The two books of Apollonius Pergaeus, concerning tangencies, as they have been restored by Franciscus Vieta and Marinus Ghetaldus Apollonius Pergaeus John Lawson 868387657
<?xml version="1.0" encoding="UTF-8"?> <div> <p xmlns="http://www.w3.org/1999/xhtml" id="false"> angle contained by AU and IU is to the ſquare on YF; hence the rect- <br /> angle AE, IU is equal to the ſquare on YF, and AE is to YF as YF is <br /> to IU. </p> <p xmlns="http://www.w3.org/1999/xhtml">Q. E. D.</p> <h2 xmlns="http://www.w3.org/1999/xhtml"> <span class="headingNumber">82.</span> <span class="head">LEMMA V.</span> </h2> <p xmlns="http://www.w3.org/1999/xhtml"> If in any ſtraight line four points A, U, E and I (Fig. 31.) be aſſigned, <br /> and if the point O be ſo taken by <span class="caps" style="font-size: 75%">Lemma</span> II, that the ratio of the rect- <br /> angle contained by AO and UO to that contained by EO and IO may be <br /> the leaſt poſſible; alſo if through O the indefinite perpendicular FG be <br /> drawn; and laſtly, if from E and I, EG and IF be applied to FG, the <br /> former equal to a mean proportional between AE and UE, and the latter <br /> to one between AI and UI: then ſhall FG be equal to the ſum of two <br /> mean proportionals between AE and UI, AI and UE. </p> <p xmlns="http://www.w3.org/1999/xhtml"> <span class="caps" style="font-size: 75%">Demonstration</span> . Draw AF and AG, and, through U, FV and GY, <br /> produce GE to meet FV in H, and let fall on FV the perpendicular XI, <br /> cutting FG in N; moreover draw UM through N, and NP through E, <br /> and theſe two laſt will be reſpectively perpendiculars to IF and UG, be- <br /> cauſe the three perpendiculars of every plane triangle meet in a point. Since by conſtruction and <span class="caps" style="font-size: 75%">Eu</span> . VI. 17, the ſquare on EG is equal to the <br /> rectangle contained by AE and UE, and the ſquare on IF to that con- <br /> tained by AI and UI, and becauſe ( <span class="caps" style="font-size: 75%">Lem</span> . II.) the ſquare on EO is to the <br /> ſquare on IO as the rectangle AE, UE is to the rectangle AI, UI; the <br /> ſquare on EO is to the ſquare on IO as the ſquare on EG is to the ſquare <br /> on IF, and ( <span class="caps" style="font-size: 75%">Eu</span> . VI. 22.) EO is to IO as EG is to IF; from whence it <br /> appears that the triangles EOG and IOF are ſimilar, and HG parallel to <br /> IF, and the angle UHE equal to the angle UFI. Again, becauſe AI is <br /> to IF as IF is to UI, the triangles AIF, and UFI are ſimilar, and ſo, for <br /> like reaſons, are the triangles AEG and GEU, wherefore the angles UFI <br /> and FAE are equal, and alſo the angles UGE and UAG; hence there- <br /> fore ( <span class="caps" style="font-size: 75%">Eu</span> . I 32.) the angle YUF is equal to the angle UAF (UHE) toge- <br /> ther with the angle UAG (UGE) and conſequently, the angles VAY and <br /> YUV are together equal to two right angles; wherefore the points AYUV <br /> are in a circle: hence, and becauſe AO is perpendicular to FG, GY will </p> </div>