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

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          <p>
            <s xml:id="echoid-s2308" xml:space="preserve">
              <pb o="[30]" file="0100" n="107"/>
            be perpendicular to AF, and FV to AG; </s>
            <s xml:id="echoid-s2309" xml:space="preserve">whence it follows that GA is
              <lb/>
            parallel to XI, and the angle NIU equal to the angle UAG; </s>
            <s xml:id="echoid-s2310" xml:space="preserve">but UAG is
              <lb/>
            equal to UGE, which is equal to CNE; </s>
            <s xml:id="echoid-s2311" xml:space="preserve">wherefore, in the triangles UNI,
              <lb/>
            UNE, the angles at I and N being equal, and that at U common, they
              <lb/>
            are ſimilar, and UN is to UI as UE is to UN, conſequently the ſquare on
              <lb/>
            UN is equal to the rectangle contained by UI and UE. </s>
            <s xml:id="echoid-s2312" xml:space="preserve">Moreover, ſince
              <lb/>
            IX paſſes through N, and is perpendicular to FU, by
              <emph style="sc">Eu</emph>
            . </s>
            <s xml:id="echoid-s2313" xml:space="preserve">I. </s>
            <s xml:id="echoid-s2314" xml:space="preserve">47, the dif-
              <lb/>
            ference of the ſquares on IF and IU is equal to the difference of the
              <lb/>
            ſquares on NF and NU: </s>
            <s xml:id="echoid-s2315" xml:space="preserve">now the ſquare on IF being equal to the rect-
              <lb/>
            angle contained by AI and UI, that is (
              <emph style="sc">Eu</emph>
            . </s>
            <s xml:id="echoid-s2316" xml:space="preserve">II. </s>
            <s xml:id="echoid-s2317" xml:space="preserve">I.) </s>
            <s xml:id="echoid-s2318" xml:space="preserve">to the rectangle con-
              <lb/>
            tained by AU and UI together with the ſquare on UI, the difference of
              <lb/>
            the ſquares on IF and UI, and conſequently the difference of the ſquares on
              <lb/>
            NF and NU is equal to the rectangle contained by AU and UI; </s>
            <s xml:id="echoid-s2319" xml:space="preserve">but the
              <lb/>
            ſquare on NU has been proved equal to the rectangle contained by UI
              <lb/>
            and UE, therefore the ſquare on NF is equal to the rectangle contained
              <lb/>
            by EU and IU together with that contained by AU and IU, that is (
              <emph style="sc">Eu</emph>
            .
              <lb/>
            </s>
            <s xml:id="echoid-s2320" xml:space="preserve">II. </s>
            <s xml:id="echoid-s2321" xml:space="preserve">1.) </s>
            <s xml:id="echoid-s2322" xml:space="preserve">to the rectangle contained by AE and UI; </s>
            <s xml:id="echoid-s2323" xml:space="preserve">wherefore AE is to NF
              <lb/>
            as NF is to UI.</s>
            <s xml:id="echoid-s2324" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s2325" xml:space="preserve">Laſtly, for the like reaſons which were urged above, the difference of
              <lb/>
            the ſquares on NU and NG is equal to the difference of thoſe on GE and
              <lb/>
            UE: </s>
            <s xml:id="echoid-s2326" xml:space="preserve">now the ſquare on GE is equal to the rectangle contained by AE
              <lb/>
            and UE, that is, to the rectangle contained by AU and UE together
              <lb/>
            with the ſquare on UE; </s>
            <s xml:id="echoid-s2327" xml:space="preserve">therefore the difference of the ſquares on GE
              <lb/>
            and UE, or the difference of thoſe on NU and NG, is equal to the rect-
              <lb/>
            angle contained by AU and UE; </s>
            <s xml:id="echoid-s2328" xml:space="preserve">but the ſquare on NU is equal to the
              <lb/>
            rectangle contained by UI and UE, therefore the ſquare on NG is
              <lb/>
            equal to the rectangle contained by AU and UE together with that
              <lb/>
            contained by UI and UE; </s>
            <s xml:id="echoid-s2329" xml:space="preserve">that is, to the rectangle contained by AI and
              <lb/>
            UE, and ſo AI is to NG as NG is to UE. </s>
            <s xml:id="echoid-s2330" xml:space="preserve">Now FG is equal to the ſum
              <lb/>
            of NF and NG; </s>
            <s xml:id="echoid-s2331" xml:space="preserve">therefore FG is equal to the ſum of two mean propor-
              <lb/>
            tionals between AE and UI, AI and UE.</s>
            <s xml:id="echoid-s2332" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s2333" xml:space="preserve">Q. </s>
            <s xml:id="echoid-s2334" xml:space="preserve">E. </s>
            <s xml:id="echoid-s2335" xml:space="preserve">D.</s>
            <s xml:id="echoid-s2336" xml:space="preserve"/>
          </p>
        </div>
      </text>
    </echo>
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