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-div93" type="section" level="1" n="86">
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            <s xml:id="echoid-s2698" xml:space="preserve">
              <pb o="[40]" file="0110" n="117"/>
            rectangle contained by AB and AC; </s>
            <s xml:id="echoid-s2699" xml:space="preserve">but the ſquare on HO is equal to
              <lb/>
            the rectangle contained by AB and EC: </s>
            <s xml:id="echoid-s2700" xml:space="preserve">now EC is, by ſuppoſition,
              <lb/>
            greater than AC, therefore the rectangle AB, EC is greater than the
              <lb/>
            rectangle AB, AC, and the ſquare on HO greater than the ſquare on AG,
              <lb/>
            conſequently HO is itſelf greater than AG; </s>
            <s xml:id="echoid-s2701" xml:space="preserve">but this could not be the
              <lb/>
            Caſe unleſs O fell beyond A. </s>
            <s xml:id="echoid-s2702" xml:space="preserve">In the ſame manner my it be proved that O
              <lb/>
            will fall beyond U in Fig. </s>
            <s xml:id="echoid-s2703" xml:space="preserve">59 and 60.</s>
            <s xml:id="echoid-s2704" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s2705" xml:space="preserve">
              <emph style="sc">Limitation</emph>
            . </s>
            <s xml:id="echoid-s2706" xml:space="preserve">In the above four Caſes the given ratio of R to S muſt
              <lb/>
            not exceed that which the ſquare on AU bears to the ſquare on the ſum of
              <lb/>
            two mean proportionals between AI and UE, AE and UI. </s>
            <s xml:id="echoid-s2707" xml:space="preserve">For (Fig. </s>
            <s xml:id="echoid-s2708" xml:space="preserve">30.)
              <lb/>
            </s>
            <s xml:id="echoid-s2709" xml:space="preserve">demit from A, on KO produced, the perpendicular AH. </s>
            <s xml:id="echoid-s2710" xml:space="preserve">Now it has been
              <lb/>
            proved (Lem. </s>
            <s xml:id="echoid-s2711" xml:space="preserve">III.) </s>
            <s xml:id="echoid-s2712" xml:space="preserve">that the ratio of the rectangle continued by AO and UO
              <lb/>
            to that contained by EO and IO, or which is the ſame thing, the given
              <lb/>
            ratio of R to S is the greateſt poſſible; </s>
            <s xml:id="echoid-s2713" xml:space="preserve">and (Lem. </s>
            <s xml:id="echoid-s2714" xml:space="preserve">IV.) </s>
            <s xml:id="echoid-s2715" xml:space="preserve">that KF is a mean
              <lb/>
            proportional between AI and UE, alſo that YF is a mean proportional
              <lb/>
            between AE and UI: </s>
            <s xml:id="echoid-s2716" xml:space="preserve">but HK is equal to YF, therefore HF is equal to
              <lb/>
            the ſum of two mean proportionals between AI and UE, AE and UI; </s>
            <s xml:id="echoid-s2717" xml:space="preserve">
              <lb/>
            it only then remains to prove, that the rectangle contained by AO and UO
              <lb/>
            is to that contained by EO and IO as the ſquare on AU is to the ſquare
              <lb/>
            on HF. </s>
            <s xml:id="echoid-s2718" xml:space="preserve">The triangles OEK, OHA, OIY and OUF are all ſimilar; </s>
            <s xml:id="echoid-s2719" xml:space="preserve">con-
              <lb/>
            ſequently OK is to OE as OA is to OH, as OY is to OI, and therefore
              <lb/>
            by compound ratio, the rectangle contained by AO and UO (OK and
              <lb/>
            OY) is to that contained by EO and IO as the ſquare on AO is to the
              <lb/>
            ſquare on OH; </s>
            <s xml:id="echoid-s2720" xml:space="preserve">but alſo AO is to UO as HO is to EO, and by compoſi-
              <lb/>
            tion and permutation, AU is to HF as AO is to HO, or (Eu. </s>
            <s xml:id="echoid-s2721" xml:space="preserve">VI. </s>
            <s xml:id="echoid-s2722" xml:space="preserve">22.) </s>
            <s xml:id="echoid-s2723" xml:space="preserve">
              <lb/>
            the ſquare on AU is to the ſquare on HF as the ſquare on AO is to the
              <lb/>
            ſquare on HO, and ſo by equality of ratios, the rectangle contained by AO
              <lb/>
            and UO is to that contained by EO and IO as the ſquare on AU is to the
              <lb/>
            ſquare on HF.</s>
            <s xml:id="echoid-s2724" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s2725" xml:space="preserve">Q.</s>
            <s xml:id="echoid-s2726" xml:space="preserve">E.</s>
            <s xml:id="echoid-s2727" xml:space="preserve">D.</s>
            <s xml:id="echoid-s2728" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s2729" xml:space="preserve">
              <emph style="sc">Scholium</emph>
            . </s>
            <s xml:id="echoid-s2730" xml:space="preserve">In the four Caſes wherein the points A and U are means,
              <lb/>
            the limiting ratio will be a minimum, and the ſame with that which the
              <lb/>
            ſquare on HF bears to the ſquare on EI.</s>
            <s xml:id="echoid-s2731" xml:space="preserve"/>
          </p>
        </div>
        <div xml:id="echoid-div94" type="section" level="1" n="87">
          <head xml:id="echoid-head102" xml:space="preserve">THE END.</head>
        </div>
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