Salusbury, Thomas, Mathematical collections and translations (Tome I), 1667

Table of figures

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    <archimedes>
      <text>
        <body>
          <chap>
            <pb xlink:href="040/01/1060.jpg" pagenum="366"/>
            <figure id="id.040.01.1060.1.jpg" xlink:href="040/01/1060/1.jpg" number="258"/>
            <p type="main">
              <s>
                <emph type="italics"/>
              For if it be poſſible, let it fall ſhort of it: and let R T be pro­
                <lb/>
              longed as farre as to A C in V: and then thorow V draw V X pa­
                <lb/>
              rallel to F D. Now, by the thing we have last demonſtrated, A X
                <lb/>
              ſhall have the ſame proportion unto A R, as A F hath to A E.
                <lb/>
              </s>
              <s>But A S hath alſo the ſame proportion to A R: Wherefore
                <emph.end type="italics"/>
              (a)
                <lb/>
                <arrow.to.target n="marg1226"/>
                <lb/>
              A S
                <emph type="italics"/>
              is equall to A X, the part to the whole, which is impoſſi­
                <lb/>
              ble. </s>
              <s>The ſame abſurdity will follow if we ſuppoſe the Toint
                <lb/>
              T to fall beyond the Line A C: It is therefore neceſſary that
                <lb/>
              it do fall in the ſaid A C. </s>
              <s>Which we propounded to be demonstrated.
                <emph.end type="italics"/>
              </s>
            </p>
            <p type="margin">
              <s>
                <margin.target id="marg1226"/>
              (a)
                <emph type="italics"/>
              By 9. of the
                <lb/>
              fifth.
                <emph.end type="italics"/>
              </s>
            </p>
            <p type="head">
              <s>LEMMA III.</s>
            </p>
            <p type="main">
              <s>Let there be a Parabola, whoſe Diameter
                <lb/>
                <arrow.to.target n="marg1227"/>
                <lb/>
              let be A B; and let the Right Lines A C and B D be ^{*} con­
                <lb/>
              tingent to it, A C in the Point C, and B D in B: And two
                <lb/>
              Lines being drawn thorow C, the one C E, parallel unto
                <lb/>
              the Diameter; the other C F, parallel to B D; take any
                <lb/>
              Point in the Diameter, as G; and as F B is to B G, ſo let B
                <lb/>
              G be to B H: and thorow G and H draw G K L, and H E
                <lb/>
              M, parallel unto B D; and thorow M draw M N O parallel
                <lb/>
              to
                <emph type="italics"/>
              A C,
                <emph.end type="italics"/>
              and cutting the Diameter in O: and the Line
                <emph type="italics"/>
              N P
                <emph.end type="italics"/>
                <lb/>
              being drawn thorow
                <emph type="italics"/>
              N
                <emph.end type="italics"/>
              unto the Diameter let it be parallel
                <lb/>
              to B D. </s>
              <s>I ſay that H O is double to G B.</s>
            </p>
            <p type="margin">
              <s>
                <margin.target id="marg1227"/>
              * Or touch it.</s>
            </p>
            <p type="main">
              <s>
                <emph type="italics"/>
              For the Line M N O cutteth the Diameter either in G, or in other Points: and if it do
                <lb/>
              cut it in G, one and the ſame Point ſhall be noted by the two letters G and O. </s>
              <s>Therfore F C,
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              P N, and H E M being Parallels, and A C being Parallels to M N O, they ſhall make the
                <emph.end type="italics"/>
                <lb/>
                <figure id="id.040.01.1060.2.jpg" xlink:href="040/01/1060/2.jpg" number="259"/>
                <lb/>
                <emph type="italics"/>
              Triangles A F C, O P N and O H M like to
                <emph.end type="italics"/>
                <lb/>
                <arrow.to.target n="marg1228"/>
                <lb/>
                <emph type="italics"/>
              each other: Wherefore
                <emph.end type="italics"/>
              (a)
                <emph type="italics"/>
              O H ſhall be to
                <lb/>
              H M, as A F to FC: and
                <emph.end type="italics"/>
              ^{*} Permutando,
                <lb/>
                <arrow.to.target n="marg1229"/>
                <lb/>
                <emph type="italics"/>
              O H ſhall be to A F, as H M to F C: But
                <lb/>
              the Square H M is to the Square G L as the Line
                <lb/>
              H B is to the Line B G, by 20. of our firſt Book
                <lb/>
              of
                <emph.end type="italics"/>
              Conicks;
                <emph type="italics"/>
              and the Square G L is unto the
                <lb/>
              Square F C, as the Line G B is to the Line B F:
                <lb/>
              and the Lines H B, B G and B F are thereupon
                <emph.end type="italics"/>
                <lb/>
                <arrow.to.target n="marg1230"/>
                <lb/>
                <emph type="italics"/>
              Proportionals: Therefore the
                <emph.end type="italics"/>
              (b)
                <emph type="italics"/>
              Squares
                <lb/>
              H M, G L and F C and there Sides, ſhall alſo be
                <lb/>
              Proportionals: And, therefore, as the (c)
                <lb/>
              Square H M is to the Square G L, ſo is the Line
                <emph.end type="italics"/>
                <lb/>
                <arrow.to.target n="marg1231"/>
                <lb/>
                <emph type="italics"/>
              H M to the Line F C: But as H M is to F C, ſo
                <lb/>
              is O H to A F; and as the Square H M is to
                <lb/>
              the Square G L, ſo is the Line H B to B G; that
                <lb/>
              is, B G to B F: From whence it followeth that
                <lb/>
              O H is to A F, as B G to B F: And
                <emph.end type="italics"/>
              Permu­
                <lb/>
              tando,
                <emph type="italics"/>
              O H is to B G, as A F to F B; But A F is double to F B: Therefore A B and B F
                <lb/>
              are equall, by 35. of our firſt Book of
                <emph.end type="italics"/>
              Conicks:
                <emph type="italics"/>
              And therefore N O is double to G B:
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              Which was to be demonſtrated.
                <emph.end type="italics"/>
              </s>
            </p>
          </chap>
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    </archimedes>
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