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 handwritten notes

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        <div xml:id="echoid-div86" type="section" level="1" n="81">
          <p>
            <s xml:id="echoid-s2276" xml:space="preserve">
              <pb o="[29]" file="0099" n="106"/>
            angle contained by AU and IU is to the ſquare on YF; </s>
            <s xml:id="echoid-s2277" xml:space="preserve">hence the rect-
              <lb/>
            angle AE, IU is equal to the ſquare on YF, and AE is to YF as YF is
              <lb/>
            to IU.</s>
            <s xml:id="echoid-s2278" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s2279" xml:space="preserve">Q. </s>
            <s xml:id="echoid-s2280" xml:space="preserve">E. </s>
            <s xml:id="echoid-s2281" xml:space="preserve">D.</s>
            <s xml:id="echoid-s2282" xml:space="preserve"/>
          </p>
        </div>
        <div xml:id="echoid-div88" type="section" level="1" n="82">
          <head xml:id="echoid-head97" xml:space="preserve">LEMMA V.</head>
          <p>
            <s xml:id="echoid-s2283" xml:space="preserve">If in any ſtraight line four points A, U, E and I (Fig. </s>
            <s xml:id="echoid-s2284" xml:space="preserve">31.) </s>
            <s xml:id="echoid-s2285" xml:space="preserve">be aſſigned,
              <lb/>
            and if the point O be ſo taken by
              <emph style="sc">Lemma</emph>
            II, that the ratio of the rect-
              <lb/>
            angle contained by AO and UO to that contained by EO and IO may be
              <lb/>
            the leaſt poſſible; </s>
            <s xml:id="echoid-s2286" xml:space="preserve">alſo if through O the indefinite perpendicular FG be
              <lb/>
            drawn; </s>
            <s xml:id="echoid-s2287" xml:space="preserve">and laſtly, if from E and I, EG and IF be applied to FG, the
              <lb/>
            former equal to a mean proportional between AE and UE, and the latter
              <lb/>
            to one between AI and UI: </s>
            <s xml:id="echoid-s2288" xml:space="preserve">then ſhall FG be equal to the ſum of two
              <lb/>
            mean proportionals between AE and UI, AI and UE.</s>
            <s xml:id="echoid-s2289" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s2290" xml:space="preserve">
              <emph style="sc">Demonstration</emph>
            . </s>
            <s xml:id="echoid-s2291" xml:space="preserve">Draw AF and AG, and, through U, FV and GY,
              <lb/>
            produce GE to meet FV in H, and let fall on FV the perpendicular XI,
              <lb/>
            cutting FG in N; </s>
            <s xml:id="echoid-s2292" xml:space="preserve">moreover draw UM through N, and NP through E,
              <lb/>
            and theſe two laſt will be reſpectively perpendiculars to IF and UG, be-
              <lb/>
            cauſe the three perpendiculars of every plane triangle meet in a point.
              <lb/>
            </s>
            <s xml:id="echoid-s2293" xml:space="preserve">Since by conſtruction and
              <emph style="sc">Eu</emph>
            . </s>
            <s xml:id="echoid-s2294" xml:space="preserve">VI. </s>
            <s xml:id="echoid-s2295" xml:space="preserve">17, the ſquare on EG is equal to the
              <lb/>
            rectangle contained by AE and UE, and the ſquare on IF to that con-
              <lb/>
            tained by AI and UI, and becauſe (
              <emph style="sc">Lem</emph>
            . </s>
            <s xml:id="echoid-s2296" xml:space="preserve">II.) </s>
            <s xml:id="echoid-s2297" xml:space="preserve">the ſquare on EO is to the
              <lb/>
            ſquare on IO as the rectangle AE, UE is to the rectangle AI, UI; </s>
            <s xml:id="echoid-s2298" xml:space="preserve">the
              <lb/>
            ſquare on EO is to the ſquare on IO as the ſquare on EG is to the ſquare
              <lb/>
            on IF, and (
              <emph style="sc">Eu</emph>
            . </s>
            <s xml:id="echoid-s2299" xml:space="preserve">VI. </s>
            <s xml:id="echoid-s2300" xml:space="preserve">22.) </s>
            <s xml:id="echoid-s2301" xml:space="preserve">EO is to IO as EG is to IF; </s>
            <s xml:id="echoid-s2302" xml:space="preserve">from whence it
              <lb/>
            appears that the triangles EOG and IOF are ſimilar, and HG parallel to
              <lb/>
            IF, and the angle UHE equal to the angle UFI. </s>
            <s xml:id="echoid-s2303" xml:space="preserve">Again, becauſe AI is
              <lb/>
            to IF as IF is to UI, the triangles AIF, and UFI are ſimilar, and ſo, for
              <lb/>
            like reaſons, are the triangles AEG and GEU, wherefore the angles UFI
              <lb/>
            and FAE are equal, and alſo the angles UGE and UAG; </s>
            <s xml:id="echoid-s2304" xml:space="preserve">hence there-
              <lb/>
            fore (
              <emph style="sc">Eu</emph>
            . </s>
            <s xml:id="echoid-s2305" xml:space="preserve">I 32.) </s>
            <s xml:id="echoid-s2306" xml:space="preserve">the angle YUF is equal to the angle UAF (UHE) toge-
              <lb/>
            ther with the angle UAG (UGE) and conſequently, the angles VAY and
              <lb/>
            YUV are together equal to two right angles; </s>
            <s xml:id="echoid-s2307" xml:space="preserve">wherefore the points AYUV
              <lb/>
            are in a circle: </s>
            <s xml:id="echoid-s2308" xml:space="preserve">hence, and becauſe AO is perpendicular to FG, GY
              <note symbol="*" position="foot" xlink:label="note-0099-01" xlink:href="note-0099-01a" xml:space="preserve">See
                <emph style="sc">Pap</emph>
              . Math. Collect. B. vii. prop. 60.</note>
            </s>
          </p>
        </div>
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