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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[31.] PROBLEM II.
[32.] PROBLEM III.
[33.] PROBLEM IV.
[34.] PROBLEM V.
[35.] PROBLEM VI.
[36.] General Solution.
[37.] A SECOND SUPPLEMENT, BEING Monſ. DE FERMAT’S Treatiſe on Spherical Tangencies. PROBLEM I.
[38.] PROBLEM II.
[39.] PROBLEM III.
[40.] PROBLEM IV.
[41.] PROBLEM V.
[42.] PROBLEM VI.
[43.] PROBLEM VII.
[44.] LEMMA I.
[45.] LEMMA II.
[46.] LEMMA III.
[47.] LEMMA IV.
[48.] LEMMA V.
[49.] PROBLEM VIII.
[50.] PROBLEM IX.
[51.] PROBLEM X.
[52.] PROBLEM XI.
[53.] PROBLEM XII.
[54.] PROBLEM XIII.
[55.] PROBLEM XIV.
[56.] PROBLEM XV.
[57.] Synopſis of the PROBLEMS.
[58.] THE TWO BOOKS OF APOLLONIUS PERGÆUS, CONCERNING DETERMINATE SECTION, As they have been Reſtored by WILLEBRORDUS SNELLIUS. By JOHN LAWSON, B. D. Rector of Swanſcombe, Kent. TO WHICH ARE ADDED, THE SAME TWO BOOKS, BY WILLIAM WALES, BEING AN ENTIRE NEW WORK. LONDON: Printed by G. BIGG, Succeſſor to D. LEACH. And ſold by B. White, in Fleet-Street; L. Davis, in Holborne; J. Nourse, in the Strand; and T. Payne, near the Mews-Gate. MDCC LXXII.
[59.] ADVERTISEMENT.
[60.] EXTRACT from PAPPUS's Preface to his Seventh Book in Dr. HALLEY's Tranſlation. DE SECTIONE DETERMINATA II.
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          <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
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            EO and IO be the leaſt poſſible.</s>
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            <s xml:id="echoid-s2204" xml:space="preserve">*** This being demonſtrated in the preceeding Tract of
              <emph style="sc">Snellius</emph>
            , I
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            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>
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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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