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

Table of contents

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[81.] LEMMA IV.
[82.] LEMMA V.
[83.] PROBLEM VII. (Fig. 32, 33, 34, &c.)
[84.] PROBLEM I. (Fig. 32 to 45.)
[85.] PROBLEM II. (Fig. 46 to 57.)
[86.] PROBLEM III.
[87.] THE END.
[88.] A SYNOPSIS OF ALL THE DATA FOR THE Conſtruction of Triangles, FROM WHICH GEOMETRICAL SOLUTIONS Have hitherto been in Print.
[89.] By JOHN LAWSON, B. D. Rector of Swanscombe, in KENT. ROCHESTER:
[90.] MDCCLXXIII. [Price One Shilling.]
[91.] ADVERTISEMENT.
[92.] AN EXPLANATION OF THE SYMBOLS made uſe of in this SYNOPSIS.
[93.] INDEX OF THE Authors refered to in the SYNOPSIS.
[94.] Lately was publiſhed by the ſame Author; [Price Six Shillings in Boards.]
[95.] SYNOPSIS.
[96.] Continuation of the Synopsis, Containing ſuch Data as cannot readily be expreſſed by the Symbols before uſed without more words at length.
[97.] SYNOPSIS
[98.] FINIS.
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          <p>
            <s xml:id="echoid-s1606" xml:space="preserve">
              <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>
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