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-s351" xml:space="preserve">
              <emph style="sc">From</emph>
            A draw two perpendiculars to the right lines DB, ZC; </s>
            <s xml:id="echoid-s352" xml:space="preserve">viz. </s>
            <s xml:id="echoid-s353" xml:space="preserve">ADF and
              <lb/>
            AZX; </s>
            <s xml:id="echoid-s354" xml:space="preserve">and in theſe perpendiculars take DF, ZX, on “either ſide of D and Z,
              <lb/>
            equal to the Radius of the given circle: </s>
            <s xml:id="echoid-s355" xml:space="preserve">and through F
              <unsure/>
            and X draw lines pa-
              <lb/>
            rallel to DB, ZC; </s>
            <s xml:id="echoid-s356" xml:space="preserve">viz. </s>
            <s xml:id="echoid-s357" xml:space="preserve">FG, XH; </s>
            <s xml:id="echoid-s358" xml:space="preserve">and then by the preceding Problem draw
              <lb/>
            a circle which ſhall paſs through the given center A and touch the two lines
              <lb/>
            FG, XH; </s>
            <s xml:id="echoid-s359" xml:space="preserve">and E the center of this circle will alſo be the center of the circle
              <lb/>
            required, as appears by ſubtracting equals from equals in Figure 1: </s>
            <s xml:id="echoid-s360" xml:space="preserve">and by adding
              <lb/>
            equals to equals in Figure 2.</s>
            <s xml:id="echoid-s361" xml:space="preserve"/>
          </p>
        </div>
        <div xml:id="echoid-div21" type="section" level="1" n="21">
          <head xml:id="echoid-head26" xml:space="preserve">LEMMA II.</head>
          <p>
            <s xml:id="echoid-s362" xml:space="preserve">
              <emph style="sc">If</emph>
            the two circles CEB and CED cut one another C, then I ſay a line drawn
              <lb/>
            from the point of ſection CBD cutting both circles, will cut off diſſimilar ſeg-
              <lb/>
            ments from thoſe circles.</s>
            <s xml:id="echoid-s363" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s364" xml:space="preserve">1ſt. </s>
            <s xml:id="echoid-s365" xml:space="preserve">
              <emph style="sc">Suppose</emph>
            CB to be the Diameter of one of them: </s>
            <s xml:id="echoid-s366" xml:space="preserve">then draw to the
              <lb/>
            other point of ſection E the line CE, and joining EB, ED, the angle CEB will
              <lb/>
            be a right one, and the angle CED either greater or leſs than a right one, and
              <lb/>
            conſequently CD cannot be a Diameter of the other.</s>
            <s xml:id="echoid-s367" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s368" xml:space="preserve">2dly. </s>
            <s xml:id="echoid-s369" xml:space="preserve">
              <emph style="sc">Suppose</emph>
            CBD not to paſs through the center of either: </s>
            <s xml:id="echoid-s370" xml:space="preserve">then through
              <lb/>
            C draw a Diameter CAG, and continue it to meet the other circle in F, and
              <lb/>
            join BG, DF: </s>
            <s xml:id="echoid-s371" xml:space="preserve">then the angle CBG is a right one, and the angle CDF is either
              <lb/>
            greater or leſs than a right one: </s>
            <s xml:id="echoid-s372" xml:space="preserve">and therefore the lines BG and DF are not
              <lb/>
            parallel: </s>
            <s xml:id="echoid-s373" xml:space="preserve">let H be the center of the other circle, and let a Diameter CHI be
              <lb/>
            drawn: </s>
            <s xml:id="echoid-s374" xml:space="preserve">draw DI and continue it meet to meet CG in K: </s>
            <s xml:id="echoid-s375" xml:space="preserve">then DIK will be pa-
              <lb/>
            rallel to BG: </s>
            <s xml:id="echoid-s376" xml:space="preserve">hence CB: </s>
            <s xml:id="echoid-s377" xml:space="preserve">CD:</s>
            <s xml:id="echoid-s378" xml:space="preserve">: CG: </s>
            <s xml:id="echoid-s379" xml:space="preserve">CK. </s>
            <s xml:id="echoid-s380" xml:space="preserve">But CI and CK are unequal,
              <lb/>
            (being both applied from the ſame point in a right angle) and therefore it cannot
              <lb/>
            be 2s CB: </s>
            <s xml:id="echoid-s381" xml:space="preserve">CD:</s>
            <s xml:id="echoid-s382" xml:space="preserve">: CG: </s>
            <s xml:id="echoid-s383" xml:space="preserve">CI: </s>
            <s xml:id="echoid-s384" xml:space="preserve">and hence it appears that the Segments CB and
              <lb/>
            CD are diſſimilar.</s>
            <s xml:id="echoid-s385" xml:space="preserve"/>
          </p>
        </div>
        <div xml:id="echoid-div22" type="section" level="1" n="22">
          <head xml:id="echoid-head27" xml:space="preserve">LEMMA III.</head>
          <p>
            <s xml:id="echoid-s386" xml:space="preserve">
              <emph style="sc">If</emph>
            through the legs of any triangle EDF (ſee Figure to Problem 10.) </s>
            <s xml:id="echoid-s387" xml:space="preserve">a line
              <lb/>
            BI be
              <unsure/>
            drawn parallel to the baſe DF, ſo that there be conſtituted two ſimilar
              <lb/>
            triangles about the ſame vertex; </s>
            <s xml:id="echoid-s388" xml:space="preserve">and a circle be circumſcribed about each of
              <lb/>
            theſe triangles; </s>
            <s xml:id="echoid-s389" xml:space="preserve">theſe circles will touch one another in the common vertex E.</s>
            <s xml:id="echoid-s390" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s391" xml:space="preserve">
              <emph style="sc">It</emph>
            is plain that they will either touch or cut each other in the point E: </s>
            <s xml:id="echoid-s392" xml:space="preserve">if
              <lb/>
            they cut each other, then by the preceding Lemma the Segments BE and DE
              <lb/>
            would be diſſimilar; </s>
            <s xml:id="echoid-s393" xml:space="preserve">but they are ſimilar, and they muſt therefore touch each
              <lb/>
            other.</s>
            <s xml:id="echoid-s394" xml:space="preserve"/>
          </p>
        </div>
        <div xml:id="echoid-div23" type="section" level="1" n="23">
          <head xml:id="echoid-head28" xml:space="preserve">PROBLEM X.</head>
          <p>
            <s xml:id="echoid-s395" xml:space="preserve">
              <emph style="sc">Having</emph>
            a point A, and alſo a right line BC, given in poſition; </s>
            <s xml:id="echoid-s396" xml:space="preserve">together
              <lb/>
            with a circle whoſe center is G given both in m
              <gap/>
            de and
              <unsure/>
            poſition; </s>
            <s xml:id="echoid-s397" xml:space="preserve">to </s>
          </p>
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