Archimedes
,
Natation of bodies
,
1662
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that A Z hath to Z D; by the fourth Propoſition of
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Archimedes, De quadratura Para
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bolæ:
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But A Z is Seſquialter of Z D; for it is as three to two, as we ſhallanon demon-
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ſtrate: Therefore D B is Seſquialter of B V; but D B and B K are Seſquialter:
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And, therefore, the Lines
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(c)
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B V and B K are equall: Which is imposſible:
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Therefore the Section of the Right-angled Cone A E I, ſhall paſs thorow the Point K; which
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we would demonstrate.
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(c)
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By 9 of the
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fifth,
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>In regard, therefore, that the three
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P
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ortions A P O L, A E I
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and A T D are contained betwixt Right Lines and the Sections
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of Right-angled Cones, and are Right, alike and unequall,
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touching one another, upon one and the ſame Baſe.]
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After theſe words,
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upon one and the ſame Baſe,
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we may ſee that ſomething is obliterated, that is to be
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deſired: and for the Demonſtration of theſe particulars, it is requiſite in this place to
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premiſe ſome things: which will alſo be neceſſary unto the things that follow.
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M</
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>LEMMA. I.</
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>Let there be a Right
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L
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ine A B; and let it be cut by two
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L
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ines,
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parallel to one another, A C and D E, ſo, that as
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A B
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is to
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B D. ſo
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A C
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may be to D E. </
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>I ſay that the Line that con
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joyneth the Points C and B ſhall likewiſe paſs by E.</
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For, if poſſible, let it not paſs by E, but either
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above or below it. </
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>Let it first paſs below it,
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as by F. </
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>The Triangles A B C and D B F ſhall
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be alike: And, therefore, as
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(a)
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A B is to B D,
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ſo is A C to D F: But as A B is to B D, ſo was
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A C to D E: Therefore
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(b)
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D F ſhall be equall to
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D E: that is, the part to the whole: Which is
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abſurd. </
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>The ſame abſurditie will follow, if the
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Line C B be ſuppoſed to paſs above the Point E:
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And, therefore, C B muſt of necesſity paſs thorow
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E: Which was required to be demonſtrated.
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(a)
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By 4. of the
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ſixth.
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(b)
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By 9. of the
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fifth.
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<
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>LEMMA. II.</
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<
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>Let there be two like
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ortions, contained betwixt Right Lines,
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and the Sections of Right-angled Cones; A B C the great
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er, whoſe Diameter let be B D; and E F C the leſser, whoſe
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Diameter let be F G: and, let them be ſo applyed to one
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another, that the greater include the leſser; and let their
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Baſes A C and E C be in the ſame Right Line, that the ſame
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Point C, may be the term or bound of them both: And,
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then in the Section A B C, take any Point, as H; and draw
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a Line from H to C. </
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<
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>I ſay, that the Line H C, hath to that
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part of it ſelf, that lyeth betwixt C and the Section E F C, the
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ſame proportion that A C hath to C E.</
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Draw B C, which ſhall paſs thorow F, For, in regard, that the Portions are alike, the
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Diameters with the Baſes contain equall Angles: And, therefore, B D and F G are parallel
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to one another: and B D is to A C, as F G it to E C: and,
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Permutando,
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B D is to F G, as
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A C is to C E; that is,
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(a)
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as their halfes D C to C G; therefore, it followeth, by the
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preceding Lemma, that the Line B C ſhall paſs by the Point F. Moreover, from the Point
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H unto the Diameter B D, draw the Line H K, parallel to the Baſe A C: and, draw a Line
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