Salusbury, Thomas
,
Mathematical collections and translations (Tome I)
,
1667
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1 - 30
31 - 60
61 - 90
91 - 120
121 - 150
151 - 180
181 - 210
211 - 240
241 - 270
271 - 300
301 - 330
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421 - 450
451 - 480
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the ſaid K
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in H, and A S is parallel unto the Line that toucheth
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in P; It is neceſſary that P I hath unto P H either the ſame propor
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tion that N
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hath to
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O, or greater; for this hath already been
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demonſtrated.]
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Where this is demonſtrated either by
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Archimedes
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himſelf, or by
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any other, doth not appear; touching which we will here inſert a Demonſtration, after that
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we have explained ſome things that pertaine thereto.
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C</
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<
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>LEMMA I.</
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<
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>Let the Lines A B and A C contain the Angle B A C; and from
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the point D, taken in the Line A C, draw D E and D F at
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pleaſure unto A B: and in the ſame Line any Points G and L
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being taken, draw G H & L M parallel to D E, & G K and
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L N parallel unto F D: Then from the Points D & G as farre
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as to the Line M L draw D O P, cutting G H in O, and G Q
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parallel unto B A. </
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<
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>I ſay that the Lines that lye betwixt the Pa
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rallels unto F D have unto thoſe that lye betwixt the Par
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allels unto D E (namely K N to G Q or to O P; F K to D O;
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and F N to D P) the ſame mutuall proportion: that is to ſay,
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the ſame that A F hath to A E.</
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For in regard that the Triangles A F D, A K G, and A N L
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are alike, and E F D, H K G, and M N L are alſo alike: There-
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fore,
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(a)
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as A F is to F D, ſo ſhall A K be to K G; and as F D is to
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F E, ſo ſhall K G be to K H: Wherefore,
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ex equali,
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as A F is to F
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E, ſo ſhall A K be to K H: And, by Converſion of proportion, as
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A F is to A E, ſo ſhall A K be to K H. </
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<
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>It is in the ſame manner
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proved that, as A F is to A E, ſo ſhall A N be to A M. </
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<
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>Now A
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N being to A M, as A K is to A H; The
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(b)
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Remainder K N ſhall
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be unto the Remainder H M, that is unto G Q, or unto O P, as
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A N is to A M; that is, as A F is to A E: Again, A K is to
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A H, as A F is to A E; Therefore the Remainder F K ſhall be to
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the Remainder E H, namely to D O, as A F is to A E. </
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<
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>We might in
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like manner demonstrate that ſo is F N to D P: Which is that that
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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 5. of the
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fifth.
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<
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<
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>LEMMA II.</
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<
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>In the ſame Line A B let there be two Points R and S, ſo diſpo
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ſed, that A S may have the ſame Proportion to A R that
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A F hath to A E; and thorow R draw R T parallel to E D,
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and thorow S draw S T parallel to F D, ſo, as that it may
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meet with R T in the Point T. </
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<
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>I ſay that the Point T fall
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eth in the Line A C.</
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</
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