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-div17" type="section" level="1" n="17">
          <head xml:id="echoid-head22" xml:space="preserve">LEMMA I.</head>
          <p>
            <s xml:id="echoid-s315" xml:space="preserve">
              <emph style="sc">Apoint</emph>
            A being given between the two right lines BC and DE, it is requir-
              <lb/>
            ed through the point A to draw a line cutting the two given ones at equal
              <lb/>
            angles.</s>
            <s xml:id="echoid-s316" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s317" xml:space="preserve">
              <emph style="sc">If</emph>
            the given lines be parallel, then a perpendicular to them through the point
              <lb/>
            A is the line required. </s>
            <s xml:id="echoid-s318" xml:space="preserve">But if not, then let them be produced to meet in the
              <lb/>
            point F: </s>
            <s xml:id="echoid-s319" xml:space="preserve">and let FG be drawn biſecting the Angle BFD, and through A draw
              <lb/>
            a perpendicular to FG, and it will be the line required by Euc. </s>
            <s xml:id="echoid-s320" xml:space="preserve">I. </s>
            <s xml:id="echoid-s321" xml:space="preserve">26.</s>
            <s xml:id="echoid-s322" xml:space="preserve"/>
          </p>
        </div>
        <div xml:id="echoid-div18" type="section" level="1" n="18">
          <head xml:id="echoid-head23" xml:space="preserve">PROBLEM VIII.</head>
          <p>
            <s xml:id="echoid-s323" xml:space="preserve">
              <emph style="sc">Having</emph>
            a point A given, and alſo two right lines BC and DE, to draw a circle
              <lb/>
            which ſhall paſs through the given point, and touch both the given right lines.</s>
            <s xml:id="echoid-s324" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s325" xml:space="preserve">
              <emph style="sc">By</emph>
            the preceding Lemma draw a line IAH’ through the point A, which ſhall
              <lb/>
            make equal angles with the two given lines BC and DE: </s>
            <s xml:id="echoid-s326" xml:space="preserve">biſect IH in K; </s>
            <s xml:id="echoid-s327" xml:space="preserve">and
              <lb/>
            taking KL = KA, by means of the preceeding Problem draw a circle which ſhall
              <lb/>
            paſs through the points A and L, and likewiſe touch one of the given lines, BC
              <lb/>
            for inſtance, in the point M. </s>
            <s xml:id="echoid-s328" xml:space="preserve">I ſay this circle will alſo touch the other given
              <lb/>
            line DE: </s>
            <s xml:id="echoid-s329" xml:space="preserve">for from the center N letting fall the perpendicular NO, and joining
              <lb/>
            NI, NH, NM; </s>
            <s xml:id="echoid-s330" xml:space="preserve">in the triangles NKH, NKI, NK being common, and HK =
              <lb/>
            KI, and the Angles at K right ones, by Euc. </s>
            <s xml:id="echoid-s331" xml:space="preserve">I. </s>
            <s xml:id="echoid-s332" xml:space="preserve">4. </s>
            <s xml:id="echoid-s333" xml:space="preserve">NH = NI likewiſe the
              <lb/>
            angle NHK = angle NIK, from hence it follows: </s>
            <s xml:id="echoid-s334" xml:space="preserve">that the angle NHM = angle
              <lb/>
            NIO; </s>
            <s xml:id="echoid-s335" xml:space="preserve">and the angles at M and O, being both right, and NH being proved
              <lb/>
            equal to NI, NM will be equal to NO by Euc. </s>
            <s xml:id="echoid-s336" xml:space="preserve">I. </s>
            <s xml:id="echoid-s337" xml:space="preserve">26.</s>
            <s xml:id="echoid-s338" xml:space="preserve"/>
          </p>
        </div>
        <div xml:id="echoid-div19" type="section" level="1" n="19">
          <head xml:id="echoid-head24" xml:space="preserve">Mr. Simpſon conſtructs the Problem thus.</head>
          <p>
            <s xml:id="echoid-s339" xml:space="preserve">
              <emph style="sc">Let</emph>
            BD and BC be the given lines meeting in B, and A the given point,
              <lb/>
            join AB, and draw BN biſecting the given angle DBC: </s>
            <s xml:id="echoid-s340" xml:space="preserve">and from any point E
              <lb/>
            in BN upon BC let fall the perpendicular EF, and to BA apply EG = EF, pa-
              <lb/>
            rallel to which draw AH meeting BN in H: </s>
            <s xml:id="echoid-s341" xml:space="preserve">then from center H with Interval
              <lb/>
            AH let a circle be deſcribed, and the thing is done. </s>
            <s xml:id="echoid-s342" xml:space="preserve">Upon BC and BD let fall
              <lb/>
            the perpendiculars HI, HK, which are manifeſtly equal, becauſe by Conſtruction
              <lb/>
            the angle HBI = HBK; </s>
            <s xml:id="echoid-s343" xml:space="preserve">moreover as EF: </s>
            <s xml:id="echoid-s344" xml:space="preserve">EG:</s>
            <s xml:id="echoid-s345" xml:space="preserve">: HI: </s>
            <s xml:id="echoid-s346" xml:space="preserve">HA: </s>
            <s xml:id="echoid-s347" xml:space="preserve">but EF and EG
              <lb/>
            are equal, therefore alſo HI and HA.</s>
            <s xml:id="echoid-s348" xml:space="preserve"/>
          </p>
        </div>
        <div xml:id="echoid-div20" type="section" level="1" n="20">
          <head xml:id="echoid-head25" xml:space="preserve">PROBLEM IX.</head>
          <p>
            <s xml:id="echoid-s349" xml:space="preserve">
              <emph style="sc">Having</emph>
            a circle whoſe center is A given in magnitude and poſition, and alſo
              <lb/>
            two right lines BD and ZC given in poſition, to draw a circle which ſhall touch
              <lb/>
            all three.</s>
            <s xml:id="echoid-s350" xml:space="preserve"/>
          </p>
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