Salusbury, Thomas
,
Mathematical collections and translations (Tome I)
,
1667
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be ſuppoſed to paſſe by the Point E, which ſhall make in the Cone
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a Circular Section, whoſe Diameter is G E H. </
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<
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>And becauſe upon
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the Diameter I K of the Circle I B K, B D is a Perpendicular, the
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Square of B D ſhall be equal to the Rectangle made by the parts
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I D and D K: And likewiſe in the upper Circle which is underſtood
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to paſſe by the points G F H, the Square of the Line F E is equal
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to the Rectangle of the parts G E H: Therefore the Square of B D
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hath the ſame proportion to the Square of F E, that the Rectangle
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I D K hath to the Rectangle G E H. </
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<
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>And becauſe the Line E D is
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Parallel to H K, E H ſhall be equal to D K, which alſo are Parallels:
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And therefore the Rectangle I D K ſhall have the ſame proportion
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to the Rectangle G E H, as I D hath to G E; that is, that D A hath
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to A E: Therefore the Rectangle I D K to the Rectangle G E H,
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that is, the Square B D to the Square F E, hath the ſame proportion
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that the Axis D A hath to the part A E: Which was to be de
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monſtrated.</
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>The other Propoſition, likewiſe neceſſary to the preſent Tract,
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we will thus make out. </
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<
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>Let us deſcribe the Parabola, of which let the
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Axis C A be prolonged out unto D; and taking any point B, let the
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Line B C be ſuppoſed to be continued out by the ſame Parallel un
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to the Baſe of the ſaid Parabola;
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and let D A be ſuppoſed equal
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to the part of the Axis C A. </
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>I ſay,
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that the Right-Line drawn by
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the points D and B, falleth not
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within the Parabola, but without,
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ſo as that it only toucheth the
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ſame in the ſaid point B: For, if
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it be poſſible for it to fall within,
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it cutteth it above, or being pro
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longed, it cutteth it below. </
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<
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>And
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in that Line let any point G be
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taken, by which paſſeth the Right
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Line F G E. </
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<
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>And becauſe the
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Square F E is greater than the
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Square G E, the ſaid Square F E
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ſhall have greater proportion to
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the Square B C, than the ſaid Square G E hath to the ſaid B C. </
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<
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>And
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becauſe, by the precedent, the Square F E is to the Square B C as
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E A is to A C; therefore E A hath greater proportion to A C, than
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the Square G E hath to the Square B C; that is, than the Square
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E D hath to the Square D C: (becauſe in the Triangle D G E as
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G E is to the Parallel B C, ſo is E
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D
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to
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D
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C:) But the Line E A to
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A C, that is, to A
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D
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hath the ſame proportion that four Rectangles
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E A
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D
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hath to four Squares of A
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D,
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that is, to the Square C
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D,
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