Apollonius <Pergaeus>; Lawson, John, 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

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        <div xml:id="echoid-div75" type="section" level="1" n="71">
          <pb o="[16]" file="0086" n="93"/>
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        <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>
          </p>
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