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-div73" type="section" level="1" n="69">
          <pb o="[9]" file="0079" n="86"/>
          <p>
            <s xml:id="echoid-s1453" xml:space="preserve">
              <emph style="sc">Case</emph>
            VI. </s>
            <s xml:id="echoid-s1454" xml:space="preserve">Let the given ratio of EL to LI be inæqualitatis majoris, and
              <lb/>
            let the point ſought be required to lie beyond either extreme. </s>
            <s xml:id="echoid-s1455" xml:space="preserve">Here we muſt
              <lb/>
            uſe the IIId
              <emph style="sc">Case</emph>
            of the IId
              <emph style="sc">Problem</emph>
            ; </s>
            <s xml:id="echoid-s1456" xml:space="preserve">and the
              <emph style="sc">Determination</emph>
            is that
              <lb/>
            UN (found in the ſame ratio to AI as IL is to IE) muſt not be leſs than
              <lb/>
            IE + EL + √4 IEL*.</s>
            <s xml:id="echoid-s1457" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1458" xml:space="preserve">
              <emph style="sc">Case</emph>
            VII. </s>
            <s xml:id="echoid-s1459" xml:space="preserve">Let the ſituation of O be required the ſame as in the two laſt
              <lb/>
            Caſes, but let the given ratio be that of equality, which was there ſuppoſed
              <lb/>
            of inequality. </s>
            <s xml:id="echoid-s1460" xml:space="preserve">Here the IId
              <emph style="sc">Problem</emph>
            will be of no uſe, and this Caſe
              <lb/>
            requires a particular conſtruction.</s>
            <s xml:id="echoid-s1461" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1462" xml:space="preserve">Let then the three Points be A, I, E, and I the middle one; </s>
            <s xml:id="echoid-s1463" xml:space="preserve">and let it
              <lb/>
            be required to find a fourth O beyond E, ſuch that AO x OE may equal
              <lb/>
              <emph style="ol">OI</emph>
              <emph style="sub">2</emph>
            .</s>
            <s xml:id="echoid-s1464" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1465" xml:space="preserve">
              <emph style="sc">Construction</emph>
            . </s>
            <s xml:id="echoid-s1466" xml:space="preserve">Upon AE diameter deſcribe a circle,, and let another YS
              <lb/>
            cut the former at right angles. </s>
            <s xml:id="echoid-s1467" xml:space="preserve">Join SI, and continue it to meet the circum-
              <lb/>
            ference in R. </s>
            <s xml:id="echoid-s1468" xml:space="preserve">From R draw a tangent to meet the given line in O, and I
              <lb/>
            ſay O is the point required.</s>
            <s xml:id="echoid-s1469" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1470" xml:space="preserve">
              <emph style="sc">Demonstration</emph>
            . </s>
            <s xml:id="echoid-s1471" xml:space="preserve">Joining YR, the triangles SUI and SYR will be ſimi-
              <lb/>
            lar, and the angle UIS or RIO = SYR. </s>
            <s xml:id="echoid-s1472" xml:space="preserve">But the angle IRO made by the
              <lb/>
            tangent and ſecant = SYR in the alternate ſegment. </s>
            <s xml:id="echoid-s1473" xml:space="preserve">Therefore RIO = IRO,
              <lb/>
            and OR = OI. </s>
            <s xml:id="echoid-s1474" xml:space="preserve">But by the property of the circle AOE =
              <emph style="ol">OR</emph>
              <emph style="sub">2</emph>
            . </s>
            <s xml:id="echoid-s1475" xml:space="preserve">And
              <lb/>
            therefore AOE =
              <emph style="ol">OI</emph>
              <emph style="sub">2</emph>
            .</s>
            <s xml:id="echoid-s1476" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1477" xml:space="preserve">Q. </s>
            <s xml:id="echoid-s1478" xml:space="preserve">E. </s>
            <s xml:id="echoid-s1479" xml:space="preserve">D.</s>
            <s xml:id="echoid-s1480" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1481" xml:space="preserve">The
              <emph style="sc">Determination</emph>
            is that AI muſt be greater than IE.</s>
            <s xml:id="echoid-s1482" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1483" xml:space="preserve">
              <emph style="sc">Case</emph>
            VIII. </s>
            <s xml:id="echoid-s1484" xml:space="preserve">Whereas in the Iſt and IId
              <emph style="sc">Cases</emph>
            the given ratio was that of
              <lb/>
            inequality, let us now ſuppoſe it that of equality; </s>
            <s xml:id="echoid-s1485" xml:space="preserve">and let the three points be
              <lb/>
            A, E, I, and E the middle one; </s>
            <s xml:id="echoid-s1486" xml:space="preserve">and let a fourth O be ſought between E and
              <lb/>
            I, ſuch that AOE may equal
              <emph style="ol">OI</emph>
              <emph style="sub">2</emph>
            .</s>
            <s xml:id="echoid-s1487" xml:space="preserve"/>
          </p>
          <p>
            <s xml:id="echoid-s1488" xml:space="preserve">The
              <emph style="sc">Construction</emph>
            and
              <emph style="sc">Demonstration</emph>
            of this Caſe is in every reſpect
              <lb/>
            the ſame as that of the preceeding, as will appear by comparing the figures.</s>
            <s xml:id="echoid-s1489" xml:space="preserve"/>
          </p>
        </div>
        <div xml:id="echoid-div74" type="section" level="1" n="70">
          <head xml:id="echoid-head83" xml:space="preserve">PROBLEM IV.</head>
          <p>
            <s xml:id="echoid-s1490" xml:space="preserve">To cut a given indefinite right line ſo in one point that, of the four ſeg-
              <lb/>
            ments intercepted between the ſame and four points given in the indefinite
              <lb/>
            line, the rectangle under any two aſſigned ones may be to the rectangle under
              <lb/>
            the two remaining ones in a given ratio.</s>
            <s xml:id="echoid-s1491" xml:space="preserve"/>
          </p>
        </div>
      </text>
    </echo>
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