Archimedes
,
Natation of bodies
,
1662
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Q V. And, ſuppoſe that as the Square C R is to the Square C P, ſo is the Line B N unto
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another Line; which let be S X: And, as the Square C T is to the Square C Q ſo let F O
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be to V Y. </
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>Now it is manifeſt, by the things which we have demonſtrated, in our Commentaries,
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upon the fourth Propoſition of
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Archimedes, De Conoidibus & Spheæroidibus,
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that the
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Square C P is equall to the Rectangle P S X; and alſo, that the Square C Q is equall to
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the Rectangle Q V Y; that is, the Lines S X and V Y, are the Parameters of the Sections H S C
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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
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ſame Proportion that C T hath to C Q: And, therefore, the
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(a)
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Square C R ſhall have
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to the Square C P, the ſame proportion that the
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Square C T hath to the Square C Q: There
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fore, alſo, the Line B N ſhall be to the Line
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S X, as the Line F O is to V Y: But H C was
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to C M, as A C to C E: And, therefore, alſo,
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their halves C P and C Q, are alſo to one
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another, as A D and E G: And.
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Permu
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tando,
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C P is to A D, as C Q is to E G:
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But it hath been proved, that A D is to B N,
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as E G to F O; and B N to S X, as F O to
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V Y: Therefore,
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exæquali,
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C P ſhall be
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to S X, as C Q is to V Y. And, ſince the
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Square C P is equall to the Rectangle P S X, and the Square C Q to the Rectangle Q V Y,
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the three Lines S P, PC and S X ſhall be proportionalls, and V Q, Q C and V Y ſhal be
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Proportionalls alſo: And therefore alſo S P ſhall be to P C as V Q to Q C And as P C
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is to C H, ſo ſhall Q C. be to C M: Therefore,
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ex æquali,
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as S P the Diameter of the
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Portion H S C is to its Baſe C H, ſo is V Q the Diameter of the portion M V S the
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Baſe C M; and the Angles which the Diameter with the Baſes do contain, are equall; and the
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Lines S P and V Q are parallel: Therefore the Portions, alſo, H S C and M V C ſhall be alike:
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Which was propoſed to be demonſtrated
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(a)
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By 22. of the
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ſixth.
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<
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>LEMMA. IV.</
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L
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et there be two
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L
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ines A
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B
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and C D; and let them be cut in the
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Points E and F, ſo that as A E is to E B, C F may be to F D:
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and let them be cut again in two other Points G and H; and
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let C H be to H D, as A G is to G B. </
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>I ſay that C F ſhall be to
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F H as A E is E G.</
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For in regard that as A E is to E B, ſo is C F to F D; it followeth that, by Compounding,
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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
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H D; it followeth that, by Compounding and Converting, as G B is to A B, ſo ſhall H D be
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C D: Therefore,
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ex æquali,
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and Converting as E B
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is to G B, ſo ſhall F D be to H D; And, by Conver
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ſion of Propoſition, as E B is to E G, ſo ſhall F D
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be to F H: But as A E is to E B, ſo is C F to F D:
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Ex æquali,
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therefore, as A E is to E G, ſo
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ſhall CF be to F H.
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Again, another way.
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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
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draw Lines from D to B, from H to G, and from F to E. </
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>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
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; (a)
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and likewiſe
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H G ſhall be parallel to D B; for that A H is to H D, as A G to G B
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: (b)
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Therefore F E
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and H G are parallel to each other: And conſequently, as A E is to E G, ſo is A H, that is,
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C F to F H: Which was to be demonſtrated.
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