Archimedes, Natation of bodies, 1662

Table of figures

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    <archimedes>
      <text>
        <body>
          <chap>
            <pb xlink:href="073/01/054.jpg" pagenum="384"/>
            <p type="margin">
              <s>
                <margin.target id="marg1338"/>
              (a)
                <emph type="italics"/>
              By 2. of the
                <lb/>
              ſixth.
                <emph.end type="italics"/>
              </s>
            </p>
            <p type="margin">
              <s>
                <margin.target id="marg1339"/>
              (b)
                <emph type="italics"/>
              By 30 of the
                <lb/>
              firſt.
                <emph.end type="italics"/>
              </s>
            </p>
            <p type="head">
              <s>LEMMA. V.</s>
            </p>
            <p type="main">
              <s>Again, let there be two like Portions, contained betwixt Right
                <lb/>
              Lines and the Sections of Right-angled Cones, as in the fore­
                <lb/>
              going figure, A B C, whoſe Diameter is B D; and E F C,
                <lb/>
              whoſe Diameter is F G; and from the Point E, draw the
                <lb/>
              Line E H parallel to the Diameters B D and F G; and let it
                <lb/>
              cut the Section A B C in K: and from the Point C draw C H
                <lb/>
              touching the Section A B C in C, and meeting with the Line
                <lb/>
              E H in H; which alſo toucheth the Section E F C in the ſame
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              Point C, as ſhall be demonſtrated: I ſay that the Line drawn
                <lb/>
              from C
                <emph type="italics"/>
              H
                <emph.end type="italics"/>
              unto the Section E F C ſo as that it be parallel to
                <lb/>
              the Line E H, ſhall be divided in the ſame proportion by the
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              Section A B C, in which the
                <emph type="italics"/>
              L
                <emph.end type="italics"/>
              ine C A is divided by the Section
                <lb/>
              E F C; and the part of the
                <emph type="italics"/>
              L
                <emph.end type="italics"/>
              ine C A which is betwixt the
                <lb/>
              two Sections, ſhall anſwer in proportion to the part of the Line
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              drawn, which alſo falleth betwixt the ſame Sections: that is,
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              as in the foregoing Figure, if D B be produced untill it meet
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              with C H in L, that it may interſect the Section E F C in the
                <lb/>
              Point M, the
                <emph type="italics"/>
              L
                <emph.end type="italics"/>
              ine
                <emph type="italics"/>
              L
                <emph.end type="italics"/>
              B ſhall have to B M the ſame proportion
                <lb/>
              that C E hath to E A.</s>
            </p>
            <p type="main">
              <s>
                <emph type="italics"/>
              For let G F be prolonged untill it meet the ſame Line C H in N, cutting the Section A B C
                <lb/>
              in O; and drawing a Line from B to C, which ſhall paſſe by F, as hath been ſhewn, the
                <emph.end type="italics"/>
                <lb/>
                <figure id="id.073.01.054.1.jpg" xlink:href="073/01/054/1.jpg" number="52"/>
                <lb/>
                <emph type="italics"/>
              Triangles C G F and C D B ſhall be alike; as
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              alſo the Triangles C F N and C B L: Wherefore
                <emph.end type="italics"/>
                <lb/>
              (a)
                <emph type="italics"/>
              as G F is to D B, ſo ſhall C F b to C B:
                <emph.end type="italics"/>
                <lb/>
                <arrow.to.target n="marg1340"/>
                <lb/>
                <emph type="italics"/>
              And as
                <emph.end type="italics"/>
              (b)
                <emph type="italics"/>
              C F is to C B, ſo ſhall F N be
                <lb/>
              to B L: Therefore G F ſhall be to D B, as F N
                <emph.end type="italics"/>
                <lb/>
                <arrow.to.target n="marg1341"/>
                <lb/>
                <emph type="italics"/>
              to B L: And,
                <emph.end type="italics"/>
              Permutando,
                <emph type="italics"/>
              G F ſhall be to
                <lb/>
              F N, as D B to B L: But D B is equall to
                <lb/>
              B L, by 35 of our Firſt Book of
                <emph.end type="italics"/>
              Conicks:
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                <emph type="italics"/>
              Therefore
                <emph.end type="italics"/>
              (c)
                <emph type="italics"/>
              G F alſo ſhall be equall to F N:
                <emph.end type="italics"/>
                <lb/>
                <arrow.to.target n="marg1342"/>
                <lb/>
                <emph type="italics"/>
              And by 33 of the ſame, the Line C H touch­
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              eth the Section E F C in the ſame Point. </s>
              <s>There­
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              fore, drawing a Line from C to M, prolong it
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              untill it meet with the Section A B C in P; and
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              from P unto A C draw P Q parallel to B D.
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              Becauſe, now, that the Line C H toucheth the
                <lb/>
              Section E F C in the Point C; L M ſhall have
                <lb/>
              the ſame proportion to M D that C D hath to D E,
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              by the Fifth Propoſition of
                <emph.end type="italics"/>
              Archimedes
                <emph type="italics"/>
              in his
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              Book
                <emph.end type="italics"/>
              De Quadratura Patabolæ:
                <emph type="italics"/>
              And by
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              reaſon of the Similitude of the Triangles C M D
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              and C P Q, as C M is to C D, ſo ſhall C P
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              be to C Q: And,
                <emph.end type="italics"/>
              Permutando,
                <emph type="italics"/>
              as C M is to
                <lb/>
              C P, ſo ſhall C D be to C Q: But as C M is to C P, ſo is C E to C A,; as we have but
                <lb/>
              even now demonſtrated: And therefore, as C E is to C A, ſo is C D to C
                <expan abbr="q;">que</expan>
              that is as the
                <lb/>
              whole is to the whole, ſo is the part to the part: The remainder, therefore, D E is to the
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              Remainder Q A, as C E is to C A; that is, as C D is to C Q: And,
                <emph.end type="italics"/>
              Permutando,
                <emph type="italics"/>
              C D
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              is to D E, as C Q is to Q A: And L M is alſo to M D, as C D to D E: Therefore L M is
                <emph.end type="italics"/>
              </s>
            </p>
          </chap>
        </body>
      </text>
    </archimedes>
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