to A; <lb/></s>
  <s xml:id="echoid-s1991" xml:space="preserve">ſecondly, where A is the middle point; </s>
  <s xml:id="echoid-s1992" xml:space="preserve">and thirdly, when A being again <lb/>an extreme, O is ſought beyond it, or between the other two points E and <lb/>I: </s>
  <s xml:id="echoid-s1993" xml:space="preserve">and each of theſe is ſubdiviſible into four more particular ones.</s>
  <s xml:id="echoid-s1994" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s1995" xml:space="preserve"><emph style="sc">Epitagma</emph> I. </s>
  <s xml:id="echoid-s1996" xml:space="preserve">Here the four Caſes are when E being the middle point, <lb/>O is required between A and E, or beyond I; </s>
  <s xml:id="echoid-s1997" xml:space="preserve">and theſe are both con-<lb/>ſtructed at once by Fig. </s>
  <s xml:id="echoid-s1998" xml:space="preserve">9: </s>
  <s xml:id="echoid-s1999" xml:space="preserve">when I is the middle point and O ſought between <lb/>A and I or beyond E; </s>
  <s xml:id="echoid-s2000" xml:space="preserve">and theſe are both conſtructed at once by Fig. </s>
  <s xml:id="echoid-s2001" xml:space="preserve">10. <lb/></s>
  <s xml:id="echoid-s2002" xml:space="preserve">and in both of theſe IQ is ſet off from I contrary to that direction which A
<pb o="[21]" file="0091" n="98"/>
bears therefrom, and DH drawn through F, the center of the circle on EQ: <lb/></s>
  <s xml:id="echoid-s2003" xml:space="preserve">none of theſe Caſes are ſubject to any Limitations.</s>
  <s xml:id="echoid-s2004" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2005" xml:space="preserve"><emph style="sc">Epitagma</emph> II. </s>
  <s xml:id="echoid-s2006" xml:space="preserve">Wherein A is the middle point, and the Caſes, when O <lb/>is ſought beyond E, between E and A, between A and I or beyond I. </s>
  <s xml:id="echoid-s2007" xml:space="preserve">The <lb/>firſt and third of which are conſtructed at once by Fig. </s>
  <s xml:id="echoid-s2008" xml:space="preserve">11, wherein IQ is <lb/>ſet off from I towards A and DH drawn through F, the center of the circle <lb/>on EQ. </s>
  <s xml:id="echoid-s2009" xml:space="preserve">The ſecond and fourth are conſtructed at once, alſo, by Fig. </s>
  <s xml:id="echoid-s2010" xml:space="preserve">12. <lb/></s>
  <s xml:id="echoid-s2011" xml:space="preserve">where IQ is ſet off from I the contrary way to that which A lies, and DH <lb/>drawn parallel to EQ. </s>
  <s xml:id="echoid-s2012" xml:space="preserve">There are no Limitations to any of theſe Caſes.</s>
  <s xml:id="echoid-s2013" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2014" xml:space="preserve"><emph style="sc">Epitagma</emph> III. </s>
  <s xml:id="echoid-s2015" xml:space="preserve">Here, E being the middle point, the Caſes are, when O <lb/>muſt lie beyond A, or between E and I; </s>
  <s xml:id="echoid-s2016" xml:space="preserve">and the ſame Caſes occur when <lb/>I is made the middle point. </s>
  <s xml:id="echoid-s2017" xml:space="preserve">The firſt is conſtructed by Fig. </s>
  <s xml:id="echoid-s2018" xml:space="preserve">13, the ſecond <lb/>by Fig. </s>
  <s xml:id="echoid-s2019" xml:space="preserve">14, the third by Fig. </s>
  <s xml:id="echoid-s2020" xml:space="preserve">15, and the fourth by Fig. </s>
  <s xml:id="echoid-s2021" xml:space="preserve">16: </s>
  <s xml:id="echoid-s2022" xml:space="preserve">in every one <lb/>of which IQ is ſet off from I towards A, and DH drawn parallel to EQ. <lb/></s>
  <s xml:id="echoid-s2023" xml:space="preserve">The Limits are that the given ratio of R to S, muſt not be leſs than the ratio <lb/>which the rectangle AE, P bears to the ſquare on half the Sum, or half the <lb/>difference of AE, and a fourth propor tional to R, S and P; </s>
  <s xml:id="echoid-s2024" xml:space="preserve">that is, to the <lb/>ſquare on half EQ: </s>
  <s xml:id="echoid-s2025" xml:space="preserve">ſince if it ſhould, the rectangle contained by AE and <lb/>the ſaid fourth proportional will be greater than the ſquare on half EQ; </s>
  <s xml:id="echoid-s2026" xml:space="preserve"><lb/>and of courſe ED (a mean proportional between them) greater than half <lb/>EQ, in which Caſe DH can neither cut nor touch the circle on EQ, and <lb/>ſo the problem be impoſſible. </s>
  <s xml:id="echoid-s2027" xml:space="preserve">It is farther obſervable in the two laſt caſes, <lb/>that to have the former of them poſſible, AE muſt be leſs, and to have <lb/>the latter poſſible, EI muſt be greater than the above-mentioned half <lb/>ſum; </s>
  <s xml:id="echoid-s2028" xml:space="preserve">for if this latter part of the Limitation be not obſerved, theſe caſes <lb/>are changed into one another.</s>
  <s xml:id="echoid-s2029" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div80" type="section" level="1" n="76">
<head xml:id="echoid-head91" xml:space="preserve">PROBLEM VI. <lb/>(Fig. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26.)</head>
<p>
  <s xml:id="echoid-s2030" xml:space="preserve">In any indefinite ſtraight line let there be aſſigned the points A, E and I; <lb/></s>
  <s xml:id="echoid-s2031" xml:space="preserve">it is required to cut it in another point O, ſo that the rectangle contained <lb/>by the ſegments AO, EO may be to the ſquare on IO in the ratio of two <lb/>given ſtraight lines, R and S.</s>
  <s xml:id="echoid-s2032" xml:space="preserve"/>
</p>
<pb o="[22]" file="0092" n="99"/>
<p>
  <s xml:id="echoid-s2033" xml:space="preserve"><emph style="sc">Analysis</emph>. </s>
  <s xml:id="echoid-s2034" xml:space="preserve">Let us conceive the thing effected, and that O is really the <lb/>point ſought. </s>
  <s xml:id="echoid-s2035" xml:space="preserve">Then, by ſuppoſition, the rectangle AO, EO is to the <lb/>ſquare on IO as R to S. </s>
  <s xml:id="echoid-s2036" xml:space="preserve">Make EC to IC as R is to S; </s>
  <s xml:id="echoid-s2037" xml:space="preserve">and the rectangle <lb/>AO, EO is to the ſquare on IO as EC to IC. </s>
  <s xml:id="echoid-s2038" xml:space="preserve">Let now OB be taken a <lb/>fourth proportional to EO, EC and IO; </s>
  <s xml:id="echoid-s2039" xml:space="preserve">then (<emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s2040" xml:space="preserve">V. </s>
  <s xml:id="echoid-s2041" xml:space="preserve">15.) </s>
  <s xml:id="echoid-s2042" xml:space="preserve">the rectangle <lb/>AO, EO is to the ſquare on IO as the rectangle EC, OB is to the rectangle <lb/>IC, OB; </s>
  <s xml:id="echoid-s2043" xml:space="preserve">and ſo by permutation, the rectangle AO, EO is to the rectangle <lb/>EC, OB as the ſquare on IO is to the rectangle IC, OB; </s>
  <s xml:id="echoid-s2044" xml:space="preserve">and becauſe EO is <lb/>to EC as IO to BO, AO will be to OB as IO to IC, and ſo by compoſition, <lb/>or diviſion CO is to EC as IB to OB, and AB is to OB as CO to IC; <lb/></s>
  <s xml:id="echoid-s2045" xml:space="preserve">whence, ex æquo perturb. </s>
  <s xml:id="echoid-s2046" xml:space="preserve">et permut. </s>
  <s xml:id="echoid-s2047" xml:space="preserve">AB is to IB as EC to IC; </s>
  <s xml:id="echoid-s2048" xml:space="preserve">that is in the <lb/>given ratio, and hence is given BC, the ſum or difference of CO and BO, <lb/>as alſo the rectangle contained by them, equal to the rectangle AB, IC, <lb/>wherefore theſe lines themſelves are given by the 85th or 86th of the Data.</s>
  <s xml:id="echoid-s2049" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2050" xml:space="preserve"><emph style="sc">Synthesis</emph>. </s>
  <s xml:id="echoid-s2051" xml:space="preserve">Make AB to IB and EC to IC in the given ratio, and <lb/>deſcribe on BC a circle; </s>
  <s xml:id="echoid-s2052" xml:space="preserve">erect, at B, the indefinite perpendicular BK, and <lb/>take therein BD a mean proportional between AB and IC, or between IB <lb/>and EC: </s>
  <s xml:id="echoid-s2053" xml:space="preserve">from D draw DH parallel to CB, if O muſt fall between B and C; <lb/></s>
  <s xml:id="echoid-s2054" xml:space="preserve">but through F, the center of the circle on BC, if it muſt fall without them, <lb/>cutting the ſ@id circle in H; </s>
  <s xml:id="echoid-s2055" xml:space="preserve">then draw HO perpendicular to DH, which <lb/>will cut the indefinite line in O, the point required.</s>
  <s xml:id="echoid-s2056" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2057" xml:space="preserve">For it is plain from the conſtruction that BD and HO are equal, and <lb/>(<emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s2058" xml:space="preserve">IV. </s>
  <s xml:id="echoid-s2059" xml:space="preserve">17.) </s>
  <s xml:id="echoid-s2060" xml:space="preserve">the rectangle AB, IC, or the rectangle IB, EC is equal to the <lb/>ſquare on BD, and therefore equal to the ſquare on HO, which (<emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s2061" xml:space="preserve">III. <lb/></s>
  <s xml:id="echoid-s2062" xml:space="preserve">35. </s>
  <s xml:id="echoid-s2063" xml:space="preserve">36.) </s>
  <s xml:id="echoid-s2064" xml:space="preserve">is equal to the rectangle BO, CO: </s>
  <s xml:id="echoid-s2065" xml:space="preserve">conſequently (<emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s2066" xml:space="preserve">VI. </s>
  <s xml:id="echoid-s2067" xml:space="preserve">16.) </s>
  <s xml:id="echoid-s2068" xml:space="preserve">AB <lb/>is to BO as CO to IC; </s>
  <s xml:id="echoid-s2069" xml:space="preserve">alſo EC is to CO as BO is to IB; </s>
  <s xml:id="echoid-s2070" xml:space="preserve">wherefore, by <lb/>compoſition or diviſion, AO is to BO as IO to IC, and EO to EC as IO <lb/>to BO: </s>
  <s xml:id="echoid-s2071" xml:space="preserve">conſequently by compound ratio, the rectangle contained by AO and <lb/>EO is to the rectangle contained by BO and EC, as the ſquare on IO is to <lb/>the rectangle contained by BO and IC; </s>
  <s xml:id="echoid-s2072" xml:space="preserve">by permutation, the rectangle <lb/>contained by AO and EO is to the ſquare on IO as the rectangle contained <lb/>by BO and EC is to the rectangle contained by BO and IC, that is (Euc. </s>
  <s xml:id="echoid-s2073" xml:space="preserve"><lb/>v. </s>
  <s xml:id="echoid-s2074" xml:space="preserve">15.) </s>
  <s xml:id="echoid-s2075" xml:space="preserve">as EC is to IC, or as R to S.</s>
  <s xml:id="echoid-s2076" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2077" xml:space="preserve">Q. </s>
  <s xml:id="echoid-s2078" xml:space="preserve">E. </s>
  <s xml:id="echoid-s2079" xml:space="preserve">D.</s>
  <s xml:id="echoid-s2080" xml:space="preserve"/>
</p>
<pb o="[23]" file="0093" n="100"/>
<p>
  <s xml:id="echoid-s2081" xml:space="preserve"><emph style="sc">Scholium</emph>. </s>
  <s xml:id="echoid-s2082" xml:space="preserve">This Problem alſo hath three Epitagmas; </s>
  <s xml:id="echoid-s2083" xml:space="preserve">ſirſt, when O is <lb/>ſought between I, the point which bounds the ſegment whoſe ſquare is <lb/>concerned, and either of the other given ones; </s>
  <s xml:id="echoid-s2084" xml:space="preserve">ſecondly, the ſaid point <lb/>being an extreme one, when O is ſought beyond it, or beyond both the <lb/>other given points with reſpect to it; </s>
  <s xml:id="echoid-s2085" xml:space="preserve">and thirdly, when O is required be-<lb/>yond the next in order to the abovementioned point I: </s>
  <s xml:id="echoid-s2086" xml:space="preserve">theſe are each of <lb/>them ſubdiviſible into other more particular caſes.</s>
  <s xml:id="echoid-s2087" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2088" xml:space="preserve"><emph style="sc">Epitagma</emph> I. </s>
  <s xml:id="echoid-s2089" xml:space="preserve">Here O is ſought between I, the point which bounds the <lb/>ſegment whoſe ſquare is concerned, and the next in order to it: </s>
  <s xml:id="echoid-s2090" xml:space="preserve">and there <lb/>are four caſes, viz. </s>
  <s xml:id="echoid-s2091" xml:space="preserve">when I is an extreme point; </s>
  <s xml:id="echoid-s2092" xml:space="preserve">and the given ratio of R <lb/>to S the ratio of a greater to a leſs; </s>
  <s xml:id="echoid-s2093" xml:space="preserve">when I, remaining as before, the given <lb/>ratio is of a leſs to a greater; </s>
  <s xml:id="echoid-s2094" xml:space="preserve">again when I is the middle point, and O <lb/>ſought between it and either of the other given ones.</s>
  <s xml:id="echoid-s2095" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2096" xml:space="preserve"><emph style="sc">Case</emph> I. </s>
  <s xml:id="echoid-s2097" xml:space="preserve">Here the points B and C are both made to fall beyond I <lb/>(Fig. </s>
  <s xml:id="echoid-s2098" xml:space="preserve">17.) </s>
  <s xml:id="echoid-s2099" xml:space="preserve">and DH is drawn through the center of the circle on BC, and O <lb/>will fall between the point I, and the next in order thereto; </s>
  <s xml:id="echoid-s2100" xml:space="preserve">becauſe by <lb/>conſtruction, EC is to CO as BO is to IB, and therefore when EC is greater <lb/>than CO, BO will be greater than IB, and when leſs, leſs; </s>
  <s xml:id="echoid-s2101" xml:space="preserve">but it is plain <lb/>that if O ſhould fall either beyond E or I, this could not be the Caſe. </s>
  <s xml:id="echoid-s2102" xml:space="preserve">It <lb/>is farther manifeſt that ſhould the points A and E change places, the con-<lb/>ſtruction would be no otherwiſe altered than that AB would then be greater <lb/>than IC.</s>
  <s xml:id="echoid-s2103" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2104" xml:space="preserve"><emph style="sc">Case</emph> II. </s>
  <s xml:id="echoid-s2105" xml:space="preserve">If the given points retain their poſition, but the ratio be made <lb/>of a leſs to a greater, the conſtruction will then be by Fig. </s>
  <s xml:id="echoid-s2106" xml:space="preserve">18, where B <lb/>muſt be made to fall beyond A, and C beyond E with reſpect to I; </s>
  <s xml:id="echoid-s2107" xml:space="preserve">but <lb/>DH is ſtill drawn through the center of the circle on BC: </s>
  <s xml:id="echoid-s2108" xml:space="preserve">and that O will <lb/>fall as required may be made appear by reaſonings ſimilar to thoſe uſed in <lb/>Caſe I. </s>
  <s xml:id="echoid-s2109" xml:space="preserve">Moreover no change will enſue in the conſtruction when the Points <lb/>A and E change places, except that B and C will change ſituations alſo.</s>
  <s xml:id="echoid-s2110" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2111" xml:space="preserve"><emph style="sc">Cases</emph> III and IV, Are conſtructed at once by Fig. </s>
  <s xml:id="echoid-s2112" xml:space="preserve">19, when B muſt <lb/>fall between A and I, C between I and E, and DH be drawn as before: <lb/></s>
  <s xml:id="echoid-s2113" xml:space="preserve">and it is here evident that the conſtruction will be the ſame let the given <lb/>ratio be what it will. </s>
  <s xml:id="echoid-s2114" xml:space="preserve">None of thoſe Caſes admit of any Limitations.</s>
  <s xml:id="echoid-s2115" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2116" xml:space="preserve"><emph style="sc">Epitagma</emph> II. </s>
  <s xml:id="echoid-s2117" xml:space="preserve">There are here only two Caſes, viz. </s>
  <s xml:id="echoid-s2118" xml:space="preserve">when O is required <lb/>beyond I, the point which bounds the ſegment whoſe ſquare is concerned;</s>
  <s xml:id="echoid-s2119" xml:space="preserve">
<pb o="[24]" file="0094" n="101"/>
and ſecondly when it is ſought beyond both the other given points: </s>
  <s xml:id="echoid-s2120" xml:space="preserve">in the <lb/>firſt, the ratio of R to S muſt be of a greater to a leſs, and in the latter of <lb/>a leſs to a greater. </s>
  <s xml:id="echoid-s2121" xml:space="preserve">The former is conſtructed in Fig. </s>
  <s xml:id="echoid-s2122" xml:space="preserve">17. </s>
  <s xml:id="echoid-s2123" xml:space="preserve">at the ſame time <lb/>with Caſe I of Epitagma I. </s>
  <s xml:id="echoid-s2124" xml:space="preserve">and is repreſented by the ſmall letters h and o: <lb/></s>
  <s xml:id="echoid-s2125" xml:space="preserve">the latter in Fig. </s>
  <s xml:id="echoid-s2126" xml:space="preserve">18, and pointed out by the ſame letters. </s>
  <s xml:id="echoid-s2127" xml:space="preserve">That O will <lb/>fall in both as is required needs not inſiſting on.</s>
  <s xml:id="echoid-s2128" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2129" xml:space="preserve"><emph style="sc">Epitagma</emph> III. </s>
  <s xml:id="echoid-s2130" xml:space="preserve">In which there are ſix Caſes, viz. </s>
  <s xml:id="echoid-s2131" xml:space="preserve">I being the middle <lb/>point, when O is ſought beyond A, or beyond E; </s>
  <s xml:id="echoid-s2132" xml:space="preserve">and that whether the <lb/>given ratio be of a leſs to a greater, or of a greater to a leſs; </s>
  <s xml:id="echoid-s2133" xml:space="preserve">and again, I <lb/>being an extreme point, when O is ſought between A and E, and that let <lb/>the order of the points A and E be what it will.</s>
  <s xml:id="echoid-s2134" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2135" xml:space="preserve"><emph style="sc">Cases</emph> I and II. </s>
  <s xml:id="echoid-s2136" xml:space="preserve">Are when I is a mean point and the given ratio of a <lb/>leſs to a greater; </s>
  <s xml:id="echoid-s2137" xml:space="preserve">and theſe are both conſtructed at once by Fig. </s>
  <s xml:id="echoid-s2138" xml:space="preserve">20, <lb/>wherein B is made to fall beyond A, and C beyond E with reſpect to the <lb/>middle point I, and DH is drawn through the center of the circle on BC.</s>
  <s xml:id="echoid-s2139" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2140" xml:space="preserve"><emph style="sc">Cases</emph> III and IV. </s>
  <s xml:id="echoid-s2141" xml:space="preserve">Here, the points remaining as before, the given <lb/>ratio is of a greater to a leſs; </s>
  <s xml:id="echoid-s2142" xml:space="preserve">and the conſtruction will be effected by <lb/>making B fall beyond E, and C beyond A, and drawing DH parallel to <lb/>BC, as in Fig. </s>
  <s xml:id="echoid-s2143" xml:space="preserve">21 and 22.</s>
  <s xml:id="echoid-s2144" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2145" xml:space="preserve"><emph style="sc">Case</emph> V. </s>
  <s xml:id="echoid-s2146" xml:space="preserve">Wherein I is one extreme point and A the other, and O is <lb/>ſought between A and E: </s>
  <s xml:id="echoid-s2147" xml:space="preserve">in conſtructing this Caſe, B muſt be made to <lb/>fall between A and I, C between E and I, and DH drawn parallel to BC, <lb/>as is done in Fig. </s>
  <s xml:id="echoid-s2148" xml:space="preserve">23. </s>
  <s xml:id="echoid-s2149" xml:space="preserve">The directions for Conſtructing Caſe VI. </s>
  <s xml:id="echoid-s2150" xml:space="preserve">are exactly <lb/>the ſame, as will appear by barely inſpecting Fig. </s>
  <s xml:id="echoid-s2151" xml:space="preserve">24.</s>
  <s xml:id="echoid-s2152" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2153" xml:space="preserve"><emph style="sc">Limitation</emph>. </s>
  <s xml:id="echoid-s2154" xml:space="preserve">It is plain that in the four laſt Caſes, the ratio which the <lb/>rectangle contained by AO and EO bears to the ſquare on IO, or which is <lb/>the ſame thing, the given ratio of R to S cannot exceed a certain limit; <lb/></s>
  <s xml:id="echoid-s2155" xml:space="preserve">and it is farther obvious that the ſaid limit will be when the ſtraight line <lb/>DH becomes a tangent to the circle on BC, as in Fig. </s>
  <s xml:id="echoid-s2156" xml:space="preserve">25. </s>
  <s xml:id="echoid-s2157" xml:space="preserve">26, for after <lb/>that the problem is manifeſtly impoſſible. </s>
  <s xml:id="echoid-s2158" xml:space="preserve">Now when DH is a tangent to <lb/>the circle on BC, HO will be equal to half BC; </s>
  <s xml:id="echoid-s2159" xml:space="preserve">but the ſquare on HO <lb/>is equal to the rectangle contained by IB and EC, wherefore the ſquare on <lb/>half BC will then be equal to the rectangle contained by IB and EC. </s>
  <s xml:id="echoid-s2160" xml:space="preserve"><lb/>Moreover, by the conſtruction, R is to S as AB is to IB, and as EC is to <lb/>IC; </s>
  <s xml:id="echoid-s2161" xml:space="preserve">therefore by compoſition or diviſion, the ſum or difference of R and <lb/>S is to R as EI to EC, and the ſaid ſum or difference is to S as AI is to
<pb o="[25]" file="0095" n="102"/>
IB, as EI is to IC: </s>
  <s xml:id="echoid-s2162" xml:space="preserve">and hence, by compound ratio, the ſquare on the <lb/>abovementioned ſum or difference is to the rectangle contained by R and S <lb/>as the rectangle contained by AI and EI is to the rectangle contained by <lb/>IB and EC, alſo by permutation, AI is to EI as IB is to IC; </s>
  <s xml:id="echoid-s2163" xml:space="preserve">wherefore, <lb/>by compoſition or diviſion, AE is to AI as BC is to IB, by permutation, <lb/>AE is to BC as AI is to IB, therefore by equality, the ſum or difference <lb/>of R and S is to S as AE is to BC; </s>
  <s xml:id="echoid-s2164" xml:space="preserve">or (<emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s2165" xml:space="preserve">V. </s>
  <s xml:id="echoid-s2166" xml:space="preserve">15.) </s>
  <s xml:id="echoid-s2167" xml:space="preserve">as half AE is to <lb/>half BC; </s>
  <s xml:id="echoid-s2168" xml:space="preserve">conſequently (<emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s2169" xml:space="preserve">VI. </s>
  <s xml:id="echoid-s2170" xml:space="preserve">22.) </s>
  <s xml:id="echoid-s2171" xml:space="preserve">the ſquare on the above mentioned <lb/>ſum or diſſerence is to the ſquare on S as the ſquare on half AE is to the <lb/>ſquare on half BC, or to the rectangle contained by IB and EC. </s>
  <s xml:id="echoid-s2172" xml:space="preserve">Hence <lb/>exæquo perturbaté, the rectangle contained by R and S is to the ſquare on <lb/>S as the ſquare on half AE is to the rectangle contained by AI and EI, <lb/>or (<emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s2173" xml:space="preserve">V. </s>
  <s xml:id="echoid-s2174" xml:space="preserve">15.) </s>
  <s xml:id="echoid-s2175" xml:space="preserve">R is to S as the ſquare on half AE is to the rectangle <lb/>contained by AI and EI; </s>
  <s xml:id="echoid-s2176" xml:space="preserve">and which is therefore the greateſt ratio which <lb/>R can have to S in thoſe Caſes.</s>
  <s xml:id="echoid-s2177" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2178" xml:space="preserve">It ought farther to be remarked, that to have Caſe III poſſible, where <lb/>O is ſought beyond A, and the ratio of a greater to a leſs, it is neceſſary <lb/>that AI be leſs than IE, and to have Caſe IV. </s>
  <s xml:id="echoid-s2179" xml:space="preserve">poſſible, that it be greater. <lb/></s>
  <s xml:id="echoid-s2180" xml:space="preserve">For it is plain from the Conſtruction, that IB muſt in the former caſe be <lb/>leſs, and in the latter greater than I C; </s>
  <s xml:id="echoid-s2181" xml:space="preserve">but as R is to S ſo is AB to <lb/>IB, and ſo is EC to IC, wherefore by diviſion, the exceſs of R above S <lb/>is to S as AI is to IB, and as EI is to IC; </s>
  <s xml:id="echoid-s2182" xml:space="preserve">and ſo by permutation AI is <lb/>to EI as IB is to IC: </s>
  <s xml:id="echoid-s2183" xml:space="preserve">conſequently when IB is greater than IC, AI will <lb/>be greater than EI; </s>
  <s xml:id="echoid-s2184" xml:space="preserve">and when leſs, leſs.</s>
  <s xml:id="echoid-s2185" xml:space="preserve"><anchor type="note" xlink:href="" symbol="*"/></s>
</p>
<p>
  <s xml:id="echoid-s2186" xml:space="preserve">With reſpect to thoſe caſes wherein the given ratio is that of equality, <lb/>it may be ſufficient to remark, that none of the Caſes of Epitagma II. <lb/></s>
  <s xml:id="echoid-s2187" xml:space="preserve">are poſſible under that ratio: </s>
  <s xml:id="echoid-s2188" xml:space="preserve">that one of Caſes III. </s>
  <s xml:id="echoid-s2189" xml:space="preserve">and IV. </s>
  <s xml:id="echoid-s2190" xml:space="preserve">Epitagma III. </s>
  <s xml:id="echoid-s2191" xml:space="preserve"><lb/>is always impoſſible when the given ratio of R to S is the ratio of equality; </s>
  <s xml:id="echoid-s2192" xml:space="preserve"><lb/>and both are ſo if AI be at the ſame time equal to IE. </s>
  <s xml:id="echoid-s2193" xml:space="preserve">Laſtly Caſes V. </s>
  <s xml:id="echoid-s2194" xml:space="preserve"><lb/>and VI. </s>
  <s xml:id="echoid-s2195" xml:space="preserve">are never poſſible under the ratio of equality, unleſs the ſquare on <lb/>half AE be equal to, or exceed the rectangle contained by AI and EI; </s>
  <s xml:id="echoid-s2196" xml:space="preserve"><lb/>all which naturally follows from what has been delivered above.</s>
  <s xml:id="echoid-s2197" xml:space="preserve"/>
</p>
<note symbol="*" position="foot" xml:space="preserve">See Prop. A. Book V. of Dr. Simſon’s Euclid.</note>
</div>
<div xml:id="echoid-div81" type="section" level="1" n="77">
<head xml:id="echoid-head92" xml:space="preserve">THE END OF BOOK I.</head>
<pb o="[26]" file="0096" n="103"/>
</div>
<div xml:id="echoid-div82" type="section" level="1" n="78">
<head xml:id="echoid-head93" xml:space="preserve">DETERMINATE SECTION. <lb/>BOOK II. <lb/>LEMMA I.</head>
<p>
  <s xml:id="echoid-s2198" xml:space="preserve">If from two points E and I in the diameter AU of a circle AYUV (Fig. <lb/></s>
  <s xml:id="echoid-s2199" xml:space="preserve">27.) </s>
  <s xml:id="echoid-s2200" xml:space="preserve">two perpendiculars EV, IY be drawn contrary ways to terminate in <lb/>the Circumference; </s>
  <s xml:id="echoid-s2201" xml:space="preserve">and if their extremes V and Y be joined by a ſtraight <lb/>line VY, cutting the faid diameter in O; </s>
  <s xml:id="echoid-s2202" xml:space="preserve">then will the ratio which the <lb/>rectangle contained by AO and UO bears to the rectangle contained by <lb/>EO and IO be the leaſt poſſible.</s>
  <s xml:id="echoid-s2203" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2204" xml:space="preserve">*** This being demonſtrated in the preceeding Tract of <emph style="sc">Snellius</emph>, I <lb/>ſhall not attempt it here.</s>
  <s xml:id="echoid-s2205" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div83" type="section" level="1" n="79">
<head xml:id="echoid-head94" xml:space="preserve">LEMMA II.</head>
<p>
  <s xml:id="echoid-s2206" xml:space="preserve">If to a circle deſcribed on AU, tangents EV, IY (Fig. </s>
  <s xml:id="echoid-s2207" xml:space="preserve">28. </s>
  <s xml:id="echoid-s2208" xml:space="preserve">29.) </s>
  <s xml:id="echoid-s2209" xml:space="preserve">be drawn <lb/>from E and I, two points in the diameter AU produced, and through the <lb/>points of contact V, and Y, a ſtraight line YVO be drawn to cut the line <lb/>AI in O; </s>
  <s xml:id="echoid-s2210" xml:space="preserve">then will the ratio which the rectangle contained by AO and <lb/>UO bears to that contained by EO and IO be the leaſt poſſible: </s>
  <s xml:id="echoid-s2211" xml:space="preserve">and <lb/>moreover, the ſquare on EO will be the ſquare on IO as the rectangle con-<lb/>tained by AE and UE is to the rectangle contained by AI and UI.</s>
  <s xml:id="echoid-s2212" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2213" xml:space="preserve"><emph style="sc">Demonstration</emph>. </s>
  <s xml:id="echoid-s2214" xml:space="preserve">If the ſaid ratio be not then a minimum, let it be <lb/>when the ſegments are bounded by ſome other point S, through which and <lb/>the point V, let the ſtraight line SV be drawn, meeting the circle again in <lb/>R; </s>
  <s xml:id="echoid-s2215" xml:space="preserve">draw SM parallel to OY, meeting the tangents EV and IY in L and <lb/>M, and through R and Y draw the ſtraight line RY meeting SM produced <lb/>in N: </s>
  <s xml:id="echoid-s2216" xml:space="preserve">the triangles ESL and EOV, ISM and IOY are ſimilar; </s>
  <s xml:id="echoid-s2217" xml:space="preserve">wherefore <lb/>LS is to SE as VO is to EO, and SM is to SI as YO is to OI; </s>
  <s xml:id="echoid-s2218" xml:space="preserve">conſe-
<pb o="[27]" file="0097" n="104"/>
quently, by compound ratio, the rectangle contained by LS and SM is to <lb/>that contained by SE and SI as the rectangle contained by VO and OY, <lb/>or its equal, the rectangle contained by AO and OU is to that contained <lb/>by EO and IO. </s>
  <s xml:id="echoid-s2219" xml:space="preserve">Now the triangles VSL and NSR having the angles at <lb/>R and L equal, and the angle at S common, are ſimilar; </s>
  <s xml:id="echoid-s2220" xml:space="preserve">and therefore SR <lb/>is to SN as SL is to SV; </s>
  <s xml:id="echoid-s2221" xml:space="preserve">conſequently, the rectangle contained by SR and <lb/>SV, or its equal, the rectangle contained by SA and SU is equal to that <lb/>contained by SN and SL: </s>
  <s xml:id="echoid-s2222" xml:space="preserve">but SN is neceſſarily greater than SM, in con-<lb/>ſequence whereof the rectangle contained by SN and SL, or its equal, the <lb/>rectangle contained by AS and SU is greater than that contained by SM <lb/>and SL; </s>
  <s xml:id="echoid-s2223" xml:space="preserve">wherefore the ratio which the rectangle AS, SU bears to the rect-<lb/>angle ES, SI is greater than that which the rectangle SM, SL bears to it, <lb/>and of courſe, greater than the ratio which the rectangle AO, UO bears to <lb/>the rectangle EO, IO; </s>
  <s xml:id="echoid-s2224" xml:space="preserve">and that, on which ſide ſoever of the point O, S is <lb/>taken.</s>
  <s xml:id="echoid-s2225" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2226" xml:space="preserve">Again, on YO produced, let fall the perpendiculars EB and IC: </s>
  <s xml:id="echoid-s2227" xml:space="preserve">the <lb/>triangles EBV and ICY, EBO and ICO are ſimilar, becauſe the angles EVO <lb/>and IYO are equal, and ſo EO is to IO as EB is to IC, alſo EV is to IY <lb/>as EB is to IC; </s>
  <s xml:id="echoid-s2228" xml:space="preserve">therefore by equality of ratios EO is to IO as EV is to IY, <lb/>and (<emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s2229" xml:space="preserve">VI. </s>
  <s xml:id="echoid-s2230" xml:space="preserve">22.) </s>
  <s xml:id="echoid-s2231" xml:space="preserve">the ſquare on EO is to the ſquare on IO as the ſquare <lb/>on EV is to the ſquare on IY; </s>
  <s xml:id="echoid-s2232" xml:space="preserve">that is (<emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s2233" xml:space="preserve">III. </s>
  <s xml:id="echoid-s2234" xml:space="preserve">36.) </s>
  <s xml:id="echoid-s2235" xml:space="preserve">as the rectangle con-<lb/>tained by AE and UE is to that contained by AI and UI.</s>
  <s xml:id="echoid-s2236" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2237" xml:space="preserve">Q. </s>
  <s xml:id="echoid-s2238" xml:space="preserve">E. </s>
  <s xml:id="echoid-s2239" xml:space="preserve">D.</s>
  <s xml:id="echoid-s2240" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div84" type="section" level="1" n="80">
<head xml:id="echoid-head95" xml:space="preserve">LEMMA III.</head>
<p>
  <s xml:id="echoid-s2241" xml:space="preserve">If from two points E and I, in the diameter AU, of a circle, AVYU <lb/>(Fig. </s>
  <s xml:id="echoid-s2242" xml:space="preserve">30.) </s>
  <s xml:id="echoid-s2243" xml:space="preserve">two perpendiculars EV, IY be drawn on the ſame ſide thereof to <lb/>
<anchor type="handwritten" xlink:label="hd-0097-01a" xlink:href="hd-0097-01"/>
terminate in the periphery, and if their extremes V and Y be joined by a <lb/>ſtraight line VY, cutting the ſaid diameter, produced, in O; </s>
  <s xml:id="echoid-s2244" xml:space="preserve">then will the <lb/>ratio which the rectangle contained by AO and UO bears to the rectangle <lb/>contained by EO and IO be the greateſt poſſible.</s>
  <s xml:id="echoid-s2245" xml:space="preserve"/>
</p>
<div xml:id="echoid-div84" type="float" level="2" n="1">
  <handwritten xlink:label="hd-0097-01" xlink:href="hd-0097-01a"/>
</div>
<p>
  <s xml:id="echoid-s2246" xml:space="preserve">*** This, like the ſirſt, is demonſtrated by <emph style="sc">Snellius</emph>, and needs not be <lb/>repeated.</s>
  <s xml:id="echoid-s2247" xml:space="preserve"/>
</p>
<pb o="[28]" file="0098" n="105"/>
</div>
<div xml:id="echoid-div86" type="section" level="1" n="81">
<head xml:id="echoid-head96" xml:space="preserve">LEMMA IV.</head>
<p>
  <s xml:id="echoid-s2248" xml:space="preserve">If EK and IY (Fig. </s>
  <s xml:id="echoid-s2249" xml:space="preserve">27.) </s>
  <s xml:id="echoid-s2250" xml:space="preserve">be any perpendiculars to the diameter AU of <lb/>a circle AYUV, terminating in the circumference, and if KY be drawn, <lb/>on which, from U, the perpendicular UF is demitted; </s>
  <s xml:id="echoid-s2251" xml:space="preserve">then will KF be a <lb/>mean proportional between AI and EU, alſo YF a mean proportional <lb/>between AE and IU.</s>
  <s xml:id="echoid-s2252" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2253" xml:space="preserve"><emph style="sc">Demonstration</emph>. </s>
  <s xml:id="echoid-s2254" xml:space="preserve">Draw UY, UK, KA and AY, the angles I and F <lb/>
<anchor type="handwritten" xlink:label="hd-0098-01a" xlink:href="hd-0098-01"/>
being right by conſtruction, and the angles IDU, and FKV equal, being <lb/>
<anchor type="handwritten" xlink:label="hd-0098-01a" xlink:href="hd-0098-01"/>
both equal to the angle UAY, the triangles IYU and FKU are ſimilar, <lb/>and conſequently IY is to UY as KF is to UK; </s>
  <s xml:id="echoid-s2255" xml:space="preserve">or (<emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s2256" xml:space="preserve">VI. </s>
  <s xml:id="echoid-s2257" xml:space="preserve">22.) </s>
  <s xml:id="echoid-s2258" xml:space="preserve">the <lb/>ſquare on IY is to the ſquare on UY as the ſquare on KF is to the ſquare <lb/>on UK: </s>
  <s xml:id="echoid-s2259" xml:space="preserve">now the ſquare on IY is (<emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s2260" xml:space="preserve">VI. </s>
  <s xml:id="echoid-s2261" xml:space="preserve">8. </s>
  <s xml:id="echoid-s2262" xml:space="preserve">17.) </s>
  <s xml:id="echoid-s2263" xml:space="preserve">equal to the rectangle <lb/>contained by AI and IU, the ſquare on UY to the rectangle contained by <lb/>AU and IU, and the ſquare on UK to the rectangle contained by AU and <lb/>EU; </s>
  <s xml:id="echoid-s2264" xml:space="preserve">wherefore the rectangle contained by AI and IU is to that contained <lb/>by AU and IU as the ſquare on KF is to the rectangle contained by AU <lb/>and UE, whence (<emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s2265" xml:space="preserve">V. </s>
  <s xml:id="echoid-s2266" xml:space="preserve">15.) </s>
  <s xml:id="echoid-s2267" xml:space="preserve">AI is to AU as the ſquare on KF is to the <lb/>rectangle contained by AU and UE, or the rectangle contained by AI and <lb/>UE is to that contained by AU and UE as the ſquare on KF is to the <lb/>rectangle contained by AU and UE; </s>
  <s xml:id="echoid-s2268" xml:space="preserve">ſeeing then that the conſequents are <lb/>here the ſame, the antecedents muſt be equal, and therefore AI is to KF <lb/>as KF is to UE.</s>
  <s xml:id="echoid-s2269" xml:space="preserve"/>
</p>
<div xml:id="echoid-div86" type="float" level="2" n="1">
  <handwritten xlink:label="hd-0098-01" xlink:href="hd-0098-01a"/>
  <handwritten xlink:label="hd-0098-01" xlink:href="hd-0098-01a"/>
</div>
<p>
  <s xml:id="echoid-s2270" xml:space="preserve">Again, the angle AKE is equal to AUK, which is equal to the angle <lb/>AYK, of which the angle UYF is the complement, becauſe AYU is a <lb/>right angle; </s>
  <s xml:id="echoid-s2271" xml:space="preserve">and therefore as the angles F and E are both right, the tri-<lb/>angles AKE and YUF are ſimilar, and AK is to AE as YU is to YF, <lb/>wherefore the ſquare on AK is to the ſquare on AE as the ſquare on YU <lb/>is to the ſquare on YF: </s>
  <s xml:id="echoid-s2272" xml:space="preserve">but the ſquare on AK is equal to the rectangle <lb/>contained by AU and AE, and the ſquare on YU is equal to the rectangle <lb/>contained by AU and IU, conſequently the rectangle contained by AU <lb/>and AE is to the ſquare on AE as the rectangle contained by AU and IU <lb/>is to the ſquare on YF; </s>
  <s xml:id="echoid-s2273" xml:space="preserve">whence (<emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s2274" xml:space="preserve">V. </s>
  <s xml:id="echoid-s2275" xml:space="preserve">15.) </s>
  <s xml:id="echoid-s2276" xml:space="preserve">AU is to AE as the rect-<lb/>angle contained by AU and IU is to the ſquare on YF, or the rectangle <lb/>contained by AU and IU is to that contained by AE and IU as the rect-
<pb o="[29]" file="0099" n="106"/>
angle contained by AU and IU is to the ſquare on YF; </s>
  <s xml:id="echoid-s2277" xml:space="preserve">hence the rect-<lb/>angle AE, IU is equal to the ſquare on YF, and AE is to YF as YF is <lb/>to IU.</s>
  <s xml:id="echoid-s2278" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2279" xml:space="preserve">Q. </s>
  <s xml:id="echoid-s2280" xml:space="preserve">E. </s>
  <s xml:id="echoid-s2281" xml:space="preserve">D.</s>
  <s xml:id="echoid-s2282" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div88" type="section" level="1" n="82">
<head xml:id="echoid-head97" xml:space="preserve">LEMMA V.</head>
<p>
  <s xml:id="echoid-s2283" xml:space="preserve">If in any ſtraight line four points A, U, E and I (Fig. </s>
  <s xml:id="echoid-s2284" xml:space="preserve">31.) </s>
  <s xml:id="echoid-s2285" xml:space="preserve">be aſſigned, <lb/>and if the point O be ſo taken by <emph style="sc">Lemma</emph> II, that the ratio of the rect-<lb/>angle contained by AO and UO to that contained by EO and IO may be <lb/>the leaſt poſſible; </s>
  <s xml:id="echoid-s2286" xml:space="preserve">alſo if through O the indefinite perpendicular FG be <lb/>drawn; </s>
  <s xml:id="echoid-s2287" xml:space="preserve">and laſtly, if from E and I, EG and IF be applied to FG, the <lb/>former equal to a mean proportional between AE and UE, and the latter <lb/>to one between AI and UI: </s>
  <s xml:id="echoid-s2288" xml:space="preserve">then ſhall FG be equal to the ſum of two <lb/>mean proportionals between AE and UI, AI and UE.</s>
  <s xml:id="echoid-s2289" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2290" xml:space="preserve"><emph style="sc">Demonstration</emph>. </s>
  <s xml:id="echoid-s2291" xml:space="preserve">Draw AF and AG, and, through U, FV and GY, <lb/>produce GE to meet FV in H, and let fall on FV the perpendicular XI, <lb/>cutting FG in N; </s>
  <s xml:id="echoid-s2292" xml:space="preserve">moreover draw UM through N, and NP through E, <lb/>and theſe two laſt will be reſpectively perpendiculars to IF and UG, be-<lb/>cauſe the three perpendiculars of every plane triangle meet in a point. <lb/></s>
  <s xml:id="echoid-s2293" xml:space="preserve">Since by conſtruction and <emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s2294" xml:space="preserve">VI. </s>
  <s xml:id="echoid-s2295" xml:space="preserve">17, the ſquare on EG is equal to the <lb/>rectangle contained by AE and UE, and the ſquare on IF to that con-<lb/>tained by AI and UI, and becauſe (<emph style="sc">Lem</emph>. </s>
  <s xml:id="echoid-s2296" xml:space="preserve">II.) </s>
  <s xml:id="echoid-s2297" xml:space="preserve">the ſquare on EO is to the <lb/>ſquare on IO as the rectangle AE, UE is to the rectangle AI, UI; </s>
  <s xml:id="echoid-s2298" xml:space="preserve">the <lb/>ſquare on EO is to the ſquare on IO as the ſquare on EG is to the ſquare <lb/>on IF, and (<emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s2299" xml:space="preserve">VI. </s>
  <s xml:id="echoid-s2300" xml:space="preserve">22.) </s>
  <s xml:id="echoid-s2301" xml:space="preserve">EO is to IO as EG is to IF; </s>
  <s xml:id="echoid-s2302" xml:space="preserve">from whence it <lb/>appears that the triangles EOG and IOF are ſimilar, and HG parallel to <lb/>IF, and the angle UHE equal to the angle UFI. </s>
  <s xml:id="echoid-s2303" xml:space="preserve">Again, becauſe AI is <lb/>to IF as IF is to UI, the triangles AIF, and UFI are ſimilar, and ſo, for <lb/>like reaſons, are the triangles AEG and GEU, wherefore the angles UFI <lb/>and FAE are equal, and alſo the angles UGE and UAG; </s>
  <s xml:id="echoid-s2304" xml:space="preserve">hence there-<lb/>fore (<emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s2305" xml:space="preserve">I 32.) </s>
  <s xml:id="echoid-s2306" xml:space="preserve">the angle YUF is equal to the angle UAF (UHE) toge-<lb/>ther with the angle UAG (UGE) and conſequently, the angles VAY and <lb/>YUV are together equal to two right angles; </s>
  <s xml:id="echoid-s2307" xml:space="preserve">wherefore the points AYUV <lb/>are in a circle: </s>
  <s xml:id="echoid-s2308" xml:space="preserve">hence, and becauſe AO is perpendicular to FG, <anchor type="note" xlink:href="" symbol="*"/>GY will
<anchor type="note" xlink:label="note-0099-01a" xlink:href="note-0099-01"/>
<pb o="[30]" file="0100" n="107"/>
be perpendicular to AF, and FV to AG; </s>
  <s xml:id="echoid-s2309" xml:space="preserve">whence it follows that GA is <lb/>parallel to XI, and the angle NIU equal to the angle UAG; </s>
  <s xml:id="echoid-s2310" xml:space="preserve">but UAG is <lb/>equal to UGE, which is equal to CNE; </s>
  <s xml:id="echoid-s2311" xml:space="preserve">wherefore, in the triangles UNI, <lb/>UNE, the angles at I and N being equal, and that at U common, they <lb/>are ſimilar, and UN is to UI as UE is to UN, conſequently the ſquare on <lb/>UN is equal to the rectangle contained by UI and UE. </s>
  <s xml:id="echoid-s2312" xml:space="preserve">Moreover, ſince <lb/>IX paſſes through N, and is perpendicular to FU, by <emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s2313" xml:space="preserve">I. </s>
  <s xml:id="echoid-s2314" xml:space="preserve">47, the dif-<lb/>ference of the ſquares on IF and IU is equal to the difference of the <lb/>ſquares on NF and NU: </s>
  <s xml:id="echoid-s2315" xml:space="preserve">now the ſquare on IF being equal to the rect-<lb/>angle contained by AI and UI, that is (<emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s2316" xml:space="preserve">II. </s>
  <s xml:id="echoid-s2317" xml:space="preserve">I.) </s>
  <s xml:id="echoid-s2318" xml:space="preserve">to the rectangle con-<lb/>tained by AU and UI together with the ſquare on UI, the difference of <lb/>the ſquares on IF and UI, and conſequently the difference of the ſquares on <lb/>NF and NU is equal to the rectangle contained by AU and UI; </s>
  <s xml:id="echoid-s2319" xml:space="preserve">but the <lb/>ſquare on NU has been proved equal to the rectangle contained by UI <lb/>and UE, therefore the ſquare on NF is equal to the rectangle contained <lb/>by EU and IU together with that contained by AU and IU, that is (<emph style="sc">Eu</emph>. <lb/></s>
  <s xml:id="echoid-s2320" xml:space="preserve">II. </s>
  <s xml:id="echoid-s2321" xml:space="preserve">1.) </s>
  <s xml:id="echoid-s2322" xml:space="preserve">to the rectangle contained by AE and UI; </s>
  <s xml:id="echoid-s2323" xml:space="preserve">wherefore AE is to NF <lb/>as NF is to UI.</s>
  <s xml:id="echoid-s2324" xml:space="preserve"/>
</p>
<div xml:id="echoid-div88" type="float" level="2" n="1">
<note symbol="*" position="foot" xlink:label="note-0099-01" xlink:href="note-0099-01a" xml:space="preserve">See <emph style="sc">Pap</emph>. Math. Collect. B. vii. prop. 60.</note>
</div>
<p>
  <s xml:id="echoid-s2325" xml:space="preserve">Laſtly, for the like reaſons which were urged above, the difference of <lb/>the ſquares on NU and NG is equal to the difference of thoſe on GE and <lb/>UE: </s>
  <s xml:id="echoid-s2326" xml:space="preserve">now the ſquare on GE is equal to the rectangle contained by AE <lb/>and UE, that is, to the rectangle contained by AU and UE together <lb/>with the ſquare on UE; </s>
  <s xml:id="echoid-s2327" xml:space="preserve">therefore the difference of the ſquares on GE <lb/>and UE, or the difference of thoſe on NU and NG, is equal to the rect-<lb/>angle contained by AU and UE; </s>
  <s xml:id="echoid-s2328" xml:space="preserve">but the ſquare on NU is equal to the <lb/>rectangle contained by UI and UE, therefore the ſquare on NG is <lb/>equal to the rectangle contained by AU and UE together with that <lb/>contained by UI and UE; </s>
  <s xml:id="echoid-s2329" xml:space="preserve">that is, to the rectangle contained by AI and <lb/>UE, and ſo AI is to NG as NG is to UE. </s>
  <s xml:id="echoid-s2330" xml:space="preserve">Now FG is equal to the ſum <lb/>of NF and NG; </s>
  <s xml:id="echoid-s2331" xml:space="preserve">therefore FG is equal to the ſum of two mean propor-<lb/>tionals between AE and UI, AI and UE.</s>
  <s xml:id="echoid-s2332" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2333" xml:space="preserve">Q. </s>
  <s xml:id="echoid-s2334" xml:space="preserve">E. </s>
  <s xml:id="echoid-s2335" xml:space="preserve">D.</s>
  <s xml:id="echoid-s2336" xml:space="preserve"/>
</p>
<pb o="[31]" file="0101" n="108"/>
</div>
<div xml:id="echoid-div90" type="section" level="1" n="83">
<head xml:id="echoid-head98" xml:space="preserve">PROBLEM VII. (Fig. 32, 33, 34, &amp;c.)</head>
<p>
  <s xml:id="echoid-s2337" xml:space="preserve">In any indeſinite ſtraight line let there be aſſigned the points A, E, I <lb/>and U; </s>
  <s xml:id="echoid-s2338" xml:space="preserve">it is required to cut it in another point, O, ſo that the rectangle <lb/>contained by the ſegments AO, UO may be to that contained by the ſeg-<lb/>ments EO, IO in the ratio of two given ſtraight lines, R and S.</s>
  <s xml:id="echoid-s2339" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2340" xml:space="preserve"><emph style="sc">Analysis</emph>. </s>
  <s xml:id="echoid-s2341" xml:space="preserve">Imagine the thing done, and O the point ſought: </s>
  <s xml:id="echoid-s2342" xml:space="preserve">then will <lb/>the rectangle AO, UO be to the rectangle EO, IO as R is to S. </s>
  <s xml:id="echoid-s2343" xml:space="preserve">Make <lb/>UC to EC as R is to S; </s>
  <s xml:id="echoid-s2344" xml:space="preserve">and the rectangle AO, UO will be the rectangle <lb/>EO, IO as UC is to EC. </s>
  <s xml:id="echoid-s2345" xml:space="preserve">Let now OB be taken a fourth proportional to <lb/>UO, UC and IO: </s>
  <s xml:id="echoid-s2346" xml:space="preserve">then (<emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s2347" xml:space="preserve">V. </s>
  <s xml:id="echoid-s2348" xml:space="preserve">15.) </s>
  <s xml:id="echoid-s2349" xml:space="preserve">the rectangle AO, UO will be to <lb/>the rectangle EO, IO as the rectangle UC, OB is to the rectangle EC, OB; <lb/></s>
  <s xml:id="echoid-s2350" xml:space="preserve">or (<emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s2351" xml:space="preserve">V. </s>
  <s xml:id="echoid-s2352" xml:space="preserve">16.) </s>
  <s xml:id="echoid-s2353" xml:space="preserve">the rectangle AO, UO is to the rectangle UC, OB as the <lb/>rectangle EO, IO is to the rectangle EC, OB; </s>
  <s xml:id="echoid-s2354" xml:space="preserve">wherefore ſince UO is to UC <lb/>as IO to OB, by conſtruction, AO will be to BO as EO to EC; </s>
  <s xml:id="echoid-s2355" xml:space="preserve">and ſo by <lb/>compoſition or diviſion, CO is to CU as IB to BO, and AB is to BO as <lb/>CO to EC: </s>
  <s xml:id="echoid-s2356" xml:space="preserve">wherefore ex æquo perturb. </s>
  <s xml:id="echoid-s2357" xml:space="preserve">&amp; </s>
  <s xml:id="echoid-s2358" xml:space="preserve">permut. </s>
  <s xml:id="echoid-s2359" xml:space="preserve">AB is to IB as UC to <lb/>EC, that is, in the given ratio; </s>
  <s xml:id="echoid-s2360" xml:space="preserve">and hence is given BC, the ſum or dif-<lb/>ference of CO and BO, as alſo the rectangle contained by them, equal to <lb/>the rectangle CU, IB, whence thoſe lines themſelves are given by the 85th <lb/>or 86th of the Data.</s>
  <s xml:id="echoid-s2361" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2362" xml:space="preserve"><emph style="sc">Synthesis</emph>. </s>
  <s xml:id="echoid-s2363" xml:space="preserve">Make AB to IB, and UC to EC in the given ratio, and de-<lb/>ſcribe on BC a circle; </s>
  <s xml:id="echoid-s2364" xml:space="preserve">erect, at B the indeſinite perpendicular BK, and take <lb/>therein BD a mean proportional between AB and EC, or between IB and <lb/>and UC: </s>
  <s xml:id="echoid-s2365" xml:space="preserve">from D, draw DH, parallel to BC, if O be required any where <lb/>between B and C; </s>
  <s xml:id="echoid-s2366" xml:space="preserve">but through F, the center of the circle on BC, if it be <lb/>ſought any where without them, cutting the circle on BC in H. </s>
  <s xml:id="echoid-s2367" xml:space="preserve">Laſtly, <lb/>draw HO perpendicular to DH, which will cut the indeſinite line in O, <lb/>the point required.</s>
  <s xml:id="echoid-s2368" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2369" xml:space="preserve">For it is plain from the conſtruction that HO and BD are equal, and <lb/>(<emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s2370" xml:space="preserve">VI. </s>
  <s xml:id="echoid-s2371" xml:space="preserve">17.) </s>
  <s xml:id="echoid-s2372" xml:space="preserve">the rectangle AB, EC, or the rectangle IB, UC is equal to <lb/>the ſquare on BD, and therefore equal to the ſquare on HO, which (<emph style="sc">Eu</emph>. <lb/></s>
  <s xml:id="echoid-s2373" xml:space="preserve">III. </s>
  <s xml:id="echoid-s2374" xml:space="preserve">35. </s>
  <s xml:id="echoid-s2375" xml:space="preserve">36.) </s>
  <s xml:id="echoid-s2376" xml:space="preserve">is equal to the rectangle BO, OC. </s>
  <s xml:id="echoid-s2377" xml:space="preserve">Hence (<emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s2378" xml:space="preserve">VI. </s>
  <s xml:id="echoid-s2379" xml:space="preserve">16.) </s>
  <s xml:id="echoid-s2380" xml:space="preserve">AB <lb/>is to BO as CO is to CE, and CO is to CU as IB is to BO; </s>
  <s xml:id="echoid-s2381" xml:space="preserve">whence, <lb/>by compoſition or diviſion, AO is to BO as EO is to CE, and UO is to
<pb o="[32]" file="0102" n="109"/>
CU as IO is to BO; </s>
  <s xml:id="echoid-s2382" xml:space="preserve">and ſo, by compound ratio, the rectangle AO, UO <lb/>is to the rectangle BO, CU as the rectangle EO, IO is to the rectangle <lb/>BO, CE; </s>
  <s xml:id="echoid-s2383" xml:space="preserve">by permutation, the rectangle AO, UO is to the rectangle EO, <lb/>IO, as the rectangle BO, CU is to the rectangle BO, CE; </s>
  <s xml:id="echoid-s2384" xml:space="preserve">or (<emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s2385" xml:space="preserve">V. </s>
  <s xml:id="echoid-s2386" xml:space="preserve">15.) <lb/></s>
  <s xml:id="echoid-s2387" xml:space="preserve">as CU is to CE; </s>
  <s xml:id="echoid-s2388" xml:space="preserve">that is, by conſtruction as R to S.</s>
  <s xml:id="echoid-s2389" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2390" xml:space="preserve">Q. </s>
  <s xml:id="echoid-s2391" xml:space="preserve">E. </s>
  <s xml:id="echoid-s2392" xml:space="preserve">D.</s>
  <s xml:id="echoid-s2393" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2394" xml:space="preserve"><emph style="sc">Scholium</emph>. </s>
  <s xml:id="echoid-s2395" xml:space="preserve">In enumerating the ſeveral Caſes of this Problem I ſhall en-<lb/>deavour to follow the method which I conceive Apollonius did: </s>
  <s xml:id="echoid-s2396" xml:space="preserve">and there-<lb/>fore, notwithſtanding the preceding Analyſis and Conſtruction are general <lb/>for the whole, divide it into three Problems, each Problem into three Epi-<lb/>tagmas, or general Caſes, and theſe again into their ſeveral particular ones.</s>
  <s xml:id="echoid-s2397" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div91" type="section" level="1" n="84">
<head xml:id="echoid-head99" xml:space="preserve">PROBLEM I. (Fig. 32 to 45.)</head>
<p>
  <s xml:id="echoid-s2398" xml:space="preserve">Here O is ſought between the two mean points of the four given ones: <lb/></s>
  <s xml:id="echoid-s2399" xml:space="preserve">and the three Epitagmas are, firſt, when A and U, the points which bound <lb/>the ſegments containing the antecedent rectangle, are one an extreme, and <lb/>the other an alternate mean; </s>
  <s xml:id="echoid-s2400" xml:space="preserve">ſecondly, when thoſe points are one an ex-<lb/>treme and the other an adjacent mean; </s>
  <s xml:id="echoid-s2401" xml:space="preserve">thirdly, when they are both means, <lb/>or both extremes.</s>
  <s xml:id="echoid-s2402" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2403" xml:space="preserve"><emph style="sc">Epitagma</emph> I, Conſiſts of eight Caſes, viz. </s>
  <s xml:id="echoid-s2404" xml:space="preserve">when the order of the given <lb/>points is A, I, U, E; </s>
  <s xml:id="echoid-s2405" xml:space="preserve">U, E, A, I; </s>
  <s xml:id="echoid-s2406" xml:space="preserve">A, E, U, I; </s>
  <s xml:id="echoid-s2407" xml:space="preserve">or U, I, A, E, and the <lb/>given ratio of a leſs to a greater, and four others wherein the order of the <lb/>points is the ſame as in thoſe, but the ratio of R to S, the ratio of a greater <lb/>to a leſs.</s>
  <s xml:id="echoid-s2408" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2409" xml:space="preserve"><emph style="sc">Case</emph> I. </s>
  <s xml:id="echoid-s2410" xml:space="preserve">Let the order of the given points be A, I, U, E, and the given <lb/>ratio of a leſs to a greater; </s>
  <s xml:id="echoid-s2411" xml:space="preserve">and the Conſtruction will be as in Fig. </s>
  <s xml:id="echoid-s2412" xml:space="preserve">32, where <lb/>B is made to fall beyond A, with reſpect to I, and C beyond U with re-<lb/>ſpect to E, and DH is drawn through F, the center of the circle on BC. <lb/></s>
  <s xml:id="echoid-s2413" xml:space="preserve">That O, when this conſtruction is uſed, will fall between I and U is <lb/>plain, becauſe CO is to CU as IB is to BO; </s>
  <s xml:id="echoid-s2414" xml:space="preserve">and therefore if CU be <lb/>greater than CO, BO will be greater than IB, and if leſs, leſs; </s>
  <s xml:id="echoid-s2415" xml:space="preserve">but this, <lb/>it is manifeſt, cannot be the Caſe if O falls either beyond I or U, and <lb/>therefore it falls between them.</s>
  <s xml:id="echoid-s2416" xml:space="preserve"/>
</p>
<pb o="[33]" file="0103" n="110"/>
<p>
  <s xml:id="echoid-s2417" xml:space="preserve"><emph style="sc">Case</emph> II. </s>
  <s xml:id="echoid-s2418" xml:space="preserve">If the order oſ the given points be retained, but the ratio be of <lb/>a greater to a leſs, B muſt ſall beyond I with reſpect to A, and C beyond <lb/>E, as in Fig. </s>
  <s xml:id="echoid-s2419" xml:space="preserve">33, and DH muſt be drawn as before: </s>
  <s xml:id="echoid-s2420" xml:space="preserve">and it may be proved <lb/>by reaſonings ſimilar to thoſe uſed in the ſirſt Caſe that O will fall between <lb/>I and U, as was required.</s>
  <s xml:id="echoid-s2421" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2422" xml:space="preserve"><emph style="sc">Cases</emph> III and IV. </s>
  <s xml:id="echoid-s2423" xml:space="preserve">Theſe caſes are conſtructed exactly in the ſame man-<lb/>ner as Caſes I and II reſpectively; </s>
  <s xml:id="echoid-s2424" xml:space="preserve">and the reaſonings to prove that O will <lb/>fall as it ought arethe ſame with thoſe made uſe of in Caſe I, as will appear <lb/>by inſpecting Fig. </s>
  <s xml:id="echoid-s2425" xml:space="preserve">34. </s>
  <s xml:id="echoid-s2426" xml:space="preserve">and 35.</s>
  <s xml:id="echoid-s2427" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2428" xml:space="preserve"><emph style="sc">Case</emph> V. </s>
  <s xml:id="echoid-s2429" xml:space="preserve">Let now the order of the given points be A, E, U, I, and the <lb/>given ratio of a leſs to a greater. </s>
  <s xml:id="echoid-s2430" xml:space="preserve">Then muſt B be made to fall beyond A <lb/>and C beyond U, as in Fig. </s>
  <s xml:id="echoid-s2431" xml:space="preserve">36, and DH is here to be drawn parallel to <lb/>to BC: </s>
  <s xml:id="echoid-s2432" xml:space="preserve">and that O will fall as required may be made to appear thus. </s>
  <s xml:id="echoid-s2433" xml:space="preserve">Draw <lb/>EF and UG perpendicular to BC. </s>
  <s xml:id="echoid-s2434" xml:space="preserve">Now (<emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s2435" xml:space="preserve">VI. </s>
  <s xml:id="echoid-s2436" xml:space="preserve">13. </s>
  <s xml:id="echoid-s2437" xml:space="preserve">17.) </s>
  <s xml:id="echoid-s2438" xml:space="preserve">the ſquare on <lb/>EF is equal to the rectangle contained by EC and EB, the ſquare on UG <lb/>to the rectangle contained by UC and UB; </s>
  <s xml:id="echoid-s2439" xml:space="preserve">and, by conſtruction, the <lb/>ſquare on HO to the rectangle contained by EC and AB, or to the rect-<lb/>angle contained by UC and IB; </s>
  <s xml:id="echoid-s2440" xml:space="preserve">and ſince, by ſuppoſition, EB is greater <lb/>than AB, and UB leſs than IB, the rectangle EC, AB, or its equal, the <lb/>rectangle UC, IB will be leſs than the rectangle EC, EB, and greater than <lb/>the rectangle UC, UB; </s>
  <s xml:id="echoid-s2441" xml:space="preserve">and conſequently the ſquare on HO will be leſs <lb/>than the ſquare on EF and greater than the ſquare on UG; </s>
  <s xml:id="echoid-s2442" xml:space="preserve">and therefore <lb/>HO leſs than EF and greater than UG: </s>
  <s xml:id="echoid-s2443" xml:space="preserve">but this could not be the Caſe <lb/>unleſs O fell between E and U, as was was required.</s>
  <s xml:id="echoid-s2444" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2445" xml:space="preserve"><emph style="sc">Case</emph> VI. </s>
  <s xml:id="echoid-s2446" xml:space="preserve">If the order of the points be retained; </s>
  <s xml:id="echoid-s2447" xml:space="preserve">but the given ratio be <lb/>of a greater to a leſs, B muſt then fall beyond I, and C beyond E; </s>
  <s xml:id="echoid-s2448" xml:space="preserve">and <lb/>DH is drawn as in the preceding Caſe (See Fig. </s>
  <s xml:id="echoid-s2449" xml:space="preserve">37.) </s>
  <s xml:id="echoid-s2450" xml:space="preserve">moreover, that O <lb/>will fall between E and U may be made to appear, by reaſonings ſimilar to <lb/>thoſe there made uſe on the like occaſion.</s>
  <s xml:id="echoid-s2451" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2452" xml:space="preserve"><emph style="sc">Case</emph> VII. </s>
  <s xml:id="echoid-s2453" xml:space="preserve">Is conſtructed exactly in the ſame manner as Caſe V. </s>
  <s xml:id="echoid-s2454" xml:space="preserve">and,</s>
</p>
<p>
  <s xml:id="echoid-s2455" xml:space="preserve"><emph style="sc">Case</emph> VIII. </s>
  <s xml:id="echoid-s2456" xml:space="preserve">As Caſe VI: </s>
  <s xml:id="echoid-s2457" xml:space="preserve">the truth of which will appear by. </s>
  <s xml:id="echoid-s2458" xml:space="preserve">barely in-<lb/>ſpecting Fig. </s>
  <s xml:id="echoid-s2459" xml:space="preserve">38 and 39.</s>
  <s xml:id="echoid-s2460" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2461" xml:space="preserve"><emph style="sc">Epitagma</emph> II. </s>
  <s xml:id="echoid-s2462" xml:space="preserve">In this Epitagma there are alſo eight Caſes, viz. </s>
  <s xml:id="echoid-s2463" xml:space="preserve">when the <lb/>order of the given points is A, U, E, I; </s>
  <s xml:id="echoid-s2464" xml:space="preserve">A, U, I, E; </s>
  <s xml:id="echoid-s2465" xml:space="preserve">U, A, E, I; </s>
  <s xml:id="echoid-s2466" xml:space="preserve">or <lb/>U, A, I, E; </s>
  <s xml:id="echoid-s2467" xml:space="preserve">and the given ratio of a leſs to a greater: </s>
  <s xml:id="echoid-s2468" xml:space="preserve">and there are <lb/>four others wherein the order of the points are the ſame as in thoſe, but
<pb o="[34]" file="0104" n="111"/>
the ratio of a greater to a leſs; </s>
  <s xml:id="echoid-s2469" xml:space="preserve">but theſe, I ſhall ſhew, may be reduced <lb/>to four.</s>
  <s xml:id="echoid-s2470" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2471" xml:space="preserve"><emph style="sc">Case</emph> I. </s>
  <s xml:id="echoid-s2472" xml:space="preserve">The order of the given points being A, U, E, I; </s>
  <s xml:id="echoid-s2473" xml:space="preserve">and the given <lb/>ratio a leſs to a greater, the conſtruction will be eſſected by Fig. </s>
  <s xml:id="echoid-s2474" xml:space="preserve">40, <lb/>wherein B is made to fall beyond A with reſpect to I and C beyond U, and <lb/>DH is drawn through the center of the circle on BC. </s>
  <s xml:id="echoid-s2475" xml:space="preserve">And O will fall <lb/>between U and E for reaſons ſimilar to thoſe urged in the ſirſt Caſe of Epi-<lb/>tagma I. </s>
  <s xml:id="echoid-s2476" xml:space="preserve">It is moreover obvious that the conſtruction will not be eſſentially <lb/>different ſhould the points E and I change places, and therefore need not <lb/>here be made a new Caſe.</s>
  <s xml:id="echoid-s2477" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2478" xml:space="preserve"><emph style="sc">Case</emph> II. </s>
  <s xml:id="echoid-s2479" xml:space="preserve">The order of the points being the ſame as in the laſt Caſe, let <lb/>the given ratio be of a greater to a leſs; </s>
  <s xml:id="echoid-s2480" xml:space="preserve">then, as in Fig. </s>
  <s xml:id="echoid-s2481" xml:space="preserve">41, B muſt fall <lb/>beyond I, and C beyond E; </s>
  <s xml:id="echoid-s2482" xml:space="preserve">but DC muſt ſtill be drawn through the <lb/>center of the circle on BC. </s>
  <s xml:id="echoid-s2483" xml:space="preserve">It is manifeſt that this conſtruction will ſerve <lb/>for that Caſe wherein the points A and U change ſituations, if the ratio be, <lb/>as here, of a greater to a leſs.</s>
  <s xml:id="echoid-s2484" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2485" xml:space="preserve"><emph style="sc">Case</emph> III. </s>
  <s xml:id="echoid-s2486" xml:space="preserve">Here, let the order of the points be U, A, I, E, and the <lb/>given ratio of a leſs to a greater, and the Conſtruction will be aſſected by <lb/>Fig. </s>
  <s xml:id="echoid-s2487" xml:space="preserve">42, in which B falls beyond A, and C beyond U with reſpect to I and <lb/>E: </s>
  <s xml:id="echoid-s2488" xml:space="preserve">and the ſame conſtruction will ſerve if I and E change places, but the <lb/>ratio remain the ſame.</s>
  <s xml:id="echoid-s2489" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2490" xml:space="preserve"><emph style="sc">Case</emph> IV. </s>
  <s xml:id="echoid-s2491" xml:space="preserve">If the poſition of the points be retained, but the ratio be <lb/>made of a leſs to a greater; </s>
  <s xml:id="echoid-s2492" xml:space="preserve">then muſt B fall beyond I (Fig. </s>
  <s xml:id="echoid-s2493" xml:space="preserve">43.) </s>
  <s xml:id="echoid-s2494" xml:space="preserve">and C <lb/>beyond E; </s>
  <s xml:id="echoid-s2495" xml:space="preserve">but DH drawn as before. </s>
  <s xml:id="echoid-s2496" xml:space="preserve">That O muſt fall as was required, <lb/>in theſe three laſt caſes, is obvious enough from what has been ſaid be-<lb/>fore on the like occaſion: </s>
  <s xml:id="echoid-s2497" xml:space="preserve">and it is alſo plain that the conſtruction will <lb/>not be materially diſſerent though A and U change places.</s>
  <s xml:id="echoid-s2498" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2499" xml:space="preserve"><emph style="sc">Scholium</emph>. </s>
  <s xml:id="echoid-s2500" xml:space="preserve">That none of the Caſes of theſe two Epitagmas are ſubject <lb/>to Limitations, might be proved with the utmoſt rigour of geometrical <lb/>reaſoning was it not ſuſſiciently manifeſt from conſidering that as the point <lb/>O approaches points A, or U, the ratio of the rectangle AO, OU to the <lb/>rectangle EO, OI will become very ſmall, and as it approaches the points <lb/>E, or I the ſaid ratio will become very great: </s>
  <s xml:id="echoid-s2501" xml:space="preserve">and nothing hinders that <lb/>the ſaid point may ſall any where between thoſe.</s>
  <s xml:id="echoid-s2502" xml:space="preserve"/>
</p>
<pb o="[35]" file="0105" n="112"/>
<p>
  <s xml:id="echoid-s2503" xml:space="preserve"><emph style="sc">Epitagma</emph> III. </s>
  <s xml:id="echoid-s2504" xml:space="preserve">There are here but four Caſes, viz. </s>
  <s xml:id="echoid-s2505" xml:space="preserve">when the order of the <lb/>given points is A, E, I, U; </s>
  <s xml:id="echoid-s2506" xml:space="preserve">A, I, E, U; </s>
  <s xml:id="echoid-s2507" xml:space="preserve">E, A, U, I; </s>
  <s xml:id="echoid-s2508" xml:space="preserve">or E, U, A, I; <lb/></s>
  <s xml:id="echoid-s2509" xml:space="preserve">the two ſirſt of theſe are not poſſible unleſs the given ratio be the ratio of a <lb/>greater to a leſs; </s>
  <s xml:id="echoid-s2510" xml:space="preserve">nor the two latter, unleſs it be of a leſs to a greater, and <lb/>as theſe are reduced to the ſirſt two by reading every where E for A, I <lb/>for U, and the contrary, I ſhall omit ſpecifying them.</s>
  <s xml:id="echoid-s2511" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2512" xml:space="preserve"><emph style="sc">Case</emph> I. </s>
  <s xml:id="echoid-s2513" xml:space="preserve">If the order of the given points be A, E, I, U, the conſtruc-<lb/>tion will be effected by Fig. </s>
  <s xml:id="echoid-s2514" xml:space="preserve">44, wherein B is made to fall beyond I, and <lb/>C beyond E, and DH is drawn parallel to BC. </s>
  <s xml:id="echoid-s2515" xml:space="preserve">That O, when this con-<lb/>ſtruction is uſed, will fall between E and I, is eaſily made appear by rea-<lb/>ſoning in a manner ſimilar to what was done in Caſe V. </s>
  <s xml:id="echoid-s2516" xml:space="preserve">of Epitagma I.</s>
  <s xml:id="echoid-s2517" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2518" xml:space="preserve"><emph style="sc">Case</emph> II. </s>
  <s xml:id="echoid-s2519" xml:space="preserve">The conſtruction of this Caſe, where the order of the points <lb/>is A, I, E, U, is not materially different from that above exhibited as ap-<lb/>pears by Fig. </s>
  <s xml:id="echoid-s2520" xml:space="preserve">45, and that O will fall between I and E is manifeſt without <lb/>farther illuſtration.</s>
  <s xml:id="echoid-s2521" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2522" xml:space="preserve"><emph style="sc">Limitation</emph>. </s>
  <s xml:id="echoid-s2523" xml:space="preserve">In theſe two Caſes the given ratio of R to S cannot be <lb/>leſs than that which the ſquare on AU bears to the ſquare on a line which <lb/>is the difference of two mean proportionals between AI and EU, AE and <lb/>IU. </s>
  <s xml:id="echoid-s2524" xml:space="preserve">For by Lemma I. </s>
  <s xml:id="echoid-s2525" xml:space="preserve">the leaſt ratio which the rectangle contained by AO <lb/>and UO can have to the rectangle contained by EO and IO; </s>
  <s xml:id="echoid-s2526" xml:space="preserve">or, which <lb/>is the ſame thing, that R can have to S, will be when the point O is <lb/>the interſection of the diameter AU, of a circle AYUV, with a ſtraight <lb/>line YV. </s>
  <s xml:id="echoid-s2527" xml:space="preserve">joining the tops of two perpendiculars EV, IY to the indeſi-<lb/>nite line, on contrary ſides thereof, and terminating in the periphery of <lb/>the circle. </s>
  <s xml:id="echoid-s2528" xml:space="preserve">Produce VE (Fig. </s>
  <s xml:id="echoid-s2529" xml:space="preserve">27.) </s>
  <s xml:id="echoid-s2530" xml:space="preserve">to meet the circle again in K, and <lb/>draw the diameter KL; </s>
  <s xml:id="echoid-s2531" xml:space="preserve">join LY and KY, on which, produced, let fall <lb/>the perpendicular UF. </s>
  <s xml:id="echoid-s2532" xml:space="preserve">Now, ſince by Lemma III. </s>
  <s xml:id="echoid-s2533" xml:space="preserve">KF is a mean propor-<lb/>tional between AI and EU, and YF a mean proportional between AE <lb/>and IU: </s>
  <s xml:id="echoid-s2534" xml:space="preserve">it remains only to prove that the ratio of the rectangle con-<lb/>tained by AO and OU to the rectangle contained by EO and OI is the <lb/>ſame with the ratio which the ſquare on AU bears to the ſquare on KY, <lb/>which is the diſſerence between KF and YF. </s>
  <s xml:id="echoid-s2535" xml:space="preserve">Becauſe the angles E and <lb/>KYL are both right, and the angles EVO and KYL equal (<emph style="sc">Eu</emph>. </s>
  <s xml:id="echoid-s2536" xml:space="preserve">III. </s>
  <s xml:id="echoid-s2537" xml:space="preserve">21.) <lb/></s>
  <s xml:id="echoid-s2538" xml:space="preserve">the triangles EVO and YLK are ſimilar; </s>
  <s xml:id="echoid-s2539" xml:space="preserve">and ſo VO is to EO as AU (LK) <lb/>is to KY; </s>
  <s xml:id="echoid-s2540" xml:space="preserve">or the ſquare on VO is to the ſquare on EO as the ſquare on AU <lb/>is to the ſquare on KY. </s>
  <s xml:id="echoid-s2541" xml:space="preserve">Now the triangles EVO, IYO being alſo ſimilar,
<pb o="[36]" file="0106" n="113"/>
OY will be to OV as OI is to OE, and (Eu. </s>
  <s xml:id="echoid-s2542" xml:space="preserve">V. </s>
  <s xml:id="echoid-s2543" xml:space="preserve">15. </s>
  <s xml:id="echoid-s2544" xml:space="preserve">16.) </s>
  <s xml:id="echoid-s2545" xml:space="preserve">the rectangle <lb/>contained by VO and YO or its equal, the rectangle contained by AO and <lb/>OU, is to the rectangle contained by EO and OI as the ſquare on OV is <lb/>to the ſquare on EO, as the ſquaree on AU is to the ſquare on KY.</s>
  <s xml:id="echoid-s2546" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2547" xml:space="preserve">Q. </s>
  <s xml:id="echoid-s2548" xml:space="preserve">E. </s>
  <s xml:id="echoid-s2549" xml:space="preserve">D.</s>
  <s xml:id="echoid-s2550" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2551" xml:space="preserve"><emph style="sc">Scholium</emph>. </s>
  <s xml:id="echoid-s2552" xml:space="preserve">It might be obſerved that in the two Caſes of this Epitagma <lb/>where the points A and U are means, the limiting ratio will be a maxi-<lb/>mum inſtead of a minimum; </s>
  <s xml:id="echoid-s2553" xml:space="preserve">and that ratio will be the ſame with that which <lb/>the ſquare on KY bears to the ſquare on EI, as is plain from what hath <lb/>been advanced above.</s>
  <s xml:id="echoid-s2554" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div92" type="section" level="1" n="85">
<head xml:id="echoid-head100" xml:space="preserve">PROBLEM II. (Fig. 46 to 57.)</head>
<p>
  <s xml:id="echoid-s2555" xml:space="preserve">Where O is ſought between a mean and an extream point: </s>
  <s xml:id="echoid-s2556" xml:space="preserve">and here, <lb/>as in the firſt Problem, there are three Epitagmas. </s>
  <s xml:id="echoid-s2557" xml:space="preserve">Firſt when the points <lb/>A and U, which bound the ſegments containing the antecedent rectangle, <lb/>are one an extreme, and the other an alternate mean; </s>
  <s xml:id="echoid-s2558" xml:space="preserve">ſecondly when they <lb/>are both means, or both extremes; </s>
  <s xml:id="echoid-s2559" xml:space="preserve">thirdly when they are one an extreme, <lb/>and the other an adjacent mean.</s>
  <s xml:id="echoid-s2560" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2561" xml:space="preserve"><emph style="sc">Epitagma</emph> I. </s>
  <s xml:id="echoid-s2562" xml:space="preserve">There are here eight Caſes, but they are conſtructed at <lb/>four times, becauſe it is indifferent whether the given ratio be of a leſs to a <lb/>greater, or of a greater to a leſs.</s>
  <s xml:id="echoid-s2563" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2564" xml:space="preserve"><emph style="sc">Case</emph> I. </s>
  <s xml:id="echoid-s2565" xml:space="preserve">The order of the given points being A, E, U, I, as in Fig. </s>
  <s xml:id="echoid-s2566" xml:space="preserve">46, <lb/>make B to fall between A and I, C between U and E, and draw DH <lb/>through the center of the circle on BC; </s>
  <s xml:id="echoid-s2567" xml:space="preserve">and O will fall between A and <lb/>E, becauſe AB is to BO as CO is to CE, and therefore, if AB be greater <lb/>than BO, CO muſt be greater than CE, and if leſs, leſs; </s>
  <s xml:id="echoid-s2568" xml:space="preserve">but this cannot <lb/>be the caſe if O falls either beyond A or E: </s>
  <s xml:id="echoid-s2569" xml:space="preserve">and the like abſurdity fol. <lb/></s>
  <s xml:id="echoid-s2570" xml:space="preserve">lows if o be ſuppoſed to fall otherwiſe than between I and U.</s>
  <s xml:id="echoid-s2571" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2572" xml:space="preserve"><emph style="sc">Case</emph> II. </s>
  <s xml:id="echoid-s2573" xml:space="preserve">Wherein the order of the points is U, I, A, E; </s>
  <s xml:id="echoid-s2574" xml:space="preserve">and it is con-<lb/>ſtructed in the very ſame manner that Caſe I. </s>
  <s xml:id="echoid-s2575" xml:space="preserve">is, as appears by barely <lb/>inſpecting Fig. </s>
  <s xml:id="echoid-s2576" xml:space="preserve">47.</s>
  <s xml:id="echoid-s2577" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2578" xml:space="preserve"><emph style="sc">Case</emph> III. </s>
  <s xml:id="echoid-s2579" xml:space="preserve">If the order of the given points be A, I, U, E (Fig. </s>
  <s xml:id="echoid-s2580" xml:space="preserve">48.) </s>
  <s xml:id="echoid-s2581" xml:space="preserve">the <lb/>points B and C muſt be made to fall as in the two preceding Caſes; </s>
  <s xml:id="echoid-s2582" xml:space="preserve">but <lb/>DH muſt be drawn parallel to BC, and O will fall as required. </s>
  <s xml:id="echoid-s2583" xml:space="preserve">For erect
<pb o="[37]" file="0107" n="114"/>
at I, the perpendicular IG: </s>
  <s xml:id="echoid-s2584" xml:space="preserve">by Eu. </s>
  <s xml:id="echoid-s2585" xml:space="preserve">VI. </s>
  <s xml:id="echoid-s2586" xml:space="preserve">13. </s>
  <s xml:id="echoid-s2587" xml:space="preserve">17. </s>
  <s xml:id="echoid-s2588" xml:space="preserve">the ſquare on IG is equal <lb/>to the rectangle contained by IB and IC; </s>
  <s xml:id="echoid-s2589" xml:space="preserve">and the ſquare on HO is equal <lb/>to the rectangle contained by IB and UC. </s>
  <s xml:id="echoid-s2590" xml:space="preserve">Now IC is by ſuppoſition <lb/>greater than UC, and therefore the rectangle IB, IC is greater than the <lb/>rectangle IB, UC: </s>
  <s xml:id="echoid-s2591" xml:space="preserve">conſequently the ſquare on IG is greater than the <lb/>ſquare on HO, and IG than HO; </s>
  <s xml:id="echoid-s2592" xml:space="preserve">whence O muſt fall between I and B, <lb/>much more between I and A. </s>
  <s xml:id="echoid-s2593" xml:space="preserve">And in the ſame manner it may be proved <lb/>that the point o falls between U and E.</s>
  <s xml:id="echoid-s2594" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2595" xml:space="preserve"><emph style="sc">Case</emph> IV. </s>
  <s xml:id="echoid-s2596" xml:space="preserve">In which the order of the given points is U, E, A, I; </s>
  <s xml:id="echoid-s2597" xml:space="preserve">it is <lb/>conſtructed exactly in the ſame manner as Caſe III, and is exhibited by <lb/>Fig. </s>
  <s xml:id="echoid-s2598" xml:space="preserve">49.</s>
  <s xml:id="echoid-s2599" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2600" xml:space="preserve"><emph style="sc">Epitagma</emph> II. </s>
  <s xml:id="echoid-s2601" xml:space="preserve">There are here only four Caſes, becauſe, as in Epi-<lb/>tagma I. </s>
  <s xml:id="echoid-s2602" xml:space="preserve">it is indifferent whether the given ratio be of a leſs to a greater, <lb/>or of a greater to a leſs; </s>
  <s xml:id="echoid-s2603" xml:space="preserve">and the two laſt of thoſe, viz. </s>
  <s xml:id="echoid-s2604" xml:space="preserve">where the order <lb/>of the given points is E, A, U, I; </s>
  <s xml:id="echoid-s2605" xml:space="preserve">or E, U, A, I, being reducible to the <lb/>two former by reading every where I for A, E for U, and the contrary, <lb/>I ſhall omit ſaying any thing of their conſtructions, except that they are <lb/>exhibited by Fig. </s>
  <s xml:id="echoid-s2606" xml:space="preserve">52 and 53.</s>
  <s xml:id="echoid-s2607" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2608" xml:space="preserve"><emph style="sc">Case</emph> I. </s>
  <s xml:id="echoid-s2609" xml:space="preserve">The order of the given points, being A,E,I,U, make B to <lb/>fall between A and I, C between E and U, and draw DH through the <lb/>center of the circle on BC, as is done in Fig. </s>
  <s xml:id="echoid-s2610" xml:space="preserve">50; </s>
  <s xml:id="echoid-s2611" xml:space="preserve">and O will fall as re-<lb/>quired for reaſons ſimilar to thoſe urged in Caſe I. </s>
  <s xml:id="echoid-s2612" xml:space="preserve">of the firſt Epitagma <lb/>of this Problem.</s>
  <s xml:id="echoid-s2613" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2614" xml:space="preserve"><emph style="sc">Case</emph> II. </s>
  <s xml:id="echoid-s2615" xml:space="preserve">If the order of the given points be A, I, E, U, the conſtruc-<lb/>tion will be as in Fig. </s>
  <s xml:id="echoid-s2616" xml:space="preserve">51, where B and C are made to fall, and DH is <lb/>drawn as in Caſe I.</s>
  <s xml:id="echoid-s2617" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2618" xml:space="preserve"><emph style="sc">Epitagma</emph> III. </s>
  <s xml:id="echoid-s2619" xml:space="preserve">Here there are eight Caſes, viz. </s>
  <s xml:id="echoid-s2620" xml:space="preserve">four where in the <lb/>order of the given points is A, U, E, I; </s>
  <s xml:id="echoid-s2621" xml:space="preserve">A, U, I, E; </s>
  <s xml:id="echoid-s2622" xml:space="preserve">U, A, E, I; </s>
  <s xml:id="echoid-s2623" xml:space="preserve">and <lb/>U, A, I, E, and the given ratio of a greater to a leſs, when O will fall <lb/>between the two given points, which bound the conſequent rectangle; <lb/></s>
  <s xml:id="echoid-s2624" xml:space="preserve">and four others@ wherein the order of the given points is the ſame as <lb/>here, but the given ratio of a leſs to a greater, and in which the point <lb/>O will fall between the points that bound the antecedent rectangle; </s>
  <s xml:id="echoid-s2625" xml:space="preserve">but <lb/>as theſe laſt are reducible to the former by the ſame means which have been <lb/>uſed on former ſimilar occaſions, I ſhall not ſtop to ſpecify them.</s>
  <s xml:id="echoid-s2626" xml:space="preserve"/>
</p>
<pb o="[38]" file="0108" n="115"/>
<p>
  <s xml:id="echoid-s2627" xml:space="preserve">The former four are all conſtructed by making B to fall between A and <lb/>I, C between U and E, and drawing DH parallel to BC; </s>
  <s xml:id="echoid-s2628" xml:space="preserve">and it will ap-<lb/>pear by reaſonings ſimilar to thoſe uſed for the like purpoſe in Caſe III. </s>
  <s xml:id="echoid-s2629" xml:space="preserve">of <lb/>Epitagma I. </s>
  <s xml:id="echoid-s2630" xml:space="preserve">that O muſt fall between E and I as was propoſed. </s>
  <s xml:id="echoid-s2631" xml:space="preserve">See Fig. <lb/></s>
  <s xml:id="echoid-s2632" xml:space="preserve">54, 55, 56 and 57.</s>
  <s xml:id="echoid-s2633" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2634" xml:space="preserve"><emph style="sc">Limitation</emph>. </s>
  <s xml:id="echoid-s2635" xml:space="preserve">In theſe four Caſes, the given ratio of R to S muſt not <lb/>be leſs than that which the ſquare on the ſum of two mean proportionals <lb/>between AE and IU, AI and EU bears to the ſquare on EI. </s>
  <s xml:id="echoid-s2636" xml:space="preserve">For it has <lb/>been proved (Lem. </s>
  <s xml:id="echoid-s2637" xml:space="preserve">II.) </s>
  <s xml:id="echoid-s2638" xml:space="preserve">that when the ratio of the rectangle contained by <lb/>AO and UO to that contained by EO and IO; </s>
  <s xml:id="echoid-s2639" xml:space="preserve">or, which is the ſame thing, <lb/>the given ratio of R to S is the leaſt poſſible, the ſquare on EO will be to <lb/>the ſquare on IO as the rectangle contained by AE and UE is to that con-<lb/>tained by AI and UI; </s>
  <s xml:id="echoid-s2640" xml:space="preserve">and (<emph style="sc">Lem</emph>. </s>
  <s xml:id="echoid-s2641" xml:space="preserve">V. </s>
  <s xml:id="echoid-s2642" xml:space="preserve">Fig. </s>
  <s xml:id="echoid-s2643" xml:space="preserve">31.) </s>
  <s xml:id="echoid-s2644" xml:space="preserve">that FG will then be the <lb/>fum of two mean proportionals between AE and UI, AI and UE: </s>
  <s xml:id="echoid-s2645" xml:space="preserve">it <lb/>therefore only remains to prove that the rectangle contained by AO and <lb/>UO is to that contained by EO and IO as the ſquare on FG is to the <lb/>fquare on EI. </s>
  <s xml:id="echoid-s2646" xml:space="preserve">Now it has been proved in demonſtrating Lem. </s>
  <s xml:id="echoid-s2647" xml:space="preserve">V. </s>
  <s xml:id="echoid-s2648" xml:space="preserve">that the <lb/>triangles EOG and IOF are ſimilar, and that the angle at V is right, <lb/>whence it follows that the triangles AOG and FOU are alſo ſimilar, and <lb/>conſequently that AO is to OG as OF is to UO; </s>
  <s xml:id="echoid-s2649" xml:space="preserve">therefore the rectangle <lb/>contained by AO and UO is equal to that contained by GO and OF. </s>
  <s xml:id="echoid-s2650" xml:space="preserve">More-<lb/>over GO is to OF as EO is to IO, and ſo by compoſition and permutation, <lb/>FG is to EI as OG is to EO, and as OF is to IO: </s>
  <s xml:id="echoid-s2651" xml:space="preserve">hence by compound <lb/>ratio the ſquare on FG is to the ſquare on EI as the rectangle contained <lb/>by (OG and OF) AO and UO is to that contained by EO and IO.</s>
  <s xml:id="echoid-s2652" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2653" xml:space="preserve">Q. </s>
  <s xml:id="echoid-s2654" xml:space="preserve">E. </s>
  <s xml:id="echoid-s2655" xml:space="preserve">D.</s>
  <s xml:id="echoid-s2656" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2657" xml:space="preserve"><emph style="sc">Scholium</emph>. </s>
  <s xml:id="echoid-s2658" xml:space="preserve">In the four Caſes, wherein the given ratio is of a leſs to a <lb/>greater, and wherein the point O muſt fall between thoſe given ones which <lb/>bound the antecedent rectangle, the limiting ratio will be a maximum, and <lb/>the ſame with that which the ſquare on AU bears to the ſquare on FG.</s>
  <s xml:id="echoid-s2659" xml:space="preserve"/>
</p>
<pb o="[39]" file="0109" n="116"/>
</div>
<div xml:id="echoid-div93" type="section" level="1" n="86">
<head xml:id="echoid-head101" xml:space="preserve">PROBLEM III.</head>
<p>
  <s xml:id="echoid-s2660" xml:space="preserve">In this, the point O is ſought without all the given ones, and the three <lb/>Epitagmas are as in Problem I.</s>
  <s xml:id="echoid-s2661" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2662" xml:space="preserve"><emph style="sc">Epitagma</emph> I. </s>
  <s xml:id="echoid-s2663" xml:space="preserve">There are here eight Caſes, viz. </s>
  <s xml:id="echoid-s2664" xml:space="preserve">four when the order of <lb/>the given points is the ſame as ſpecified in Epitagma I. </s>
  <s xml:id="echoid-s2665" xml:space="preserve">of Problem I, and <lb/>O ſought beyond the given point which bounds the antecedent rectangle; <lb/></s>
  <s xml:id="echoid-s2666" xml:space="preserve">and four others when O is ſought beyond that which bounds the conſe-<lb/>quent one: </s>
  <s xml:id="echoid-s2667" xml:space="preserve">the Conſtructions of the four firſt are ſhewn by the ſmall <lb/>letters b and o in Fig. </s>
  <s xml:id="echoid-s2668" xml:space="preserve">32, 34, 36 and 38; </s>
  <s xml:id="echoid-s2669" xml:space="preserve">and the four latter ones by <lb/>the ſame letters in Fig. </s>
  <s xml:id="echoid-s2670" xml:space="preserve">33, 35, 37 and 39; </s>
  <s xml:id="echoid-s2671" xml:space="preserve">and the demonſtrations that <lb/>o will fall as required by the Problem are exactly the ſame as thoſe made <lb/>uſe of in the laſt mentioned Epitagma. </s>
  <s xml:id="echoid-s2672" xml:space="preserve">It is farther obſervable, that the <lb/>four firſt Caſes are not poſſible, unleſs the given ratio be of a leſs to a <lb/>greater; </s>
  <s xml:id="echoid-s2673" xml:space="preserve">nor the four latter, unleſs it be of a greater to a leſs, as is ma-<lb/>nifeſt without farther illuſtration.</s>
  <s xml:id="echoid-s2674" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2675" xml:space="preserve"><emph style="sc">Epitagma</emph> II. </s>
  <s xml:id="echoid-s2676" xml:space="preserve">Here, as in the ſecond Epitagma of Problem I, the points <lb/>A and U are one an extreme, and the other an adjacent mean, and there <lb/>are eight Caſes; </s>
  <s xml:id="echoid-s2677" xml:space="preserve">but it will be ſufficient to exhibit the conſtructions of <lb/>four of them, the others being not eſſentially different; </s>
  <s xml:id="echoid-s2678" xml:space="preserve">and theſe are ſhewn <lb/>by the ſmall b and o in Fig. </s>
  <s xml:id="echoid-s2679" xml:space="preserve">40, 41, 42 and 43; </s>
  <s xml:id="echoid-s2680" xml:space="preserve">the demonſtrations that o <lb/>will fall as required need not be pointed out here; </s>
  <s xml:id="echoid-s2681" xml:space="preserve">but it may be neceſſary <lb/>to remark that the firſt and third are not poſſible unleſs the given ratio be <lb/>of a leſs to a greater, nor the ſecond and fourth unleſs it be of a greater to <lb/>a leſs, as is obvious enough.</s>
  <s xml:id="echoid-s2682" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2683" xml:space="preserve"><emph style="sc">Epitagma</emph> III. </s>
  <s xml:id="echoid-s2684" xml:space="preserve">In which the points A and U are both means, or both <lb/>extremes; </s>
  <s xml:id="echoid-s2685" xml:space="preserve">and there are here eight Caſes, viz. </s>
  <s xml:id="echoid-s2686" xml:space="preserve">four wherem theſe points <lb/>are extremes, and four others wherein they are means: </s>
  <s xml:id="echoid-s2687" xml:space="preserve">but theſe laſt being <lb/>reducible to the former by the ſame method that was uſed in the third Epi-<lb/>tagmas of the two preceding Problems, I ſhall omit them.</s>
  <s xml:id="echoid-s2688" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2689" xml:space="preserve">All the Caſes of this Epitagma are conſtructed by making B fall beyond <lb/>I, and C beyond E, with reſpect to A and U; </s>
  <s xml:id="echoid-s2690" xml:space="preserve">and drawing DH parallel to <lb/>BC. </s>
  <s xml:id="echoid-s2691" xml:space="preserve">That O will fall beyond A in Fig. </s>
  <s xml:id="echoid-s2692" xml:space="preserve">58 and 60, and beyond U in Fig. <lb/></s>
  <s xml:id="echoid-s2693" xml:space="preserve">59 and 61 appears hence. </s>
  <s xml:id="echoid-s2694" xml:space="preserve">Draw AG perpendicular to BC, meeting the <lb/>circle on BC in G: </s>
  <s xml:id="echoid-s2695" xml:space="preserve">by Eu. </s>
  <s xml:id="echoid-s2696" xml:space="preserve">VI. </s>
  <s xml:id="echoid-s2697" xml:space="preserve">13. </s>
  <s xml:id="echoid-s2698" xml:space="preserve">17, the ſquare on AG is equal to the
<pb o="[40]" file="0110" n="117"/>
rectangle contained by AB and AC; </s>
  <s xml:id="echoid-s2699" xml:space="preserve">but the ſquare on HO is equal to <lb/>the rectangle contained by AB and EC: </s>
  <s xml:id="echoid-s2700" xml:space="preserve">now EC is, by ſuppoſition, <lb/>greater than AC, therefore the rectangle AB, EC is greater than the <lb/>rectangle AB, AC, and the ſquare on HO greater than the ſquare on AG, <lb/>conſequently HO is itſelf greater than AG; </s>
  <s xml:id="echoid-s2701" xml:space="preserve">but this could not be the <lb/>Caſe unleſs O fell beyond A. </s>
  <s xml:id="echoid-s2702" xml:space="preserve">In the ſame manner my it be proved that O <lb/>will fall beyond U in Fig. </s>
  <s xml:id="echoid-s2703" xml:space="preserve">59 and 60.</s>
  <s xml:id="echoid-s2704" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2705" xml:space="preserve"><emph style="sc">Limitation</emph>. </s>
  <s xml:id="echoid-s2706" xml:space="preserve">In the above four Caſes the given ratio of R to S muſt <lb/>not exceed that which the ſquare on AU bears to the ſquare on the ſum of <lb/>two mean proportionals between AI and UE, AE and UI. </s>
  <s xml:id="echoid-s2707" xml:space="preserve">For (Fig. </s>
  <s xml:id="echoid-s2708" xml:space="preserve">30.) <lb/></s>
  <s xml:id="echoid-s2709" xml:space="preserve">demit from A, on KO produced, the perpendicular AH. </s>
  <s xml:id="echoid-s2710" xml:space="preserve">Now it has been <lb/>proved (Lem. </s>
  <s xml:id="echoid-s2711" xml:space="preserve">III.) </s>
  <s xml:id="echoid-s2712" xml:space="preserve">that the ratio of the rectangle continued by AO and UO <lb/>to that contained by EO and IO, or which is the ſame thing, the given <lb/>ratio of R to S is the greateſt poſſible; </s>
  <s xml:id="echoid-s2713" xml:space="preserve">and (Lem. </s>
  <s xml:id="echoid-s2714" xml:space="preserve">IV.) </s>
  <s xml:id="echoid-s2715" xml:space="preserve">that KF is a mean <lb/>proportional between AI and UE, alſo that YF is a mean proportional <lb/>between AE and UI: </s>
  <s xml:id="echoid-s2716" xml:space="preserve">but HK is equal to YF, therefore HF is equal to <lb/>the ſum of two mean proportionals between AI and UE, AE and UI; </s>
  <s xml:id="echoid-s2717" xml:space="preserve"><lb/>it only then remains to prove, that the rectangle contained by AO and UO <lb/>is to that contained by EO and IO as the ſquare on AU is to the ſquare <lb/>on HF. </s>
  <s xml:id="echoid-s2718" xml:space="preserve">The triangles OEK, OHA, OIY and OUF are all ſimilar; </s>
  <s xml:id="echoid-s2719" xml:space="preserve">con-<lb/>ſequently OK is to OE as OA is to OH, as OY is to OI, and therefore <lb/>by compound ratio, the rectangle contained by AO and UO (OK and <lb/>OY) is to that contained by EO and IO as the ſquare on AO is to the <lb/>ſquare on OH; </s>
  <s xml:id="echoid-s2720" xml:space="preserve">but alſo AO is to UO as HO is to EO, and by compoſi-<lb/>tion and permutation, AU is to HF as AO is to HO, or (Eu. </s>
  <s xml:id="echoid-s2721" xml:space="preserve">VI. </s>
  <s xml:id="echoid-s2722" xml:space="preserve">22.) </s>
  <s xml:id="echoid-s2723" xml:space="preserve"><lb/>the ſquare on AU is to the ſquare on HF as the ſquare on AO is to the <lb/>ſquare on HO, and ſo by equality of ratios, the rectangle contained by AO <lb/>and UO is to that contained by EO and IO as the ſquare on AU is to the <lb/>ſquare on HF.</s>
  <s xml:id="echoid-s2724" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2725" xml:space="preserve">Q.</s>
  <s xml:id="echoid-s2726" xml:space="preserve">E.</s>
  <s xml:id="echoid-s2727" xml:space="preserve">D.</s>
  <s xml:id="echoid-s2728" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2729" xml:space="preserve"><emph style="sc">Scholium</emph>. </s>
  <s xml:id="echoid-s2730" xml:space="preserve">In the four Caſes wherein the points A and U are means, <lb/>the limiting ratio will be a minimum, and the ſame with that which the <lb/>ſquare on HF bears to the ſquare on EI.</s>
  <s xml:id="echoid-s2731" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div94" type="section" level="1" n="87">
<head xml:id="echoid-head102" xml:space="preserve">THE END.</head>
<pb file="0111" n="118"/>
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<pb file="0111a" n="119"/>
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<pb file="0112" n="120"/>
<pb file="0113" n="121"/>
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    <image file="0113-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0113-01"/>
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<pb file="0113a" n="122"/>
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    <image file="0113a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0113a-01"/>
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<pb file="0114" n="123"/>
<pb file="0115" n="124"/>
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    <image file="0115-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0115-01"/>
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<pb file="0115a" n="125"/>
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    <image file="0115a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0115a-01"/>
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<pb file="0116" n="126"/>
<pb file="0117" n="127"/>
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    <image file="0117-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0117-01"/>
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<pb file="0117a" n="128"/>
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    <image file="0117a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0117a-01"/>
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<pb file="0118" n="129"/>
<pb file="0119" n="130"/>
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    <image file="0119-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0119-01"/>
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<pb file="0119a" n="131"/>
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    <image file="0119a-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0119a-01"/>
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<pb file="0120" n="132"/>
<pb file="0121" n="133"/>
</div>
<div xml:id="echoid-div95" type="section" level="1" n="88">
<head xml:id="echoid-head103" xml:space="preserve">A</head>
<head xml:id="echoid-head104" xml:space="preserve">SYNOPSIS <lb/>OF ALL THE DATA FOR THE <lb/>Conſtruction of <emph style="sc">Triangles</emph>, <lb/>FROM WHICH <lb/>GEOMETRICAL SOLUTIONS</head>
<head xml:id="echoid-head105" xml:space="preserve"><emph style="sc">Have hitherto been in</emph> <emph style="sc">Print</emph>.</head>
<p>
  <s xml:id="echoid-s2732" xml:space="preserve">With References to the Authors, where thoſe <emph style="sc">Solutions</emph> are to be found:</s>
  <s xml:id="echoid-s2733" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div96" type="section" level="1" n="89">
<head xml:id="echoid-head106" xml:space="preserve">By JOHN LAWSON, B. D. <lb/>Rector of <emph style="sc">Swanscombe</emph>, in KENT.</head>
<head xml:id="echoid-head107" xml:space="preserve">ROCHESTER:</head>
<p>
  <s xml:id="echoid-s2734" xml:space="preserve">Printed by T. </s>
  <s xml:id="echoid-s2735" xml:space="preserve"><emph style="sc">Fisher</emph>; </s>
  <s xml:id="echoid-s2736" xml:space="preserve">and Sold by J. </s>
  <s xml:id="echoid-s2737" xml:space="preserve"><emph style="sc">Nourse</emph>, B. </s>
  <s xml:id="echoid-s2738" xml:space="preserve"><emph style="sc">White</emph>, T. </s>
  <s xml:id="echoid-s2739" xml:space="preserve"><emph style="sc">Payne</emph>, and J. </s>
  <s xml:id="echoid-s2740" xml:space="preserve"><emph style="sc">Wilkie</emph>, in London@</s>
</p>
</div>
<div xml:id="echoid-div97" type="section" level="1" n="90">
<head xml:id="echoid-head108" xml:space="preserve">MDCCLXXIII. <lb/>[Price <emph style="sc">One</emph> <emph style="sc">Shilling</emph>.]</head>
<pb file="0122" n="134"/>
<pb file="0123" n="135"/>
</div>
<div xml:id="echoid-div98" type="section" level="1" n="91">
<head xml:id="echoid-head109" xml:space="preserve">ADVERTISEMENT.</head>
<p>
  <s xml:id="echoid-s2741" xml:space="preserve">IT is but few years ago ſince the Compiler of this Synopſis <lb/>conceived his firſt idea of the uſeſulneſs of ſuch an undertaking, <lb/>and he exhibited a ſmall ſpecimen thereof in a periodical Work <lb/>then publiſhing under the Title of <emph style="sc">The</emph> <emph style="sc">British</emph> <emph style="sc">Oracle</emph>. </s>
  <s xml:id="echoid-s2742" xml:space="preserve">Here-<lb/>upon he received ſeveral letters from Mathematical friends, ex-<lb/>preſſing their ſenſe of the great propriety of ſuch a collection, and <lb/>ſtrongly encouraging him to purſue the undertaking. </s>
  <s xml:id="echoid-s2743" xml:space="preserve">Since that <lb/>time it has been a growing work, and would continue ſo, were <lb/>the publication delayed ever ſo long, as freſh Problems are con-<lb/>tinually propoſed to the public. </s>
  <s xml:id="echoid-s2744" xml:space="preserve">He has therefore now determined <lb/>to ſend it abroad, as complete as he can make it to the preſent <lb/>period, and leave additions to be made by future collectors.</s>
  <s xml:id="echoid-s2745" xml:space="preserve"/>
</p>
<pb file="0124" n="136"/>
<pb file="0125" n="137"/>
</div>
<div xml:id="echoid-div99" type="section" level="1" n="92">
<head xml:id="echoid-head110" xml:space="preserve">AN <lb/>EXPLANATION <lb/>OF THE <lb/>SYMBOLS made uſe of in this SYNOPSIS.</head>
<note position="right" xml:space="preserve"> <lb/>H. # repreſents the Hypothenuſe of a right-angled triangle. <lb/>V. # Vertical angle. <lb/>B. # Baſe or ſide oppoſite V. <lb/>P. # Perpendicular from V on B. <lb/>S&amp;s. # Sides about V; S the greater, s the leſs. <lb/>A &amp; a. # Angles at B; A the greater, a the leſs. <lb/>m &amp; n. # Segments of B by P; m the greater, n the leſs. <lb/>Ar. # Area. <lb/>Per. # Perimeter. <lb/>L. # Line from V to B, biſecting V. <lb/>λ. # Line from V to B, cutting B in a given ratio. <lb/>l. # Any other line ſpecified how drawn. <lb/>R. # Radius of inſcribed circle. <lb/>☉(sun). # Circle. <lb/>◻. # Square. <lb/>: # Ratio: thus, S: s ſignifies the ratio of the ſides. <lb/></note>
<pb o="[ii]" file="0126" n="138"/>
<p>
  <s xml:id="echoid-s2746" xml:space="preserve">N. </s>
  <s xml:id="echoid-s2747" xml:space="preserve">B. </s>
  <s xml:id="echoid-s2748" xml:space="preserve">Between each of the Data a full ſtop is placed. </s>
  <s xml:id="echoid-s2749" xml:space="preserve">Moreover, m and n are <lb/>ſometimes uſed for different ſegments than thoſe of B by P, but then it is ſignified in <lb/>words: </s>
  <s xml:id="echoid-s2750" xml:space="preserve">- the ſame alſo is to be obſeved of P.</s>
  <s xml:id="echoid-s2751" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2752" xml:space="preserve">Obſerve likewiſe, for the more ready finding any propoſed Problem in the Synop-<lb/>ſis, that the data are ranged in the ſame order as the Symbols here recited; </s>
  <s xml:id="echoid-s2753" xml:space="preserve">viz. <lb/></s>
  <s xml:id="echoid-s2754" xml:space="preserve">all data whereof V is one are placed firſt; </s>
  <s xml:id="echoid-s2755" xml:space="preserve">next thoſe where B is given, either ſim-<lb/>ply by itſelf, or combined with any other datum; </s>
  <s xml:id="echoid-s2756" xml:space="preserve">next P, &amp;</s>
  <s xml:id="echoid-s2757" xml:space="preserve">c.</s>
  <s xml:id="echoid-s2758" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2759" xml:space="preserve">Moreover, when in the references you find this mark*, it ſignifies that ſuch Authors <lb/>have only conſtructed the Problem partially, and not generally as propoſed in the Synop-<lb/>ſis, e. </s>
  <s xml:id="echoid-s2760" xml:space="preserve">g. </s>
  <s xml:id="echoid-s2761" xml:space="preserve">for a right-angled triangle, when it is propoſed for a triangle in general; <lb/></s>
  <s xml:id="echoid-s2762" xml:space="preserve">and again, for a line biſecting another, when it is propoſed to cut it ſo that the ſeg-<lb/>ments may be in any given ratio.</s>
  <s xml:id="echoid-s2763" xml:space="preserve"/>
</p>
<pb o="[i]" file="0127" n="139"/>
</div>
<div xml:id="echoid-div100" type="section" level="1" n="93">
<head xml:id="echoid-head111" xml:space="preserve">INDEX <lb/>OF THE <lb/>Authors refered to in the SYNOPSIS.</head>
<note position="right" xml:space="preserve"> <lb/>ANDERSONI Var. Prob. Practice, cum Supplemento Apollonii <lb/># Redivivi, 4to. # Pariſiis 1612 <lb/>Anderſoni Exercitationum Math. Deas ima. 4to. # ibid. 1619 <lb/>Aſhby’s Algebra, 2d Edit. 12mo. # Lond. 1741 <lb/>Britiſh Oracle (Vol. I. being all that was publiſhed) 12mo. # ibid. 1769 <lb/>Caſtillioneus inNewtoni Arith. Univerſalem, 4to. # Amſtelod. 1761 <lb/>Clavius in Euclidem, var. Ed. <lb/>Court Magazine. <lb/>D’Omerique (Hugonis) Analyſis Geometrica, 4to. # Gadibus 1699 <lb/>Diarian Repoſitory, Periodical Work, printed for Robinſon, 4to. # Lond. 1770, &amp;c. <lb/>Foſter’s Miſcellanies, or Math. Lugubrations, fo. # ibid. 1659 <lb/>General Magazine. <lb/>Gentleman’s Magazine. <lb/>Gentleman’s Diary. <lb/>Ghetaldi (Marini) Var. Prob. Collectio, 4to. # Venetiis 1607 <lb/>Ghetaldus de Reſolutione &amp; Compoſitione Math. fo. # Romæ 1640 <lb/>Gregorius a Sancto Vincentio, fo. # Antverp. 1647 <lb/>Herigoni Curſus Math. Lat. &amp; Gallicè, 8vo. 5 Tom. # Paris 1644 <lb/>Hutton’s Ladies Diaries. <lb/># Mathematical Miſcellany. <lb/>Imperial Magazine. <lb/>Ladies Diaries. <lb/>Martin’s Math. Correſpondence, in his Magazine. <lb/>Mathematical Magazine. <lb/>Mathematician, Periodical Work. 8vo. # Lond. 1751
<pb o="[ii]" file="0128" n="140"/>
Miſcellanea Curioſæ, Periodical Work. 8vo. 6 Nos. # rork 1734-5 <lb/>Miſcellanea Cur. Math. Period Work, by Holliday, 4to. 2 vols. # Lond. 1745 <lb/>Miſcellanea Scientiſica Curioſa, Period. Work, 4to. # ibid. 1766 <lb/>Oughtred’s Clavis, var. Editions Lat. &amp; Eng. <lb/>Palladium, Periodical Work. <lb/>Pappus Alexandrinus Commandini, fo. # Bononiæ 1660 <lb/>Regiomontanus de Triangulis, fo. # Baſiliæ 1561 <lb/>Renaldinus (Carolus) de Res. &amp; Comp. Math. fo. # Patavii 1668 <lb/>Ronayne’s Algebra, 2d Edit. 8vo. # Lond. 1727 <lb/>Rudd’s Practical Geometry in 2 Parts, 4to. # ibid. 1650 <lb/>Saunderſon’s Algebra, 2 vols. 4to. # Camb. 1740 <lb/>Schooten’s (Franciſcus à) Exercitationes Math. 4to. # Lug. Bat. 1657 <lb/>Simpſon’s Algebra, 8vo. 3d Edit. \\ 1ſt Edit. # Lond. 1767 \\ ibid. 1745 <lb/># Select Exerciſes, 8vo. # ibid. 1752 <lb/># Geometry. 8vo. 2d Edit. # ibid. 1760 <lb/>Supplement to Gentleman’s Diary, 3 Nos. 12mo. # ibid. 1743, &amp;c. <lb/>Town and Country Magazine. <lb/>Turner’s Mathematical Exerciſes, Periodical Work, 8vo. # ibid. 1750 <lb/>Univerſal Magazine. <lb/>Vietæ Opera, fo. # Lug. Bat. 1646 <lb/>Weſt’s Mathematics, 2d Edit. 8vo. # Lond. 1763 <lb/>Wolfius’s Algebra, tranſlated by Hanna, 8vo. # ibid. 1739 <lb/></note>
  <figure>
    <image file="0128-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0128-01"/>
  </figure>
</div>
<div xml:id="echoid-div101" type="section" level="1" n="94">
<head xml:id="echoid-head112" style="it" xml:space="preserve">Lately was publiſhed by the ſame <emph style="sc">Author</emph>;</head>
<head xml:id="echoid-head113" xml:space="preserve">[Price <emph style="sc">Six</emph> <emph style="sc">Shillings</emph> in Boards.]</head>
<p>
  <s xml:id="echoid-s2764" xml:space="preserve">APOLLONIUS concerning <emph style="sc">Tangencies</emph>, as reſtored by Vieta &amp; </s>
  <s xml:id="echoid-s2765" xml:space="preserve">Ghetaldus, <lb/>with a Supplement; </s>
  <s xml:id="echoid-s2766" xml:space="preserve">the 2d Edit. </s>
  <s xml:id="echoid-s2767" xml:space="preserve">To which is now added a Second Supplement, <lb/>being Fermat’s Treatiſe on <emph style="sc">Spherical</emph> <emph style="sc">Tangencies</emph>. </s>
  <s xml:id="echoid-s2768" xml:space="preserve">Likewiſe Apollonius con-<lb/>cerning <emph style="sc">Determinate</emph> <emph style="sc">Section</emph>, as reſtored by Willebrordus Snellius; </s>
  <s xml:id="echoid-s2769" xml:space="preserve">to which <lb/>is added an entire new Work, being the ſame reſtored by Mr. </s>
  <s xml:id="echoid-s2770" xml:space="preserve">W. </s>
  <s xml:id="echoid-s2771" xml:space="preserve">Wales.</s>
  <s xml:id="echoid-s2772" xml:space="preserve"/>
</p>
<pb o="[1]" file="0129" n="141"/>
</div>
<div xml:id="echoid-div102" type="section" level="1" n="95">
<head xml:id="echoid-head114" xml:space="preserve">SYNOPSIS.</head>
<note position="right" xml:space="preserve"> <lb/>1. V. B. P. # SIMPSON’s Alg. pr. 5. - Mis. Cur. Math. Vol. I. \\ page 31. - Vieta Iſt. Ap. to Apollonius Gallus, pr. 5. \\ - *Aſhby’s Alg. pag. 111. - *Rudd’s Pract Geo. part 2d, qu. \\ 4.-L. Diary, qu. 160. <lb/>2. V. B. P: m. # Town and Country Mag. Nov. and Dec. 1772. <lb/>3. V. B. P±m. # Mathematician, pr. 77. - Univerſal Mag. Mar. 1749. <lb/>4. V. B. S: s. # Simpſon’s Alg. pr. 3. - Simp. Geom. pr. 13. - Pappus Lib. \\ VII. pr. 155. - Herigon App. Geometriæ planæ, pr. 13. - \\ D’Omerique Lib. III. pr. 35. - Court Mag. July, 1762. <lb/>5. V. B. S + s. # Simpſon’s Alg. pr. 1. - Ghetaldus var. prob. 13 &amp; *7. - Ghetaldus \\ deRes. &amp; Comp. Math. Lib. V. c. 4, pr. 4, pag. 337, &amp; *Lib. II. \\ pr. 9, pag. 93. - Renaldinus pag. 318, 326, 524, 79. - \\ Saunderſon’s Alg. art. 332. - *Rudd’s Prac. Geom. part \\ 2d, qu. 38. - *Aſhby’s Alg. pag. 102. <lb/>6. V. B. S - s. # Simpſon’s Alg. pr. 2. - Ghetaldus var. prob. 12 &amp; *6. - \\ Ghetaldus de Res. &amp; Comp. Lib. V. c. 4, pr. 3, and *Lib. II. pr. 8. \\ - Renaldinus pag. 317, 326, 527, 79. - *Simpſon’s Sel. Ex. pr. \\ 2. - *Wolfius’s Alg. pr. 12 <lb/>8. - *Aſhby’s Alg. pag. 106. <lb/>7. V. B. <emph style="ol">S + s</emph> x S. # *Town and Country Mag. Jan. and Feb. 1769. <lb/>8. V. B. S + s + P. # Arith. Univ. Caſtillionei, pr. 5. <lb/>9. V. B. Ar. # *Simpſon’s Alg. pr. 33. - Simp. Geom. pr. 5. - *Saunder - \\ ſon’s Alg. art. 329. - *D’Omerique L. III. pr. 36. - *Oughtred’s <lb/></note>
<pb o="[2]" file="0130" n="142"/>
<note position="right" xml:space="preserve"> <lb/># Clavis, ch. 19, pr. 24. - *Vieta Geo. Eff. pr. 20. - *Herigon \\ App. Geo. planæ, pr. 10. - *Rudd’s Pract. Geom. part 2d, \\ qu. 47. - *Palladium, 1754, pa. 22. <lb/>10. V. B. Per. # Reducible to V. B. S + s. <lb/>11. V. B. L. # Simpſon’s Alg. pr. 72. - Simp. Geom. pr. 21. <lb/>12. V. B. λ. # *Simpſon’s Alg. pr. 58. - *Ghetaldus var. prob. 3. - Regio - \\ montanus de triangulis, Lib. II. pr. 29. <lb/>13. V. B. Direction \\ of l thro’ V. # }Simpſon’s Sel. Ex. pr. 48. <lb/>14. V. B. R. # *Simpſon’s Sel. Ex. pr. 29. - Br. Oracle, qu. 67, Cor. - \\ *Weſt’s Mathematics, 2d. Ed. pag. 45. <lb/>15. V. B. Side of \\ ins. ◻. # }Mathematician, pr. 25. - Br. Oracle, qu. 19. <lb/>16. V. B: P. S±s. # Brit. Oracle, qu. 31. <lb/>17. V. B±P. S±s. # Arith. Un. Caſtillionei, pr. 7. - *Simpſon’s Sel. Ex. pr. 31. - \\ *Simpſon’s Alg. Iſt. Ed. pr. 81. <lb/>18. V. B±P. Ar. # *Turner’s Math. Ex. pr. 18. <lb/>19. V. B±S. s. # Ghetaldus var. prob. 16, 17. - *Idem, 10, II. - Idem de \\ Res. &amp;. Comp. Lib. V. cap. 4, pr. 7, 8. - *Anderſon var. prob. 2; \\ - *Simp. Sel. Ex. pr. 1. - Twyſden in Foſter’s Miſcell. pr. I. \\ - Renaldinus pag. 526, 529, *437. - *Mathematician, pr. \\ 42. - *Aſhby’s Alg. pr. 31, 32. - and in other places. <lb/>20. V. B±S. S±s. # Br. Oracle, qu. 102. - *Court Mag. Nov. 1761. - L. Diary, \\ qu. 661. - *Hutton’s L. Diary, qu. 147. <lb/>21. V. B±S. B±s. # *Clavius’s Euclid at end of B. II. - *Ghetaldus var. prob. 23. \\ - *Schooten pr. 37. - *Oughtred ch. 19, pr. 17. - *Saun - \\ derſon, art. 327. - *Anderſon var. prob. 15, 16. - *L. Diary, \\ 1770, p. 35. <lb/>22. V. B<emph style="sub">2</emph>: m<emph style="sub">2</emph>. P. # *Mis. Scient. Cur. qu. 54. <lb/></note>
<pb o="[3]" file="0131" n="143"/>
<note position="right" xml:space="preserve"> <lb/>23. V. Point in B. \\ : \\ {S±s.\x} # }Simpſon’s Sel. Ex. pr. 43. - Simp. Geom. pr. 19, 20. <lb/>24. V. S + s - B. P. # Hutton’s Miscellany, qu. 5. <lb/>25. V. S + s - B. Ar. # Math. Mag. No. III. pr. 4. <lb/>26. V. B + S - s. \\ S + s + m - n. # }Gent. Mag. 1768, pag. 428, 519. <lb/>27. V. B±S. Per. # *Anderſon var. prob. 1, 3. - See V. B±S. s. <lb/>28. V. B ‖ to a given \\ line. Ar. # }Simpſon’s Sel. Ex. pr. 41. - Simp. Geom. pr. 4. <lb/>29. V. Pointin B. Ar. # Simpſon’s Sel. Ex. pr. 42. <lb/>30. V. ∠ of B with λ. λ. # *Mathematician, pr. II. - *York Mis. Cur. qu. 15. <lb/>31. V. P. S: s. # Simp. Alg. pr. 3. - Court Mag. Octo. 1762. <lb/>32. V. P. S±s. # Simp. Alg. pr. 80, 78. - *Oughtred ch. 19, pr. 9, 10. - \\ *Ghetaldus var. prob. 20, 21. - *De Res. &amp; Comp. Lib. III. \\ pr. 3, 4. - Diarian Repoſitory, pag. 24. - *Hutton’s L. Diaries, \\ qu. 22. - *Wolfius’s Alg. pr. 127. - *Ronayne’s Alg. B. II. c. 2, \\ pr. 2. <lb/>33. V. P. S x s. # D’Omerique Lib. I. pr. 31. - Gent. D. qu. 149. <lb/>34. V. P. m: n. # Simp. Alg. pr. 11. <lb/>35. V. P. m - n. # Simp. Alg. pr. 10. - Foſter’s Lug. pr. 2. - *Rudd’s Pr. \\ Geom. part 2d, qu. 5. <lb/>36. V. P. Per. # Simp. Alg. pr. 60. - Ronayne’s Alg. B. II. ch. 1, pr. 2. - \\ Arith. Univ. Caſtillionei, pr. 4. - Mis. Cur. Math. qu. 61. <lb/>37. V. P. λ. # *Town and C. Mag. 1769, pag. 296, 381. <lb/>38. V. P. Ifrom A or a \\ to bis. S or s. # }*Br. Oracle, qu. 74. <lb/>39. V. P. R. # *Br. Oracle, qu. 51. <lb/>40. V. P. Ra. of cir- \\ cum. ☉(sun). # }Gent. Diary, 1767, qu. 300. <lb/>41. V. P + s. S - s: \\ m - n. # }Gent. Diary, 1751, qu. 108. <lb/></note>
<pb o="[4]" file="0132" n="144"/>
<note position="right" xml:space="preserve"> <lb/>42. V. S or s. S x s: \\ m x n, ſegts. by L. # }Brit. Oracle, qu. 87. - Renaldinus, pa. 337. <lb/>43. V. S x s. m x n. # Brit. Oracle, qu. 60. <lb/>44. V. S: s. m - n. # Simpſon’s Alg. pr. 3. <lb/># ☞ When V and S: s are given, the triangle is given in \\ ſpecies, and therefore may be conſtructed with any other \\ datum which does not affect the angles. <lb/>45. V. S: s. R. # Br. Oracle, qu. 21. <lb/>46. V. S±s. m: n. # Simpſon’s Alg. pr. 7, 9. - *York Mis. Cur. qu. 46. <lb/>47. V. S±s. m - n. # Simpſon’s Alg. pr. 6, 8. - *Oughtred ch. 19, pr. 12, 13. - \\ *Ghetaldus var. pr. 8, 9. - Idem, pr. 14, 15. - Renaldinus \\ pag. 319, 529. <lb/>48. V. S±s. Ar. # Anderſon var. prob. 22. - *Simp. Alg. pr. 34. <lb/>49. V. S±s. L. # Deducible from Simpſon’s Geo. pr. 19, 20. <lb/>50. V. S - s. R. # *Br. Oracle, qu. 20. <lb/>51. V. S<emph style="sub">2</emph> + s<emph style="sub">2</emph>. λ. # Hutton’s Mis. qu. 24. <lb/>52. V. m. n. # Simpſon’s Alg. pr. 4. - York Mis. Cur. qu. 17, Caſe 3d. - \\ *Rudd’s Pract. Geo. part 2d, qu. 2. - Court Mag. Feb. 1763, \\ m and n being ſegts. by L. <lb/>53. V. Ar. Per. # *D’Omerique L. III. pr. 34. - *Simpſon’s Alg. pr. 35. - \\ *Simpſon’s Sel. Ex. pr. 30. - Arith. Uni. Caſtillionei, pr. 8, \\ *3. - *Rudd’s Prac. Geo. part 2d, qu. 7. - L. Diary, 1761, \\ qu. 480. - Mathematician, qu. 49. - Gent. Diary, 1744, qu. \\ 40. - *Wolfius’s Alg. pr. 113. <lb/>54. V. Ar. Side of \\ ins. ◻. # }L. Diary, 1763, qu. 507. - Court Mag. Dec. 1762. <lb/>55. V. Per. L. # Reducible to V. P. Per. <lb/>56. V. Per. R. # *Rudd’s Prac. Geo. part Iſt, qu. 19. - Reducible to \\ V. B. S + s. <lb/>57. V. L. m: n, \\ ſegts. by L. # }Mis. Cur. Math. V. I. qu. 26. <lb/>58. V. L. m - n, \\ ſegts. by L. # }L. Diary, 1773, qu. 662. <lb/></note>
<pb o="[5]" file="0133" n="145"/>
<note position="right" xml:space="preserve"> <lb/>59. V. L. Side of \\ ins. ◻. # }T. and Country Mag. Nov. and Dec. 1772. <lb/>60. V. λ. Ra. of \\ circ. ☉(sun). # }*Gent. Diary, 1766, qu. 282. - Reducible to V. B. λ. <lb/>61. B. P. S: s. # D’Omerique L. 1. 49. - Vieta 1ſt. App. Apoll. Galli, pr. 2. - \\ Ghetaldus de Res. &amp; Comp. L. II. pag. 48, 49, &amp;c - Simp. Alg. \\ pr. 23. - Simp. Sel. Ex. pr. 19. - Simp. Geom. pr. 13. - \\ Schooten, pr. 22. - Turner’s Math. Exer. prob. 57. - Hutton’s \\ Miſcel. qu. 58. <lb/>62. B. P. S + s. # D’Omerique L. III. 25. - Vieta ib. pr. 3. - Gregorius a S. \\ Vinc. pr. 82, pag. 48. - Anderſon var. prob. 20, Cor. - \\ Simp. Alg. pr. 77. - Simp. Geom. pr. 15. - Arith. Un. \\ Caſtillionei, pr. 9. <lb/>63. B. P. S - s. # D’Omerique L. III. 26. - Vieta ib. pr. 4. - Simp. Alg. pr. \\ 76. - Simp. Sel. Ex. pr. 20. - Simp. Geom. pr. 15. <lb/>64. B. P. S x s. # D’Omerique L. I. 31. - Vieta ib. pr. 1. - Ghetaldus de Res. \\ &amp; Comp. pag. 52. - Simpſ. Sel. Ex. pr. 21. - Simp. Geom. \\ pr. 16. <lb/>65. B. P. A - a. # Simpſ. Alg. pr. 15. <lb/>66. B. P. L. # Anderſon var. prob. 21. <lb/>67. B. P. Supp. of \\ A=Comp. of a. # }Math. Mag. No. I. pr. 1. <lb/>68. B. P: S. S + s. # D’Omerique L. III. 30. <lb/>69. B. S - P. s - P. # Ghetaldus de Res. &amp; Comp. pag. 264. <lb/>70. B. ∠ of P with S. s. # Imperial Mag. Sep. 1760. <lb/>71. B. S±s. A or a. # Reducible to V. B±S. s. <lb/>72. B. S: s. A. # Gent. Diary, 1749, qu. 81. <lb/>73. B. S: s. A=2a. # Hutton’s Miſcel. qu. 16. <lb/>74. B. S: s. A - a. # D’Omerique L. I. 50. - Simp. Alg. pr. 14. <lb/>75. B. S±s. A - a # Simp. Alg. pr. 12, 13. <lb/>76. B. S + s. m: n. # D’Omerique L. III. 32. - Renaldinus, pag. 331. <lb/>77. B. S - s. Ar. # York Mis. Cur. qu. 32. - See B. P. S-s. <lb/></note>
<pb o="[6]" file="0134" n="146"/>
<note position="right" xml:space="preserve"> <lb/>78. B. S±s. Side \\ of ins. ◻. # }Reducible to B. P. S±s. <lb/>79. B. A. Ar. # Gent. Diary, 1741, qu. 5. <lb/>80. B. A - a. λ. # *Simpſ. Algebra, pr. 59. <lb/>81. B. A or a. R. # Imperial Mag. Nov. 1760. <lb/>82. B. R. Ra. of \\ circums. ☉(sun). # }Mathematician, pr. 66. <lb/>83. B±P. A - a. m - n. # L. Diary, qu. 646. <lb/>84. B±S. All the \\ angles. # }*Britiſh Oracle, qu. 50. <lb/>85. B + s. S. n. # D’Omerique L. III. 29. <lb/>86. B - S. S±. m - n. # Oughtred ch. 19, pr. 14. - Ghetaldus var. prob. 5. - Ghe- \\ taldus de Res. &amp; Comp. pag. 66. <lb/>87. B + S - s. Ar. Per. # Gent. Diary, 1749, qu. 89. <lb/>88. ∠ of B with L. \\ S: s. # }Town and C. Mag. 1769, pag. 606, 662. <lb/>89. ∠ of B with L. \\ S: s. λ. # }*Palladium, 1752, qu. 47. <lb/>90. ∠ of B with L. \\ m. n, \\ ſegts. by L. # }Ghetaldus var. prob. 4. - Regiomontanus de triang. L. II. \\ 33. - Miſ. Scient. Cur. pr. 27. <lb/>91. B x La max. S. s. # L. Diary, 1773, qu. 656. <lb/>92. B x λ a max. S. s. # *L. Diary, 1762, qu. 495. <lb/>93. P. P. P. # Mis. Cur. Math. Vol. I. pag. 30. - Rudd’s Pract. Geom. part \\ 2d, qu. 43. <lb/>94. P. S: s. A - a. # Simp. Alg. pr. 14. <lb/>95. P. S±s. A - a. # Simp. Alg. pr. 79. <lb/>96. P. S: s. m: n. # D’Omerique L. III. 33. - Simp. Alg. pr. 25. <lb/>97. P. S: s. m - n. # Simpſ. Alg. pr. 24. <lb/>98. P. S + s. m: n. # D’Omerique L. III. 31. <lb/></note>
<pb o="[7]" file="0135" n="147"/>
<note position="right" xml:space="preserve"> <lb/>99. P. S + s. m - n. # Oughtred ch. 19, pr. 15. - Ghetaldus var. prob. 2. - Ghe- \\ taldus de Res. &amp; Comp. pag. 56. - D’Omerique L. III. 27. \\ - Renaldinus, pag. 455, 456. <lb/>100. P. S - s. m - n. # Oughtred ch. 19, pr. 16. - Ghetaldus var. prob. 1. - Ghetal- \\ dus de Res. &amp; Comp. pag. 36. - D’Omerique L. III. 28. - \\ Renaldinus, pag. 460. <lb/>101. P. A or a. Per. # Reducible to B. S + s. A or a. <lb/>102. P. A - a. m: n. # Simp. Alg. pr. 17. <lb/>103. P. A - a. m - n. # Simp. Alg. pr. 19. <lb/>104. P. m - n. Ra. \\ of circum. ☉(sun). # } Martin’s Mag. qu. 395. <lb/>105. P. L. λ. # *Mathematician, pr. 10. - *L. Diary, qu. 270. <lb/>106. P: S. P - n. m - n. # Mathematician, pr. 64. <lb/>107. P: L. S - s. \\ Ra. of cir@um. ☉(sun). # }Gent. Diary, qu. 363. <lb/>108. S. s. Ar. # Mis. Cur. Math. qu. 121. - Saunderſon, art. 333. <lb/>109. S. s. λ. # *Mathematician, pr. 9. - *Gent. Diary, 1759, qu. 186. - \\ Simp. Sel. Ex. pr. 33. - Schooten, pr. 23. - *Rudd’s Prac. \\ Geometry, part 2d, qu. 14, 16. <lb/>110. S. L. n. \\ or s. L. m, \\ ſegts. by L. # }Br. Oracle, qu. 81. <lb/>111. S: s. A. L. # Town and C. Mag. Aug. and Sep. 1770. <lb/>112. S: s. A - a. m - n. # Simp. Alg. pr. 14. <lb/>113. S±s. A - a. m: n. # Simp. Alg. pr. 17. <lb/>114. S±s. A - a. m - n. # Simp. Alg. pr. 16, 18. <lb/>115. S: s. m. n. # Simp. Alg. pr. 22. <lb/>116. S±s. m. n. # Simp. Alg. pr. 20, 21. <lb/></note>
<pb o="[8]" file="0136" n="148"/>
<note position="right" xml:space="preserve"> <lb/>117. S - s. m - n. L. \\ Segts. by L. # }Gent. Mag. 1768, pag. 471, 570. <lb/>118. Sxm. s x n. L. \\ Segts. by L. # }L. Diary, qu. 622. <lb/>119. S - s. m - n. R. # Br. Oracle, qu. 61. <lb/>120. A or a. R. Side \\ of ins. ◻. # } *Simpſ. Sel. Ex. pr. 27. <lb/>121. A - a. Per. L. # Simp. Alg. pr. 61. <lb/>122. m. n. L. # Simp. Alg. pr. 57. <lb/>123. Per. All the \\ angles. # }Mathematician, pr. 44. <lb/># ☞ When all the angles are given, the triangle is given in \\ ſpecies, and therefore may be conſtructed, by ſimilar triangles, \\ with any other datum. <lb/></note>
<pb o="[9]" file="0137" n="149"/>
</div>
<div xml:id="echoid-div103" type="section" level="1" n="96">
<head xml:id="echoid-head115" xml:space="preserve">Continuation of the <emph style="sc">Synopsis</emph>, <lb/>Containing ſuch Data as cannot readily be expreſſed by the <lb/>Symbols before uſed without more words at length.</head>
<p>
  <s xml:id="echoid-s2773" xml:space="preserve">124. </s>
  <s xml:id="echoid-s2774" xml:space="preserve">IT is required to conſtruct an iſoceles triangle ſuch, as to have its equal legs <lb/>to a given line, and moreover ſuppoſing a circle inſcribed therein, and a di-<lb/>ameter thereof drawn parallel to the baſe, and continued to meet the equal legs; <lb/></s>
  <s xml:id="echoid-s2775" xml:space="preserve">ſuch line ſhall divide the area in a given ratio.</s>
  <s xml:id="echoid-s2776" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2777" xml:space="preserve">Rudd’s Prac. </s>
  <s xml:id="echoid-s2778" xml:space="preserve">Geom. </s>
  <s xml:id="echoid-s2779" xml:space="preserve">part 2d, qu. </s>
  <s xml:id="echoid-s2780" xml:space="preserve">4<emph style="sub">6</emph>. </s>
  <s xml:id="echoid-s2781" xml:space="preserve">- Supp. </s>
  <s xml:id="echoid-s2782" xml:space="preserve">to Gent. </s>
  <s xml:id="echoid-s2783" xml:space="preserve">Diary, 1741, 1742, qu. </s>
  <s xml:id="echoid-s2784" xml:space="preserve">II.</s>
  <s xml:id="echoid-s2785" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2786" xml:space="preserve">125. </s>
  <s xml:id="echoid-s2787" xml:space="preserve">In a plane triangle, ABC, there is given the angle <lb/>
<anchor type="figure" xlink:label="fig-0137-01a" xlink:href="fig-0137-01"/>
at C, and the parts or ſegments of the baſe AD, AE; <lb/></s>
  <s xml:id="echoid-s2788" xml:space="preserve">to conſtruct the triangle ſo, that if BD be drawn, the <lb/>angle ABD may be a maximum, and BC to EC in <lb/>a given ratio.</s>
  <s xml:id="echoid-s2789" xml:space="preserve"/>
</p>
<div xml:id="echoid-div103" type="float" level="2" n="1">
  <figure xlink:label="fig-0137-01" xlink:href="fig-0137-01a">
    <image file="0137-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0137-01"/>
  </figure>
</div>
<p>
  <s xml:id="echoid-s2790" xml:space="preserve">L. </s>
  <s xml:id="echoid-s2791" xml:space="preserve">Diary, 1773, qu. </s>
  <s xml:id="echoid-s2792" xml:space="preserve">659.</s>
  <s xml:id="echoid-s2793" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2794" xml:space="preserve">126. </s>
  <s xml:id="echoid-s2795" xml:space="preserve">V. </s>
  <s xml:id="echoid-s2796" xml:space="preserve">B. </s>
  <s xml:id="echoid-s2797" xml:space="preserve">Line from A or a to divide S or s in a given ratio.</s>
  <s xml:id="echoid-s2798" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2799" xml:space="preserve">Town and Coun. </s>
  <s xml:id="echoid-s2800" xml:space="preserve">Mag. </s>
  <s xml:id="echoid-s2801" xml:space="preserve">Dec. </s>
  <s xml:id="echoid-s2802" xml:space="preserve">1772.</s>
  <s xml:id="echoid-s2803" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2804" xml:space="preserve">127. </s>
  <s xml:id="echoid-s2805" xml:space="preserve">V. </s>
  <s xml:id="echoid-s2806" xml:space="preserve">B. </s>
  <s xml:id="echoid-s2807" xml:space="preserve">Difference of two lines from the angles at the baſe to the centre of the <lb/>inſcribed ☉(sun).</s>
  <s xml:id="echoid-s2808" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2809" xml:space="preserve">*Simpſon’s Sel. </s>
  <s xml:id="echoid-s2810" xml:space="preserve">Ex. </s>
  <s xml:id="echoid-s2811" xml:space="preserve">pr. </s>
  <s xml:id="echoid-s2812" xml:space="preserve">18.</s>
  <s xml:id="echoid-s2813" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2814" xml:space="preserve">128. </s>
  <s xml:id="echoid-s2815" xml:space="preserve">V. </s>
  <s xml:id="echoid-s2816" xml:space="preserve">B. </s>
  <s xml:id="echoid-s2817" xml:space="preserve">Two lines from A and a meeting in a point O given in poſition, as alſo <lb/>the angle of a line from V to O with either of the ſides about the vertical angle.</s>
  <s xml:id="echoid-s2818" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2819" xml:space="preserve">Mis. </s>
  <s xml:id="echoid-s2820" xml:space="preserve">Cur. </s>
  <s xml:id="echoid-s2821" xml:space="preserve">Math. </s>
  <s xml:id="echoid-s2822" xml:space="preserve">qu. </s>
  <s xml:id="echoid-s2823" xml:space="preserve">107.</s>
  <s xml:id="echoid-s2824" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2825" xml:space="preserve">129. </s>
  <s xml:id="echoid-s2826" xml:space="preserve">V. </s>
  <s xml:id="echoid-s2827" xml:space="preserve">A point in B. </s>
  <s xml:id="echoid-s2828" xml:space="preserve">Rect. </s>
  <s xml:id="echoid-s2829" xml:space="preserve">of the ſegments of B made by that point.</s>
  <s xml:id="echoid-s2830" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2831" xml:space="preserve">Simp. </s>
  <s xml:id="echoid-s2832" xml:space="preserve">Sel. </s>
  <s xml:id="echoid-s2833" xml:space="preserve">Ex. </s>
  <s xml:id="echoid-s2834" xml:space="preserve">pr. </s>
  <s xml:id="echoid-s2835" xml:space="preserve">44.</s>
  <s xml:id="echoid-s2836" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2837" xml:space="preserve">130. </s>
  <s xml:id="echoid-s2838" xml:space="preserve">Poſition of a line through V. </s>
  <s xml:id="echoid-s2839" xml:space="preserve">B. </s>
  <s xml:id="echoid-s2840" xml:space="preserve">S - s.</s>
  <s xml:id="echoid-s2841" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2842" xml:space="preserve">Simp. </s>
  <s xml:id="echoid-s2843" xml:space="preserve">Sel. </s>
  <s xml:id="echoid-s2844" xml:space="preserve">Ex. </s>
  <s xml:id="echoid-s2845" xml:space="preserve">pr. </s>
  <s xml:id="echoid-s2846" xml:space="preserve">49.</s>
  <s xml:id="echoid-s2847" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2848" xml:space="preserve">131. </s>
  <s xml:id="echoid-s2849" xml:space="preserve">Poſition of a line through V. </s>
  <s xml:id="echoid-s2850" xml:space="preserve">B. </s>
  <s xml:id="echoid-s2851" xml:space="preserve">S, line biſect. </s>
  <s xml:id="echoid-s2852" xml:space="preserve">B, &amp; </s>
  <s xml:id="echoid-s2853" xml:space="preserve">s in geometrical progreſſion.</s>
  <s xml:id="echoid-s2854" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2855" xml:space="preserve">Simp. </s>
  <s xml:id="echoid-s2856" xml:space="preserve">Sel. </s>
  <s xml:id="echoid-s2857" xml:space="preserve">Ex. </s>
  <s xml:id="echoid-s2858" xml:space="preserve">pr. </s>
  <s xml:id="echoid-s2859" xml:space="preserve">51.</s>
  <s xml:id="echoid-s2860" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2861" xml:space="preserve">132. </s>
  <s xml:id="echoid-s2862" xml:space="preserve">V. </s>
  <s xml:id="echoid-s2863" xml:space="preserve">S. </s>
  <s xml:id="echoid-s2864" xml:space="preserve">The angle of a line from extreme of S with the baſe, and ſegment of <lb/>s cut off thereby adjacent to the baſe.</s>
  <s xml:id="echoid-s2865" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2866" xml:space="preserve">*Mis. </s>
  <s xml:id="echoid-s2867" xml:space="preserve">Cur. </s>
  <s xml:id="echoid-s2868" xml:space="preserve">Math. </s>
  <s xml:id="echoid-s2869" xml:space="preserve">Vol. </s>
  <s xml:id="echoid-s2870" xml:space="preserve">II. </s>
  <s xml:id="echoid-s2871" xml:space="preserve">qu. </s>
  <s xml:id="echoid-s2872" xml:space="preserve">30.</s>
  <s xml:id="echoid-s2873" xml:space="preserve"/>
</p>
<pb o="[10]" file="0138" n="150"/>
<p>
  <s xml:id="echoid-s2874" xml:space="preserve">133. </s>
  <s xml:id="echoid-s2875" xml:space="preserve">V. </s>
  <s xml:id="echoid-s2876" xml:space="preserve">The ſquare of the baſe equal to the rectangle of one ſide and a given line.</s>
  <s xml:id="echoid-s2877" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2878" xml:space="preserve">General Mag. </s>
  <s xml:id="echoid-s2879" xml:space="preserve">qu. </s>
  <s xml:id="echoid-s2880" xml:space="preserve">61.</s>
  <s xml:id="echoid-s2881" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2882" xml:space="preserve">134. </s>
  <s xml:id="echoid-s2883" xml:space="preserve">V. </s>
  <s xml:id="echoid-s2884" xml:space="preserve">P. </s>
  <s xml:id="echoid-s2885" xml:space="preserve">The angle of two lines from extremes of B to middle of P.</s>
  <s xml:id="echoid-s2886" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2887" xml:space="preserve">Mathematician, pr. </s>
  <s xml:id="echoid-s2888" xml:space="preserve">65.</s>
  <s xml:id="echoid-s2889" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2890" xml:space="preserve">135. </s>
  <s xml:id="echoid-s2891" xml:space="preserve">V. </s>
  <s xml:id="echoid-s2892" xml:space="preserve">B. </s>
  <s xml:id="echoid-s2893" xml:space="preserve">The angle of two lines from extremes of B to the middle of P.</s>
  <s xml:id="echoid-s2894" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2895" xml:space="preserve">Brit. </s>
  <s xml:id="echoid-s2896" xml:space="preserve">Oracle, qu. </s>
  <s xml:id="echoid-s2897" xml:space="preserve">91.</s>
  <s xml:id="echoid-s2898" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2899" xml:space="preserve">136. </s>
  <s xml:id="echoid-s2900" xml:space="preserve">V. </s>
  <s xml:id="echoid-s2901" xml:space="preserve">S. </s>
  <s xml:id="echoid-s2902" xml:space="preserve">The ratio of the ſquare of a line, drawn in a given direction from V to B, <lb/>to the rectangle of the ſegments of B made thereby.</s>
  <s xml:id="echoid-s2903" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2904" xml:space="preserve">Anderſoni Exercitationes Math. </s>
  <s xml:id="echoid-s2905" xml:space="preserve">No. </s>
  <s xml:id="echoid-s2906" xml:space="preserve">8.</s>
  <s xml:id="echoid-s2907" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2908" xml:space="preserve">137. </s>
  <s xml:id="echoid-s2909" xml:space="preserve">V. </s>
  <s xml:id="echoid-s2910" xml:space="preserve">S or s. </s>
  <s xml:id="echoid-s2911" xml:space="preserve">m: </s>
  <s xml:id="echoid-s2912" xml:space="preserve">n, theſe being ſegts. </s>
  <s xml:id="echoid-s2913" xml:space="preserve">by a line from V to B dividing V into <lb/>given angles.</s>
  <s xml:id="echoid-s2914" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2915" xml:space="preserve">Hutton’s La. </s>
  <s xml:id="echoid-s2916" xml:space="preserve">Diary, qu. </s>
  <s xml:id="echoid-s2917" xml:space="preserve">139.</s>
  <s xml:id="echoid-s2918" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2919" xml:space="preserve">138. </s>
  <s xml:id="echoid-s2920" xml:space="preserve">V. </s>
  <s xml:id="echoid-s2921" xml:space="preserve">A line from V to B dividing V in a given ratio. </s>
  <s xml:id="echoid-s2922" xml:space="preserve">Area a minimum.</s>
  <s xml:id="echoid-s2923" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2924" xml:space="preserve">L. </s>
  <s xml:id="echoid-s2925" xml:space="preserve">Diary, 1761, qu. </s>
  <s xml:id="echoid-s2926" xml:space="preserve">479.</s>
  <s xml:id="echoid-s2927" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2928" xml:space="preserve">139. </s>
  <s xml:id="echoid-s2929" xml:space="preserve">V. </s>
  <s xml:id="echoid-s2930" xml:space="preserve">Line from A biſecting S. </s>
  <s xml:id="echoid-s2931" xml:space="preserve">Line from a biſecting s.</s>
  <s xml:id="echoid-s2932" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2933" xml:space="preserve">*Simpſon’s Sel. </s>
  <s xml:id="echoid-s2934" xml:space="preserve">Ex. </s>
  <s xml:id="echoid-s2935" xml:space="preserve">pr. </s>
  <s xml:id="echoid-s2936" xml:space="preserve">15.</s>
  <s xml:id="echoid-s2937" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2938" xml:space="preserve">140. </s>
  <s xml:id="echoid-s2939" xml:space="preserve">V. </s>
  <s xml:id="echoid-s2940" xml:space="preserve">Per. </s>
  <s xml:id="echoid-s2941" xml:space="preserve">Line parallel to B biſecting the area.</s>
  <s xml:id="echoid-s2942" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2943" xml:space="preserve">*Gent. </s>
  <s xml:id="echoid-s2944" xml:space="preserve">Di. </s>
  <s xml:id="echoid-s2945" xml:space="preserve">1750, qu. </s>
  <s xml:id="echoid-s2946" xml:space="preserve">98. </s>
  <s xml:id="echoid-s2947" xml:space="preserve">- Reducible to V. </s>
  <s xml:id="echoid-s2948" xml:space="preserve">B. </s>
  <s xml:id="echoid-s2949" xml:space="preserve">S + s.</s>
  <s xml:id="echoid-s2950" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2951" xml:space="preserve">141. </s>
  <s xml:id="echoid-s2952" xml:space="preserve">S. </s>
  <s xml:id="echoid-s2953" xml:space="preserve">s. </s>
  <s xml:id="echoid-s2954" xml:space="preserve">Line from V to centre of ins. </s>
  <s xml:id="echoid-s2955" xml:space="preserve">☉(sun).</s>
  <s xml:id="echoid-s2956" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2957" xml:space="preserve">L. </s>
  <s xml:id="echoid-s2958" xml:space="preserve">Diary, 1771, qu. </s>
  <s xml:id="echoid-s2959" xml:space="preserve">635.</s>
  <s xml:id="echoid-s2960" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2961" xml:space="preserve">142. </s>
  <s xml:id="echoid-s2962" xml:space="preserve">Baſe. </s>
  <s xml:id="echoid-s2963" xml:space="preserve">One ſide. </s>
  <s xml:id="echoid-s2964" xml:space="preserve">Ratio of a line from V, making a given angle with ſaid ſide, <lb/>to alternate part of B.</s>
  <s xml:id="echoid-s2965" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2966" xml:space="preserve">Hutton’s Mis. </s>
  <s xml:id="echoid-s2967" xml:space="preserve">pag. </s>
  <s xml:id="echoid-s2968" xml:space="preserve">63, Cor.</s>
  <s xml:id="echoid-s2969" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2970" xml:space="preserve">143. </s>
  <s xml:id="echoid-s2971" xml:space="preserve">B. </s>
  <s xml:id="echoid-s2972" xml:space="preserve">Point of contact therein of ins. </s>
  <s xml:id="echoid-s2973" xml:space="preserve">☉(sun). </s>
  <s xml:id="echoid-s2974" xml:space="preserve">mxn.</s>
  <s xml:id="echoid-s2975" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2976" xml:space="preserve">Hutton’s L. </s>
  <s xml:id="echoid-s2977" xml:space="preserve">Diaries, 1722, qu. </s>
  <s xml:id="echoid-s2978" xml:space="preserve">94.</s>
  <s xml:id="echoid-s2979" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2980" xml:space="preserve">144. </s>
  <s xml:id="echoid-s2981" xml:space="preserve">L. </s>
  <s xml:id="echoid-s2982" xml:space="preserve">A perpendicular thereto from one of the angles at the baſe. </s>
  <s xml:id="echoid-s2983" xml:space="preserve">The other <lb/>angle at the baſe.</s>
  <s xml:id="echoid-s2984" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2985" xml:space="preserve">L. </s>
  <s xml:id="echoid-s2986" xml:space="preserve">Diary, 1769, 1770, qu. </s>
  <s xml:id="echoid-s2987" xml:space="preserve">604.</s>
  <s xml:id="echoid-s2988" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2989" xml:space="preserve">145. </s>
  <s xml:id="echoid-s2990" xml:space="preserve">L. </s>
  <s xml:id="echoid-s2991" xml:space="preserve">A perpendicular thereto from A. </s>
  <s xml:id="echoid-s2992" xml:space="preserve">Another from a.</s>
  <s xml:id="echoid-s2993" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2994" xml:space="preserve">L. </s>
  <s xml:id="echoid-s2995" xml:space="preserve">Diary 1768, 1769, qu. </s>
  <s xml:id="echoid-s2996" xml:space="preserve">588.</s>
  <s xml:id="echoid-s2997" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s2998" xml:space="preserve">146. </s>
  <s xml:id="echoid-s2999" xml:space="preserve">One of the angles at the baſe. </s>
  <s xml:id="echoid-s3000" xml:space="preserve">Perpendicular therefrom to oppoſite ſide a max-<lb/>imum. </s>
  <s xml:id="echoid-s3001" xml:space="preserve">A line from the other angle at the baſe biſecting its oppoſite ſide.</s>
  <s xml:id="echoid-s3002" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s3003" xml:space="preserve">*L. </s>
  <s xml:id="echoid-s3004" xml:space="preserve">Diary, 1769, 1770, qu. </s>
  <s xml:id="echoid-s3005" xml:space="preserve">607.</s>
  <s xml:id="echoid-s3006" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s3007" xml:space="preserve">147. </s>
  <s xml:id="echoid-s3008" xml:space="preserve">S. </s>
  <s xml:id="echoid-s3009" xml:space="preserve">s. </s>
  <s xml:id="echoid-s3010" xml:space="preserve">m: </s>
  <s xml:id="echoid-s3011" xml:space="preserve">n, theſe being ſegts. </s>
  <s xml:id="echoid-s3012" xml:space="preserve">by a line from V to B, and the ratio of this line to <lb/>m or n.</s>
  <s xml:id="echoid-s3013" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s3014" xml:space="preserve">Rudd’s Prac. </s>
  <s xml:id="echoid-s3015" xml:space="preserve">Geom. </s>
  <s xml:id="echoid-s3016" xml:space="preserve">part 2d, qu. </s>
  <s xml:id="echoid-s3017" xml:space="preserve">40.</s>
  <s xml:id="echoid-s3018" xml:space="preserve"/>
</p>
<pb o="[11]" file="0139" n="151"/>
<p>
  <s xml:id="echoid-s3019" xml:space="preserve">148. </s>
  <s xml:id="echoid-s3020" xml:space="preserve">S. </s>
  <s xml:id="echoid-s3021" xml:space="preserve">s. </s>
  <s xml:id="echoid-s3022" xml:space="preserve">Line from V making an angle with S=A.</s>
  <s xml:id="echoid-s3023" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s3024" xml:space="preserve">Diarian Repoſitory, qu. </s>
  <s xml:id="echoid-s3025" xml:space="preserve">197.</s>
  <s xml:id="echoid-s3026" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s3027" xml:space="preserve">149. </s>
  <s xml:id="echoid-s3028" xml:space="preserve">S. </s>
  <s xml:id="echoid-s3029" xml:space="preserve">m. </s>
  <s xml:id="echoid-s3030" xml:space="preserve">n, theſe being ſegments of B by l from V making a given angle with S.</s>
  <s xml:id="echoid-s3031" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s3032" xml:space="preserve">T. </s>
  <s xml:id="echoid-s3033" xml:space="preserve">and Country Mag. </s>
  <s xml:id="echoid-s3034" xml:space="preserve">1769, pag 662.</s>
  <s xml:id="echoid-s3035" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s3036" xml:space="preserve">150. </s>
  <s xml:id="echoid-s3037" xml:space="preserve">One angle at the baſe. </s>
  <s xml:id="echoid-s3038" xml:space="preserve">The ſi le adjacent. </s>
  <s xml:id="echoid-s3039" xml:space="preserve">The ratio of the other ſide to a line <lb/>drawn from V to an unknown point in B, and the length of a line drawn from <lb/>the ſaid point parallel to the given ſide to terminate in the unknown ſide.</s>
  <s xml:id="echoid-s3040" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s3041" xml:space="preserve">Hutton’s Miſc. </s>
  <s xml:id="echoid-s3042" xml:space="preserve">qu. </s>
  <s xml:id="echoid-s3043" xml:space="preserve">23.</s>
  <s xml:id="echoid-s3044" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s3045" xml:space="preserve">151. </s>
  <s xml:id="echoid-s3046" xml:space="preserve">A - a. </s>
  <s xml:id="echoid-s3047" xml:space="preserve">R. </s>
  <s xml:id="echoid-s3048" xml:space="preserve">Line from centre of ins. </s>
  <s xml:id="echoid-s3049" xml:space="preserve">circle to middle of B.</s>
  <s xml:id="echoid-s3050" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s3051" xml:space="preserve">Gent. </s>
  <s xml:id="echoid-s3052" xml:space="preserve">Diary, 1771 - 2, qu. </s>
  <s xml:id="echoid-s3053" xml:space="preserve">349.</s>
  <s xml:id="echoid-s3054" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s3055" xml:space="preserve">152. </s>
  <s xml:id="echoid-s3056" xml:space="preserve">Three lines from the angles biſecting the oppoſite ſides.</s>
  <s xml:id="echoid-s3057" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s3058" xml:space="preserve">Mathematician, pr. </s>
  <s xml:id="echoid-s3059" xml:space="preserve">48. </s>
  <s xml:id="echoid-s3060" xml:space="preserve">- Simp. </s>
  <s xml:id="echoid-s3061" xml:space="preserve">Sel. </s>
  <s xml:id="echoid-s3062" xml:space="preserve">Ex. </s>
  <s xml:id="echoid-s3063" xml:space="preserve">pr. </s>
  <s xml:id="echoid-s3064" xml:space="preserve">22. </s>
  <s xml:id="echoid-s3065" xml:space="preserve">\\ - Palladium, 1752. </s>
  <s xml:id="echoid-s3066" xml:space="preserve">qu. </s>
  <s xml:id="echoid-s3067" xml:space="preserve">43.</s>
  <s xml:id="echoid-s3068" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s3069" xml:space="preserve">153. </s>
  <s xml:id="echoid-s3070" xml:space="preserve">Sxs: </s>
  <s xml:id="echoid-s3071" xml:space="preserve">mxn, ſegts by L. </s>
  <s xml:id="echoid-s3072" xml:space="preserve">m - n. </s>
  <s xml:id="echoid-s3073" xml:space="preserve">Angle made by L &amp; </s>
  <s xml:id="echoid-s3074" xml:space="preserve">I biſecting the baſe.</s>
  <s xml:id="echoid-s3075" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s3076" xml:space="preserve">Brit. </s>
  <s xml:id="echoid-s3077" xml:space="preserve">Oracle, qu. </s>
  <s xml:id="echoid-s3078" xml:space="preserve">93.</s>
  <s xml:id="echoid-s3079" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s3080" xml:space="preserve">154. </s>
  <s xml:id="echoid-s3081" xml:space="preserve">The baſe of an iſoceles triangle, and the diſtance of the vertical angle from <lb/>the foot of a perpendicular from one of the equal angles upon the oppoſite ſide.</s>
  <s xml:id="echoid-s3082" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s3083" xml:space="preserve">Brit. </s>
  <s xml:id="echoid-s3084" xml:space="preserve">Oracle, qu. </s>
  <s xml:id="echoid-s3085" xml:space="preserve">8.</s>
  <s xml:id="echoid-s3086" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s3087" xml:space="preserve">155. </s>
  <s xml:id="echoid-s3088" xml:space="preserve">P - n. </s>
  <s xml:id="echoid-s3089" xml:space="preserve">m of an iſoceles triangle, m and n being ſegts. </s>
  <s xml:id="echoid-s3090" xml:space="preserve">by a perpendicular from <lb/>one of the angles at the baſe on one of the equal ſides.</s>
  <s xml:id="echoid-s3091" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s3092" xml:space="preserve">Court Mag. </s>
  <s xml:id="echoid-s3093" xml:space="preserve">Sep. </s>
  <s xml:id="echoid-s3094" xml:space="preserve">1761.</s>
  <s xml:id="echoid-s3095" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s3096" xml:space="preserve">156. </s>
  <s xml:id="echoid-s3097" xml:space="preserve">V. </s>
  <s xml:id="echoid-s3098" xml:space="preserve">Line biſecting A or a. </s>
  <s xml:id="echoid-s3099" xml:space="preserve">Neareſt diſtance from V to periphery of ins. </s>
  <s xml:id="echoid-s3100" xml:space="preserve">☉(sun).</s>
  <s xml:id="echoid-s3101" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s3102" xml:space="preserve">*Gent. </s>
  <s xml:id="echoid-s3103" xml:space="preserve">Diary, qu. </s>
  <s xml:id="echoid-s3104" xml:space="preserve">129.</s>
  <s xml:id="echoid-s3105" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s3106" xml:space="preserve">157. </s>
  <s xml:id="echoid-s3107" xml:space="preserve">V. </s>
  <s xml:id="echoid-s3108" xml:space="preserve">The ſegments of S made by a line drawn from A to make a given angle <lb/>with B.</s>
  <s xml:id="echoid-s3109" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s3110" xml:space="preserve">*Mis. </s>
  <s xml:id="echoid-s3111" xml:space="preserve">Cur. </s>
  <s xml:id="echoid-s3112" xml:space="preserve">Math. </s>
  <s xml:id="echoid-s3113" xml:space="preserve">Vol. </s>
  <s xml:id="echoid-s3114" xml:space="preserve">I. </s>
  <s xml:id="echoid-s3115" xml:space="preserve">qu. </s>
  <s xml:id="echoid-s3116" xml:space="preserve">79.</s>
  <s xml:id="echoid-s3117" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s3118" xml:space="preserve">158. </s>
  <s xml:id="echoid-s3119" xml:space="preserve">V. </s>
  <s xml:id="echoid-s3120" xml:space="preserve">B. </s>
  <s xml:id="echoid-s3121" xml:space="preserve">Line from A or a to the centre of inſcribed circle.</s>
  <s xml:id="echoid-s3122" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s3123" xml:space="preserve">Mis. </s>
  <s xml:id="echoid-s3124" xml:space="preserve">Sci. </s>
  <s xml:id="echoid-s3125" xml:space="preserve">Cur. </s>
  <s xml:id="echoid-s3126" xml:space="preserve">qu. </s>
  <s xml:id="echoid-s3127" xml:space="preserve">53.</s>
  <s xml:id="echoid-s3128" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s3129" xml:space="preserve">159. </s>
  <s xml:id="echoid-s3130" xml:space="preserve">B. </s>
  <s xml:id="echoid-s3131" xml:space="preserve">L. </s>
  <s xml:id="echoid-s3132" xml:space="preserve">Line from extremity of L parallel to S or s.</s>
  <s xml:id="echoid-s3133" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s3134" xml:space="preserve">Math. </s>
  <s xml:id="echoid-s3135" xml:space="preserve">Mag. </s>
  <s xml:id="echoid-s3136" xml:space="preserve">No. </s>
  <s xml:id="echoid-s3137" xml:space="preserve">III. </s>
  <s xml:id="echoid-s3138" xml:space="preserve">prob. </s>
  <s xml:id="echoid-s3139" xml:space="preserve">7.</s>
  <s xml:id="echoid-s3140" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s3141" xml:space="preserve">160. </s>
  <s xml:id="echoid-s3142" xml:space="preserve">V. </s>
  <s xml:id="echoid-s3143" xml:space="preserve">The ſegments of B made by a line dividing V into given angles.</s>
  <s xml:id="echoid-s3144" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s3145" xml:space="preserve">Mis. </s>
  <s xml:id="echoid-s3146" xml:space="preserve">Cur. </s>
  <s xml:id="echoid-s3147" xml:space="preserve">Math. </s>
  <s xml:id="echoid-s3148" xml:space="preserve">Vol. </s>
  <s xml:id="echoid-s3149" xml:space="preserve">II. </s>
  <s xml:id="echoid-s3150" xml:space="preserve">qu. </s>
  <s xml:id="echoid-s3151" xml:space="preserve">48.</s>
  <s xml:id="echoid-s3152" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s3153" xml:space="preserve">161. </s>
  <s xml:id="echoid-s3154" xml:space="preserve">V. </s>
  <s xml:id="echoid-s3155" xml:space="preserve">One ſide. </s>
  <s xml:id="echoid-s3156" xml:space="preserve">Ratio of the ſegments of the baſe made by a line dividing V <lb/>into given angles.</s>
  <s xml:id="echoid-s3157" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s3158" xml:space="preserve">Palladium, 1756, pa: </s>
  <s xml:id="echoid-s3159" xml:space="preserve">43.</s>
  <s xml:id="echoid-s3160" xml:space="preserve"/>
</p>
<pb o="[12]" file="0140" n="152"/>
</div>
<div xml:id="echoid-div105" type="section" level="1" n="97">
<head xml:id="echoid-head116" xml:space="preserve">SYNOPSIS</head>
<p>
  <s xml:id="echoid-s3161" xml:space="preserve">Of Data for Right-angled Triangles which have not yet <lb/>been conſtructed in general, the vertical angle being <lb/>ſuppoſed acute or obtuſe.</s>
  <s xml:id="echoid-s3162" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s3163" xml:space="preserve">☞ The much greater part of theſe problems are purpoſely leſt without any reference. <lb/></s>
  <s xml:id="echoid-s3164" xml:space="preserve">The Compiler has ſeen an Author from whom Conſtructions to them all may <lb/>be derived, but he forbears to name him, in order to leave them as Exerciſes <lb/>for young <emph style="sc">Geometricians</emph>.</s>
  <s xml:id="echoid-s3165" xml:space="preserve"/>
</p>
<note position="right" xml:space="preserve"> <lb/>1. <emph style="bf">H</emph>. H x P + P<emph style="sub">2</emph>. <lb/>2. H. P<emph style="sub">2</emph> - n<emph style="sub">2</emph>. <lb/>3. H. m<emph style="sub">2</emph> - P<emph style="sub">2</emph>. <lb/>4. H. S±m, or H. s±n. <lb/>5. H. S<emph style="sub">2</emph> + m<emph style="sub">2</emph>, or H. s<emph style="sub">2</emph> + n<emph style="sub">2</emph>. <lb/>6. H. S<emph style="sub">2</emph> + n<emph style="sub">2</emph>. <lb/>7. H. m<emph style="sub">2</emph> - n<emph style="sub">2</emph>. <lb/>8. H. l biſecting A or a. # Univer. Muſeum, July, 1767. - Ladies \\ Diary, 1772, qu. 633, Cor. - Mis. \\ Scient. Curioſa, pag. 196, Cor. II. <lb/></note>
<pb o="[13]" file="0141" n="153"/>
<note position="right" xml:space="preserve"> <lb/>9. H<emph style="sub">2</emph>: S x s. Any other datum. <lb/>10. H<emph style="sub">2</emph>: S<emph style="sub">2</emph> + m<emph style="sub">2</emph>, or H<emph style="sub">2</emph>: s<emph style="sub">2</emph> + n<emph style="sub">2</emph>. any other. <lb/>11. H + P. H x P. <lb/>12. H + P. H x P + P<emph style="sub">2</emph>. <lb/>13. H + P. S<emph style="sub">2</emph> + n<emph style="sub">2</emph>. <lb/>14. H + P. m<emph style="sub">2</emph> + n<emph style="sub">2</emph>. <lb/>15. H x P + P<emph style="sub">2</emph>. P. <lb/>16. H x P: S<emph style="sub">2</emph> + n<emph style="sub">2</emph>. any other. <lb/>17. H x P + P<emph style="sub">2</emph>: S<emph style="sub">2</emph> + n<emph style="sub">2</emph>. any other. <lb/>18. {1/2}H - P. S - s. # Turner’s Math. Ex. pr. 37. <lb/>19. H±S. m, or H±s. n. <lb/>20. H + S. n, or H + s. m. <lb/>21. H<emph style="sub">2</emph> + S<emph style="sub">2</emph>: P<emph style="sub">2</emph>, or H<emph style="sub">2</emph> + s<emph style="sub">2</emph>: P<emph style="sub">2</emph>. any other. <lb/>22. H<emph style="sub">2</emph> + S<emph style="sub">2</emph>: S<emph style="sub">2</emph> + m<emph style="sub">2</emph>, or H<emph style="sub">2</emph> + s<emph style="sub">2</emph>: s<emph style="sub">2</emph> + n<emph style="sub">2</emph>. any other. <lb/>23. H<emph style="sub">2</emph> + S<emph style="sub">2</emph>. m, or H<emph style="sub">2</emph> + s<emph style="sub">2</emph>. n. <lb/>24. H<emph style="sub">2</emph> + S<emph style="sub">2</emph>. n, or H<emph style="sub">2</emph> + s<emph style="sub">2</emph>. m. <lb/>25. H<emph style="sub">2</emph> + S<emph style="sub">2</emph>: m<emph style="sub">2</emph>, or H<emph style="sub">2</emph> + s<emph style="sub">2</emph>: n<emph style="sub">2</emph>. any other. <lb/>26. H x S. s. or Hxs. S. # D’Omerique L. III. pr. 37. - Vieta Geo. \\ Eff. pr. 18. - Oughtred, ch. 19, pr. \\ 25. - Herigon Geo. planæ, pr. 9. - \\ Schooten, pr. 38. <lb/>27. H x S: s<emph style="sub">2</emph>, or H x s: S<emph style="sub">2</emph>. any other <lb/>28. H + S + m. H<emph style="sub">2</emph> + S<emph style="sub">2</emph> + m<emph style="sub">2</emph>. <lb/>29. H + S + m. H x S + S x m + S<emph style="sub">2</emph>. <lb/>30. H + S + m. H x S + S x m. <lb/>31. H + S + m. H<emph style="sub">2</emph> + m<emph style="sub">2</emph>. <lb/>32. H<emph style="sub">2</emph> + S<emph style="sub">2</emph> + m<emph style="sub">2</emph>: H x S + S x m. any other. <lb/>33. H<emph style="sub">2</emph> + S<emph style="sub">2</emph> + m<emph style="sub">2</emph>: H x S + S x m + S<emph style="sub">2</emph>. any other. <lb/></note>
<pb o="[14]" file="0142" n="154"/>
<note position="right" xml:space="preserve"> <lb/>34. H<emph style="sub">2</emph> + S<emph style="sub">2</emph> + m<emph style="sub">2</emph>. H + m. <lb/>35. H<emph style="sub">2</emph> + S<emph style="sub">2</emph> + m<emph style="sub">2</emph>: H<emph style="sub">2</emph> - m<emph style="sub">2</emph>. any other. <lb/>36. H<emph style="sub">2</emph> + S<emph style="sub">2</emph> + m<emph style="sub">2</emph>: H<emph style="sub">2</emph> + 2m<emph style="sub">2</emph>. any other. <lb/>37. H<emph style="sub">2</emph> + S<emph style="sub">2</emph> + m<emph style="sub">2</emph>: P<emph style="sub">2</emph>. any other. <lb/>38. H<emph style="sub">2</emph> + S<emph style="sub">2</emph> + m<emph style="sub">2</emph>. S. <lb/>39. H<emph style="sub">2</emph> + S<emph style="sub">2</emph> + m<emph style="sub">2</emph>: s<emph style="sub">2</emph>. any other. <lb/>40. H<emph style="sub">2</emph> + S<emph style="sub">2</emph> + m<emph style="sub">2</emph>: 2S<emph style="sub">2</emph> + m<emph style="sub">2</emph>. any other. <lb/>41. H<emph style="sub">2</emph> + S<emph style="sub">2</emph> + m<emph style="sub">2</emph>.: S<emph style="sub">2</emph> + 2m<emph style="sub">2</emph>. any other. <lb/>42. H<emph style="sub">2</emph> + S<emph style="sub">2</emph> + m<emph style="sub">2</emph>: Sxn. any other. <lb/>43. H<emph style="sub">2</emph> + S<emph style="sub">2</emph> + m<emph style="sub">2</emph>. n. <lb/>44. H x S + S x m + S<emph style="sub">2</emph>. H + m. <lb/>45. H x S + S x m + S<emph style="sub">2</emph>. S. <lb/>46. H + m. H<emph style="sub">2</emph> - m<emph style="sub">2</emph>. <lb/>47. H + m. P. <lb/>48. H + m. s. <lb/>49. H<emph style="sub">2</emph> + 2m<emph style="sub">2</emph>: P<emph style="sub">2</emph>. any other. <lb/>50. H<emph style="sub">2</emph>±m<emph style="sub">2</emph>: S<emph style="sub">2</emph>. any other. <lb/>51. H<emph style="sub">2</emph> - m<emph style="sub">2</emph>: S<emph style="sub">2</emph> + 2m<emph style="sub">2</emph>. any other. <lb/>52. H<emph style="sub">2</emph> - m<emph style="sub">2</emph>. n. <lb/>53. P. m<emph style="sub">2</emph> + n<emph style="sub">2</emph>. <lb/>54. P<emph style="sub">2</emph>: m<emph style="sub">2</emph>±n<emph style="sub">2</emph>. any other. <lb/>55. P + m. n, or P + n. m. <lb/>56. P - n. m. <lb/>57. m - P. n. <lb/>58. P + m. m - n. <lb/>59. P + n. m - n. <lb/>60. P<emph style="sub">2</emph> + s<emph style="sub">2</emph>: m<emph style="sub">2</emph> - P<emph style="sub">2</emph>. any other. <lb/>61. P<emph style="sub">2</emph> + s<emph style="sub">2</emph>: S<emph style="sub">2</emph> + n<emph style="sub">2</emph>. any other. <lb/></note>
<pb o="[15]" file="0143" n="155"/>
<note position="right" xml:space="preserve"> <lb/>62. P<emph style="sub">2</emph> - n<emph style="sub">2</emph>. m<emph style="sub">2</emph> - P<emph style="sub">2</emph>. <lb/>63. P<emph style="sub">2</emph> - n<emph style="sub">2</emph>: S<emph style="sub">2</emph>. any other. <lb/>64. m<emph style="sub">2</emph> - P<emph style="sub">2</emph>: s<emph style="sub">2</emph>. any other. <lb/>65. m<emph style="sub">2</emph> - P<emph style="sub">2</emph>: S<emph style="sub">2</emph> + n<emph style="sub">2</emph>. any other. <lb/>66. P<emph style="sub">2</emph> - n<emph style="sub">2</emph>: S<emph style="sub">2</emph> + n<emph style="sub">2</emph>. any other. <lb/>67. P<emph style="sub">2</emph> - n<emph style="sub">2</emph>. m. <lb/>68. m<emph style="sub">2</emph> - P<emph style="sub">2</emph>. n. <lb/>69. P<emph style="sub">2</emph> - n<emph style="sub">2</emph>: m<emph style="sub">2</emph>. any other. <lb/>70. m<emph style="sub">2</emph> - P<emph style="sub">2</emph>: n<emph style="sub">2</emph>. any other. <lb/>71. P<emph style="sub">2</emph> - n<emph style="sub">2</emph>. m - n. <lb/>72. m<emph style="sub">2</emph> - P<emph style="sub">2</emph>. m - n. <lb/>73. P<emph style="sub">2</emph> - n<emph style="sub">2</emph>: m<emph style="sub">2</emph> + 2n<emph style="sub">2</emph>. any other. <lb/>74. Pxm - P x n: S<emph style="sub">2</emph> + n<emph style="sub">2</emph>. any other. <lb/>75. S. n, or s. m # Oughtred ch. 19, pr. II. - Ronayne’s Alg. \\ B. II. ch. 2, pr. I - Ghetaldus var. prob. 19. \\ - Idem de Res. &amp; Comp. L. III. pr. 2. - \\ Renaldinus. pag. 518. <lb/>76. S. m - n, or s. m - n. # Oughtred ch. 19, pr. 8. - D’Omerique, L. \\ III. pr. 24. - Renaldinus, pag. 412. - Ghe- \\ taldus var. prob. 18. - Idem de Res. &amp; Comp. \\ L. III. pr. I. - Foſter’s Math. Lug. pr. 18. <lb/>77. S<emph style="sub">2</emph>: n<emph style="sub">2</emph>, or s<emph style="sub">2</emph>: m<emph style="sub">2</emph>. any other. <lb/>78. S<emph style="sub">2</emph>: s<emph style="sub">2</emph> + n<emph style="sub">2</emph>, or s<emph style="sub">2</emph>: S<emph style="sub">2</emph> + m<emph style="sub">2</emph>. any other. <lb/>79. S<emph style="sub">2</emph>: 2s<emph style="sub">2</emph> + n<emph style="sub">2</emph>, or s<emph style="sub">2</emph>: 2S<emph style="sub">2</emph> + m<emph style="sub">2</emph>. any other. <lb/>80. S + m. n, or s + n. m. <lb/>81. S<emph style="sub">2</emph> + m<emph style="sub">2</emph>. n, or s<emph style="sub">2</emph> + n<emph style="sub">2</emph>. m. <lb/>82. S<emph style="sub">2</emph> + n<emph style="sub">2</emph>. m, or S<emph style="sub">2</emph> + n<emph style="sub">2</emph>. n. <lb/>83. S<emph style="sub">2</emph> + n<emph style="sub">2</emph>: s<emph style="sub">2</emph> + n<emph style="sub">2</emph>. any other. <lb/></note>
<pb o="[16]" file="0144" n="156"/>
<note position="right" xml:space="preserve"> <lb/>84. S<emph style="sub">2</emph> + n<emph style="sub">2</emph>. m - n. <lb/>85. S<emph style="sub">2</emph> + n<emph style="sub">2</emph>: m<emph style="sub">2</emph> - n<emph style="sub">2</emph>. any other. <lb/>86. S<emph style="sub">2</emph> + n<emph style="sub">2</emph>: m<emph style="sub">2</emph> + 2n<emph style="sub">2</emph>. any other. <lb/>87. s<emph style="sub">2</emph> + n<emph style="sub">2</emph>: m<emph style="sub">2</emph> - n<emph style="sub">2</emph>. any other. <lb/>88. m - n. m<emph style="sub">2</emph> - n<emph style="sub">2</emph>. <lb/></note>
<p>
  <s xml:id="echoid-s3166" xml:space="preserve">89. </s>
  <s xml:id="echoid-s3167" xml:space="preserve">Ar. </s>
  <s xml:id="echoid-s3168" xml:space="preserve">Sides in arithmetical progreſſion.</s>
  <s xml:id="echoid-s3169" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s3170" xml:space="preserve">Simp. </s>
  <s xml:id="echoid-s3171" xml:space="preserve">Alg. </s>
  <s xml:id="echoid-s3172" xml:space="preserve">pr. </s>
  <s xml:id="echoid-s3173" xml:space="preserve">36. </s>
  <s xml:id="echoid-s3174" xml:space="preserve">- Simp. </s>
  <s xml:id="echoid-s3175" xml:space="preserve">Sel. </s>
  <s xml:id="echoid-s3176" xml:space="preserve">Ex. </s>
  <s xml:id="echoid-s3177" xml:space="preserve">pr. </s>
  <s xml:id="echoid-s3178" xml:space="preserve">45.</s>
  <s xml:id="echoid-s3179" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s3180" xml:space="preserve">90. </s>
  <s xml:id="echoid-s3181" xml:space="preserve">Ar. </s>
  <s xml:id="echoid-s3182" xml:space="preserve">Sides in geom. </s>
  <s xml:id="echoid-s3183" xml:space="preserve">progreffion.</s>
  <s xml:id="echoid-s3184" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s3185" xml:space="preserve">Wolfiu’s Alg. </s>
  <s xml:id="echoid-s3186" xml:space="preserve">pr. </s>
  <s xml:id="echoid-s3187" xml:space="preserve">114. </s>
  <s xml:id="echoid-s3188" xml:space="preserve">- Simps. </s>
  <s xml:id="echoid-s3189" xml:space="preserve">Sel. </s>
  <s xml:id="echoid-s3190" xml:space="preserve">Ex. </s>
  <s xml:id="echoid-s3191" xml:space="preserve">pr. <lb/></s>
  <s xml:id="echoid-s3192" xml:space="preserve">46. </s>
  <s xml:id="echoid-s3193" xml:space="preserve">- Vieta Iſt. </s>
  <s xml:id="echoid-s3194" xml:space="preserve">App. </s>
  <s xml:id="echoid-s3195" xml:space="preserve">to Apoll. </s>
  <s xml:id="echoid-s3196" xml:space="preserve">Gallus, pr. </s>
  <s xml:id="echoid-s3197" xml:space="preserve">7.</s>
  <s xml:id="echoid-s3198" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s3199" xml:space="preserve">91. </s>
  <s xml:id="echoid-s3200" xml:space="preserve">Per. </s>
  <s xml:id="echoid-s3201" xml:space="preserve">Sides in geom. </s>
  <s xml:id="echoid-s3202" xml:space="preserve">progreſſion.</s>
  <s xml:id="echoid-s3203" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s3204" xml:space="preserve">Simps. </s>
  <s xml:id="echoid-s3205" xml:space="preserve">Alg. </s>
  <s xml:id="echoid-s3206" xml:space="preserve">pr. </s>
  <s xml:id="echoid-s3207" xml:space="preserve">39.</s>
  <s xml:id="echoid-s3208" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s3209" xml:space="preserve">92. </s>
  <s xml:id="echoid-s3210" xml:space="preserve">I biſecting an acute angle. </s>
  <s xml:id="echoid-s3211" xml:space="preserve">I from right angle biſecting the foregoing given line.</s>
  <s xml:id="echoid-s3212" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s3213" xml:space="preserve">Gent. </s>
  <s xml:id="echoid-s3214" xml:space="preserve">Diary, qu. </s>
  <s xml:id="echoid-s3215" xml:space="preserve">266.</s>
  <s xml:id="echoid-s3216" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s3217" xml:space="preserve">93. </s>
  <s xml:id="echoid-s3218" xml:space="preserve">H. </s>
  <s xml:id="echoid-s3219" xml:space="preserve">Part of S adjacent to the right angle intercepted by a perpendicular to H <lb/>from middle of H.</s>
  <s xml:id="echoid-s3220" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s3221" xml:space="preserve">L. </s>
  <s xml:id="echoid-s3222" xml:space="preserve">Diary, qu. </s>
  <s xml:id="echoid-s3223" xml:space="preserve">633.</s>
  <s xml:id="echoid-s3224" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s3225" xml:space="preserve">94. </s>
  <s xml:id="echoid-s3226" xml:space="preserve">One leg and a line parallel thereto intercepted by the hypothenuſe and <lb/>the other leg being given; </s>
  <s xml:id="echoid-s3227" xml:space="preserve">to determine the triangle ſuch, that the rectangle un-<lb/>der the hypothenuſe and a line from the acute angle, adjacent to the given leg, <lb/>to the point of interſection of the parallel and the other leg may be of a given <lb/>magnitude.</s>
  <s xml:id="echoid-s3228" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s3229" xml:space="preserve">L. </s>
  <s xml:id="echoid-s3230" xml:space="preserve">Diary. </s>
  <s xml:id="echoid-s3231" xml:space="preserve">qu. </s>
  <s xml:id="echoid-s3232" xml:space="preserve">648.</s>
  <s xml:id="echoid-s3233" xml:space="preserve"/>
</p>
<p>
  <s xml:id="echoid-s3234" xml:space="preserve">N. </s>
  <s xml:id="echoid-s3235" xml:space="preserve">B. </s>
  <s xml:id="echoid-s3236" xml:space="preserve">In all the Nos. </s>
  <s xml:id="echoid-s3237" xml:space="preserve">from 28 to 52 incluſive, S may be changed for s, and m <lb/>for n, though this be not expreſſed as in others.</s>
  <s xml:id="echoid-s3238" xml:space="preserve"/>
</p>
</div>
<div xml:id="echoid-div106" type="section" level="1" n="98">
<head xml:id="echoid-head117" style="it" xml:space="preserve">FINIS.</head>
<pb file="0145" n="157"/>
<pb file="0146" n="158"/>
<pb file="0147" n="159"/>
  <figure>
    <image file="0147-01" xlink:href="http://echo.mpiwg-berlin.mpg.de/zogilib?fn=/permanent/library/xxxxxxxx/figures/0147-01"/>
  </figure>
<pb file="0148" n="160"/>  </div></text>
</echo>