100円 A В C PROPOSITION XXXVII. THEOREM XXXII. IF four ftraight lines (A, B, C, & D) be proportionals, (that is, if, A: B = C: D): the fimilar and fimilarly described parallelepipeds, from the two first (A & B), will be proportional to the fimilar and fimilarly defcribed parallelepipeds, from the two last (C & D); and if the two fimilar and fimilarly described parallelepipeds, from the two lines (A & B); be proportional to the two other fimilar and fimilarly defcribed parallelepipeds, from the two other straight lines (C & D); the homologous fides of the first (A & B), will be proportional to the homologous fides (C & D) of the last. Hypothefis. I. A B C : D. II. From A & B there has been defcribed. III. Alfo from C & D. BECAUSE BA: DEMONSTRATION. B (Hyp. 2). ECAUSE the A is to the Thefis. 1. The SP.33. B.11. Cor. to three times D. Ax.6. B. 1. PP I. BECAU. ECAUSE the triangular prifm is the half of its parallelepiped (P. 28. B. 11.), it follows (Ax. 7. B. 1 ), that the same truth is applicable to fimilar triangular prifms. II. It may be also applied to fimilar polygon prisms; because they may be divided by planes into triangular prijms. (Remark 2. of P. 34. B. 11). IF F two planes (AZ & AX) be perpendicular to one another; and a ftraight line (CD) be drawn from the point (C) in one of the planes (AZ) perpendicular to the other (A X): this ftraight line fhall fall on the common fection (A B) of the planes. There may be drawn a Las CE, which will not fall on the Preparation. From the point C, let fall on A B, in the plane A Z, BECAUSE CD is to the common fection A B (Prep). 1. CD will be to the plane A X. 2. (Sup.). But EC is to the fame plane. 3. Which is impoffible. 4. Confequently, the LCD let fall from the point C, of the plane AZ, to the plane A X (which is perpendicular to it) paffes thro' their common fection A B. Which was to be demonftrated. IN PROPOSITION XXXIX. THEOREM XXXIV. N a parallelepiped (A E) if the fides (G D, A B; GF, AH; FE, HC; ED, & BC) of the oppofite planes, (F A & EB; FC & GB) be divided each into two equal parts, the common fection (MS) of the planes (I P & LR), paffing thro' the points of fection (K, P, O, I & L, Q, R, N) and the diameter (FB) of the parallelepiped (A E) cut each other into two equal parts in the point (T). Hypothefis. 1. In the AE, having for diam FB; the fides D G, AB, &c. are bifected in the points K, P, &c. II. The planes KO&LR, have been paffed thro' the points, K, P, O, I, & L, Q, R, N. Thefis. The common section MS of those planes, & the diam. F B, cut each other into two equal parts in the point T. Preparation. Draw SB, SH, FM, & M D. DEMONSTRATION. Pof.1. B. 1. THE fides HQ & SQ being = to the fides B R & S R (Hyp.1). P.34. B. 1. And the VHQS = V SR B. 1. The bafe HS of the AH SQ will be to the bafe SB of the ABSR, & VHSQ=VRS B. P.29 P. 4 B. 1. B. 1. But the RSH & H SQ together, are 2 L P.13. B. 1. 2. Confequently, VRSH+VRSB=2 L. Ax.1. B. 1. 3. Wherefore, VHSB is a straight line. 4. It may be demonftrated after the fame manner, that FD is a ftraight line. P.14. B. 1. Moreover, B D being & plle. to A G & A G = & plle. to F H. =& AG 5. The line BD will be = & plle. to F H. 6. And, confequently, F D is & plle. to H B. 7. From whence it follows, that F B & MS are in the fame plane FDBH. P.33. B. 1. P. 7. B.11. But in the AF MT, & TS B, the fides F M & S B are equal, (because the AF MT is = & to the AHSO, HS=SB), (Arg. 1). Moreover, VST BY FTM, & VFMTS P.15. B. 1. VTS B. P.29. B. 1. 8. Therefore, MT=TS, & FT=TB (P. 26. B.1.) that is, the common fection MS of the planes K O & LR, & the diameter FB of the parallelepiped, cut each other into two equal parts, in the point T. Which was to be demonftrated. Euclid |