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

Table of figures

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          <chap>
            <p type="main">
              <s>
                <pb xlink:href="040/01/1077.jpg" pagenum="383"/>
                <emph type="italics"/>
              Q V. And, ſuppoſe that as the Square C R is to the Square C P, ſo is the Line B N unto
                <lb/>
              another Line; which let be S X: And, as the Square C T is to the Square C Q ſo let F O
                <lb/>
              be to V Y. </s>
              <s>Now it is manifeſt, by the things which we have demonſtrated, in our Commentaries,
                <lb/>
              upon the fourth Propoſition of
                <emph.end type="italics"/>
              Archimedes, De Conoidibus & Spheæroidibus,
                <emph type="italics"/>
              that the
                <lb/>
              Square C P is equall to the Rectangle P S X; and alſo, that the Square C Q is equall to
                <lb/>
              the Rectangle Q V Y; that is, the Lines S X and V Y, are the Parameters of the Sections H S C
                <lb/>
              and M V C: But ſince the Triangles C P R and C Q T are alike; C R ſhall have to C P, the
                <lb/>
              ſame Proportion that C T hath to C Q: And, therefore, the
                <emph.end type="italics"/>
              (a)
                <emph type="italics"/>
              Square C R ſhall have
                <emph.end type="italics"/>
                <lb/>
                <arrow.to.target n="marg1337"/>
                <lb/>
                <emph type="italics"/>
              to the Square C P, the ſame proportion that the
                <emph.end type="italics"/>
                <lb/>
                <figure id="id.040.01.1077.1.jpg" xlink:href="040/01/1077/1.jpg" number="276"/>
                <lb/>
                <emph type="italics"/>
              Square C T hath to the Square C Q: There­
                <lb/>
              fore, alſo, the Line B N ſhall be to the Line
                <lb/>
              S X, as the Line F O is to V Y: But H C was
                <lb/>
              to C M, as A C to C E: And, therefore, alſo,
                <lb/>
              their halves C P and C Q, are alſo to one
                <lb/>
              another, as A D and E G: And.
                <emph.end type="italics"/>
              Permu­
                <lb/>
              tando,
                <emph type="italics"/>
              C P is to A D, as C Q is to E G:
                <lb/>
              But it hath been proved, that A D is to B N,
                <lb/>
              as E G to F O; and B N to S X, as F O to
                <lb/>
              V Y: Therefore,
                <emph.end type="italics"/>
              exæquali,
                <emph type="italics"/>
              C P ſhall be
                <lb/>
              to S X, as C Q is to V Y. And, ſince the
                <lb/>
              Square C P is equall to the Rectangle P S X, and the Square C Q to the Rectangle Q V Y,
                <lb/>
              the three Lines S P, PC and S X ſhall be proportionalls, and V Q, Q C and V Y ſhal be
                <lb/>
              Proportionalls alſo: And therefore alſo S P ſhall be to P C as V Q to Q C And as P C
                <lb/>
              is to C H, ſo ſhall Q C. be to C M: Therefore,
                <emph.end type="italics"/>
              ex æquali,
                <emph type="italics"/>
              as S P the Diameter of the
                <lb/>
              Portion H S C is to its Baſe C H, ſo is V Q the Diameter of the portion M V S the
                <lb/>
              Baſe C M; and the Angles which the Diameter with the Baſes do contain, are equall; and the
                <lb/>
              Lines S P and V Q are parallel: Therefore the Portions, alſo, H S C and M V C ſhall be alike:
                <lb/>
              Which was propoſed to be demonſtrated
                <emph.end type="italics"/>
              </s>
            </p>
            <p type="margin">
              <s>
                <margin.target id="marg1337"/>
              (a)
                <emph type="italics"/>
              By 22. of the
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              ſixth.
                <emph.end type="italics"/>
              </s>
            </p>
            <p type="head">
              <s>LEMMA. IV.</s>
            </p>
            <p type="main">
              <s>
                <emph type="italics"/>
              L
                <emph.end type="italics"/>
              et there be two
                <emph type="italics"/>
              L
                <emph.end type="italics"/>
              ines A
                <emph type="italics"/>
              B
                <emph.end type="italics"/>
              and C D; and let them be cut in the
                <lb/>
              Points E and F, ſo that as A E is to E B, C F may be to F D:
                <lb/>
              and let them be cut again in two other Points G and H; and
                <lb/>
              let C H be to H D, as A G is to G B. </s>
              <s>I ſay that C F ſhall be to
                <lb/>
              F H as A E is E G.</s>
            </p>
            <p type="main">
              <s>
                <emph type="italics"/>
              For in regard that as A E is to E B, ſo is C F to F D; it followeth that, by Compounding,
                <lb/>
              as A B is to E B, ſo ſhall C D be to F D. Again, ſince that as A G is to G B, ſo is C H, to
                <lb/>
              H D; it followeth that, by Compounding and Converting, as G B is to A B, ſo ſhall H D be
                <emph.end type="italics"/>
                <lb/>
                <figure id="id.040.01.1077.2.jpg" xlink:href="040/01/1077/2.jpg" number="277"/>
                <lb/>
                <emph type="italics"/>
              C D: Therefore,
                <emph.end type="italics"/>
              ex æquali,
                <emph type="italics"/>
              and Converting as E B
                <lb/>
              is to G B, ſo ſhall F D be to H D; And, by Conver­
                <lb/>
              ſion of Propoſition, as E B is to E G, ſo ſhall F D
                <lb/>
              be to F H: But as A E is to E B, ſo is C F to F D:
                <emph.end type="italics"/>
                <lb/>
              Ex æquali,
                <emph type="italics"/>
              therefore, as A E is to E G, ſo
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              ſhall CF be to F H.
                <emph.end type="italics"/>
              Again, another way.
                <emph type="italics"/>
              Let
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              the Lines A B and C D be applyed to one another,
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              ſo as that they doe make an Angle at the parts A and C;
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              and let A and C be in one and the ſame Point: then
                <lb/>
              draw Lines from D to B, from H to G, and from F to E. </s>
              <s>And ſince that as A E is to E B,
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              ſo is C F, that is A F to F D; therefore F E ſhall be parallel to D B
                <emph.end type="italics"/>
              ; (a)
                <emph type="italics"/>
              and likewiſe
                <emph.end type="italics"/>
                <lb/>
                <arrow.to.target n="marg1338"/>
                <lb/>
                <emph type="italics"/>
              H G ſhall be parallel to D B; for that A H is to H D, as A G to G B
                <emph.end type="italics"/>
              : (b)
                <emph type="italics"/>
              Therefore F E
                <lb/>
              and H G are parallel to each other: And conſequently, as A E is to E G, ſo is A H, that is,
                <emph.end type="italics"/>
                <lb/>
                <arrow.to.target n="marg1339"/>
                <lb/>
                <emph type="italics"/>
              C F to F H: Which was to be demonſtrated.
                <emph.end type="italics"/>
              </s>
            </p>
          </chap>
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    </archimedes>
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