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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[61.] THE PREFACE.
[62.] PROBLEMS CONCERNING DETERMINATE SECTION. PROBLEM I.
[63.] LEMMA I.
[64.] LEMMA II.
[65.] LEMMA III.
[66.] PROBLEM II.
[67.] LEMMA IV.
[68.] LEMMA V.
[69.] PROBLEM III.
[70.] PROBLEM IV.
[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.]
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        <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>
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          <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>
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        <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-
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            tained by AE and UE is to the rectangle contained by AI and UI.</s>
            <s xml:id="echoid-s2212" xml:space="preserve"/>
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          <p>
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              <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>
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