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

< >
[71.] DETERMINATE SECTION. BOOK I. PROBLEM I. (Fig. 1.)
[72.] PROBLEM II. (Fig. 2 and 3.)
[73.] PROBLEM III. (Fig. 4. and 5.)
[74.] PROBLEM IV. (Fig. 6. 7. and 8.)
[75.] PROBLEM V. (Fig. 9. 10. 11. 12. 13. 14. 15. 16.)
[76.] PROBLEM VI. (Fig. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26.)
[77.] THE END OF BOOK I.
[78.] DETERMINATE SECTION. BOOK II. LEMMA I.
[79.] LEMMA II.
[80.] LEMMA III.
[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.
< >
page |< < ([2]) of 161 > >|
    <echo version="1.0RC">
      <text xml:lang="en" type="free">
        <div xml:id="echoid-div66" type="section" level="1" n="62">
          <p>
            <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>
        </div>
      </text>
    </echo>