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-div25" type="section" level="1" n="25">
          <pb o="(11)" file="0023" n="23"/>
        </div>
        <div xml:id="echoid-div26" type="section" level="1" n="26">
          <head xml:id="echoid-head31" xml:space="preserve">LEMMA IV.</head>
          <p>
            <s xml:id="echoid-s477" xml:space="preserve">
              <emph style="sc">Having</emph>
            two circles ABCI and EFGH given, it is required to find a point
              <lb/>
            M, in the line joining their centers, or in that line continued, ſuch, that any
              <lb/>
            line drawn through the ſaid point M, cutting both the circles, ſhall always cut
              <lb/>
            off ſimilar ſegments.</s>
            <s xml:id="echoid-s478" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s479" xml:space="preserve">
              <emph style="sc">Let</emph>
            the line KL joining the centers be ſo cut or produced, that KM may be
              <lb/>
            to LM in the given ratio of the Radii AK to EL [in the caſe of producing
              <lb/>
            KL we muſt make it as AK-EL: </s>
            <s xml:id="echoid-s480" xml:space="preserve">EL:</s>
            <s xml:id="echoid-s481" xml:space="preserve">: KL: </s>
            <s xml:id="echoid-s482" xml:space="preserve">LM, for then by compoſition it
              <lb/>
            will be AK: </s>
            <s xml:id="echoid-s483" xml:space="preserve">EL:</s>
            <s xml:id="echoid-s484" xml:space="preserve">: KM: </s>
            <s xml:id="echoid-s485" xml:space="preserve">LM] and then I ſay that the point M will be the
              <lb/>
            point required. </s>
            <s xml:id="echoid-s486" xml:space="preserve">For from it drawing any line MGFCB cutting the circle
              <lb/>
            ABCI in B and C, and the circle EFGH in F and G; </s>
            <s xml:id="echoid-s487" xml:space="preserve">and let B and F be the
              <lb/>
            correſpondent points moſt diſtant from M, and C and G the correſpondent
              <lb/>
            points that are nearer; </s>
            <s xml:id="echoid-s488" xml:space="preserve">and let be joined BK, CK, FL, GL, and thereby two
              <lb/>
            triangles will be formed BKC, FLG. </s>
            <s xml:id="echoid-s489" xml:space="preserve">Now becauſe by Conſtri
              <unsure/>
            ction KM: </s>
            <s xml:id="echoid-s490" xml:space="preserve">LM
              <lb/>
            :</s>
            <s xml:id="echoid-s491" xml:space="preserve">: KB: </s>
            <s xml:id="echoid-s492" xml:space="preserve">LF, KB and LF will be parallel by Euc. </s>
            <s xml:id="echoid-s493" xml:space="preserve">VI. </s>
            <s xml:id="echoid-s494" xml:space="preserve">7. </s>
            <s xml:id="echoid-s495" xml:space="preserve">and for the ſame rea-
              <lb/>
            ſon KC and LG will be parallel; </s>
            <s xml:id="echoid-s496" xml:space="preserve">and therefore the triangles BKC and FLG
              <lb/>
            will be equiangled, and hence the ſegment BC will be ſimilar to the ſegment FG.</s>
            <s xml:id="echoid-s497" xml:space="preserve"/>
          </p>
        </div>
        <div xml:id="echoid-div27" type="section" level="1" n="27">
          <head xml:id="echoid-head32" xml:space="preserve">LEMMA V.</head>
          <p>
            <s xml:id="echoid-s498" xml:space="preserve">
              <emph style="sc">The</emph>
            point M being found as in the preceding Lemma, I ſay that it is a pro-
              <lb/>
            perty of the ſaid point, that MG x MB = MH x MA: </s>
            <s xml:id="echoid-s499" xml:space="preserve">as alſo that MF x
              <lb/>
            MC = ME x MI.</s>
            <s xml:id="echoid-s500" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s501" xml:space="preserve">
              <emph style="sc">For</emph>
            joining CI and GH, it is evident that theſe lines will alſo be parallel.
              <lb/>
            </s>
            <s xml:id="echoid-s502" xml:space="preserve">Hence MI: </s>
            <s xml:id="echoid-s503" xml:space="preserve">MC:</s>
            <s xml:id="echoid-s504" xml:space="preserve">: MH: </s>
            <s xml:id="echoid-s505" xml:space="preserve">MG, but MI: </s>
            <s xml:id="echoid-s506" xml:space="preserve">MC:</s>
            <s xml:id="echoid-s507" xml:space="preserve">: MB: </s>
            <s xml:id="echoid-s508" xml:space="preserve">MA, therefore
              <lb/>
            MH: </s>
            <s xml:id="echoid-s509" xml:space="preserve">MG:</s>
            <s xml:id="echoid-s510" xml:space="preserve">: MB: </s>
            <s xml:id="echoid-s511" xml:space="preserve">MA, and MG x MB = MH x MA. </s>
            <s xml:id="echoid-s512" xml:space="preserve">Again MH: </s>
            <s xml:id="echoid-s513" xml:space="preserve">MG
              <lb/>
            :</s>
            <s xml:id="echoid-s514" xml:space="preserve">: MF: </s>
            <s xml:id="echoid-s515" xml:space="preserve">ME, but MH: </s>
            <s xml:id="echoid-s516" xml:space="preserve">MG:</s>
            <s xml:id="echoid-s517" xml:space="preserve">: MI: </s>
            <s xml:id="echoid-s518" xml:space="preserve">MC, therefore MF: </s>
            <s xml:id="echoid-s519" xml:space="preserve">ME:</s>
            <s xml:id="echoid-s520" xml:space="preserve">: MI: </s>
            <s xml:id="echoid-s521" xml:space="preserve">MC
              <lb/>
            and MF x MC = ME x MI.</s>
            <s xml:id="echoid-s522" xml:space="preserve"/>
          </p>
        </div>
        <div xml:id="echoid-div28" type="section" level="1" n="28">
          <head xml:id="echoid-head33" xml:space="preserve">PROBLEM XIII.</head>
          <p>
            <s xml:id="echoid-s523" xml:space="preserve">
              <emph style="sc">Having</emph>
            two circles given in magnitude and poſition, whoſe centers are K
              <lb/>
            and L, and alſo a point D; </s>
            <s xml:id="echoid-s524" xml:space="preserve">to draw a circle which ſhall touch the two given
              <lb/>
            ones, and alſo paſs through the point D.</s>
            <s xml:id="echoid-s525" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s526" xml:space="preserve">
              <emph style="sc">Join</emph>
            the given centers by drawing KL, and in KL or KL produced ſind the
              <lb/>
            point M (by Lemma 4.) </s>
            <s xml:id="echoid-s527" xml:space="preserve">ſuch, that all the lines drawn from it cutting the given
              <lb/>
            circles ſhall cut off ſimilar ſegments; </s>
            <s xml:id="echoid-s528" xml:space="preserve">and let KL cut the circumferences, one
              <lb/>
            of them in the points A and I, and the other in the points E and H; </s>
            <s xml:id="echoid-s529" xml:space="preserve">and join-
              <lb/>
            ing MD, make it as MD: </s>
            <s xml:id="echoid-s530" xml:space="preserve">MA:</s>
            <s xml:id="echoid-s531" xml:space="preserve">: MH: </s>
            <s xml:id="echoid-s532" xml:space="preserve">MN. </s>
            <s xml:id="echoid-s533" xml:space="preserve">Then through the points D and
              <lb/>
            N draw a circle which ſhall alſo touch the given circle whoſe center is K, by
              <lb/>
            Prob. </s>
            <s xml:id="echoid-s534" xml:space="preserve">XII. </s>
            <s xml:id="echoid-s535" xml:space="preserve">and I ſay that this circle will alſo touch the other given circle whoſe
              <lb/>
            center is L. </s>
            <s xml:id="echoid-s536" xml:space="preserve">For let B be the point of contact and BM be drawn cutting the
              <lb/>
            circle K in C, and the circle L in F and G; </s>
            <s xml:id="echoid-s537" xml:space="preserve">then by Lemma 5. </s>
            <s xml:id="echoid-s538" xml:space="preserve">MB x MG </s>
          </p>
        </div>
      </text>
    </echo>
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