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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        <div xml:id="echoid-div67" type="section" level="1" n="63">
          <pb o="[3]" file="0073" n="80"/>
        </div>
        <div xml:id="echoid-div68" type="section" level="1" n="64">
          <head xml:id="echoid-head77" xml:space="preserve">LEMMA II.</head>
          <p>
            <s xml:id="echoid-s1120" xml:space="preserve">If of four proportionals the ſum of two, being either extremes or means, be
              <lb/>
            greater than the ſum of the other two; </s>
            <s xml:id="echoid-s1121" xml:space="preserve">then I ſay theſe will be greateſt and
              <lb/>
            leaſt of all.</s>
            <s xml:id="echoid-s1122" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1123" xml:space="preserve">This is the converſe of Euc. </s>
            <s xml:id="echoid-s1124" xml:space="preserve">V. </s>
            <s xml:id="echoid-s1125" xml:space="preserve">25. </s>
            <s xml:id="echoid-s1126" xml:space="preserve">and may be thus demonſtrated. </s>
            <s xml:id="echoid-s1127" xml:space="preserve">Draw
              <lb/>
            a circle whoſe diameter may be equal to the greater ſum; </s>
            <s xml:id="echoid-s1128" xml:space="preserve">and in it inſcribe the
              <lb/>
            leſſer ſum IO, which will therefore not paſs through the center, and let the
              <lb/>
            parts be IU and UO; </s>
            <s xml:id="echoid-s1129" xml:space="preserve">then through U draw a diameter AUE, and the other
              <lb/>
            two terms will be AU and EU, of which AU is greateſt of all and EU leaſt of
              <lb/>
            all, and IU and UO of intermediate magnitude, by Euc. </s>
            <s xml:id="echoid-s1130" xml:space="preserve">III. </s>
            <s xml:id="echoid-s1131" xml:space="preserve">7.</s>
            <s xml:id="echoid-s1132" xml:space="preserve"/>
          </p>
        </div>
        <div xml:id="echoid-div69" type="section" level="1" n="65">
          <head xml:id="echoid-head78" xml:space="preserve">LEMMA III.</head>
          <p>
            <s xml:id="echoid-s1133" xml:space="preserve">If of four proportionals the difference of two, being either extremes or
              <lb/>
            means, be greater than the difference of the other two, then I ſay theſe will be
              <lb/>
            the greateſt and leaſt of all.</s>
            <s xml:id="echoid-s1134" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1135" xml:space="preserve">This is demonſtrated in the ſame manner as the preceding by Euc. </s>
            <s xml:id="echoid-s1136" xml:space="preserve">III. </s>
            <s xml:id="echoid-s1137" xml:space="preserve">8.</s>
            <s xml:id="echoid-s1138" xml:space="preserve"/>
          </p>
        </div>
        <div xml:id="echoid-div70" type="section" level="1" n="66">
          <head xml:id="echoid-head79" xml:space="preserve">PROBLEM II.</head>
          <p>
            <s xml:id="echoid-s1139" xml:space="preserve">To cut a given indefinite right line in one point, ſo that, of the three ſeg-
              <lb/>
            ments intercepted between the ſaid point and three points given in the ſame in-
              <lb/>
            definite right line, the rectangle under one of them and a given external right
              <lb/>
            line may be to the rectangle under the other two in a given ratio.</s>
            <s xml:id="echoid-s1140" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1141" xml:space="preserve">In the given indefinite line let the aſſigned points be A, E, I. </s>
            <s xml:id="echoid-s1142" xml:space="preserve">It is then
              <lb/>
            required to cut it again in the point O, ſo that AO into a given external line
              <lb/>
            R may be to EO x IO as R to S. </s>
            <s xml:id="echoid-s1143" xml:space="preserve">If A be an extreme point and E the
              <lb/>
            middle one, then ſet off IU = AE the contrary way from A; </s>
            <s xml:id="echoid-s1144" xml:space="preserve">but if A be the
              <lb/>
            middle point, then ſet it off towards A. </s>
            <s xml:id="echoid-s1145" xml:space="preserve">Then from U ſet off UN = S the
              <lb/>
            conſequent of the given ratio, either towards A, or the contrary way; </s>
            <s xml:id="echoid-s1146" xml:space="preserve">for as
              <lb/>
            the Caſes vary, it’s poſition will vary. </s>
            <s xml:id="echoid-s1147" xml:space="preserve">From A and N erect perpendiculars
              <lb/>
            AY and NM to the given indefinite right line equal to AE and AI re-
              <lb/>
            ſpectively, and theſe bomotactical if A be an extreme point, but antitactical if
              <lb/>
            A be the middle point of the three given ones. </s>
            <s xml:id="echoid-s1148" xml:space="preserve">Join the extremes of theſe
              <lb/>
            perpendiculars Y and M, and upon YM as a Diameter deſcribe a circle. </s>
            <s xml:id="echoid-s1149" xml:space="preserve">I ſay
              <lb/>
            that the interſection of this circle with the given indefinite line ſolves the
              <lb/>
            Problem. </s>
            <s xml:id="echoid-s1150" xml:space="preserve">If it interſects the line in two points, then the Problem admits </s>
          </p>
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