If LI A , PROPOSITION XL. THEOREM XXXV. M F two triangular prisms (FL & EC) have the same altitude (L I & A E), and the base of one (as C L) is a parallelogram (F I), and the base of the other (E C) a triangle (A B C): if the parallelogram be double of the triangle, the first prism (L F) will be equal to the second (E C). Hypothesis. Thesis. 1. In the prisms F L & EC; the alt. LI The prison FL is = to the prism EC. is = to the alt. A E. II. The base of the prism L F is a pgr. F I, & the base of the prism EC a D'ABC. III. The pgr. F l is double of the A A BC. Preparation Complete the ENI & BD. DEMONSTRATION. Because ECAUSE the pgr. F I, base of the prism F L, is double of the Pgr. P.41. B. 1. 1. The pgr. F I is = to the Pgr. во. Moreover, the altitude Ll being to the altitude A F (Hyp. 1), 2. The EBD is = to the ENI. The given prism L F is the half of the END. P.31. B.11. And the prism E C is the half of the eBD. P.28. B.u. 3. Consequently, the prism F L is to the prifin eć. Ax.7. B. 1. 2 PROPOSITION 1. THEOREM I. IMILAR polygons (ABCDE & F GHI K), inscribed in circles Thesis. 1. The polygons À BCD E&FGHI K. Polyg. : ACE: polyg. FIH=the o are u. of the diam. EL: 0 of the diam. G M, II. They are inscribed in circles. or as diam. EL2 : diam. G M2. Preparations. In the FMH, draw the homologous lines FM & Pof.1. B. 1. DEMONSTRATION. And the A or EAB is to VGFK, & AE:ABFG:FK (D. 1. B. 6). P. 6. B. 6. 2. Wherefore, A A B E is o to AGFK, & VaV b, allo y c Y d. But Y EL A is = VEBA, or a, & V GMF=VGKF or b. P.21. B. 6. 3. Consequently, VEL A is = to VGMF. Ax.1. B. I. 4 Likewite, VEAL = VGFM. P.31. B. 3. And, because, in the two A ALE & G FM, the two WELA & EAL of the first are = to the two Y GMF & G FM of the second (Arg. 3. & 4). The third Ý A E L of the AEAL will be to the third P.32. B. 1. P. 4 EL: GM = AE: GF. polygons are inscribed. Which was to be demonstrated. B. 6. P.16. B. 5. L E M M A. M F from the greater (A B), of two unequal magnitudes (A B & C), there be taken more than its half (viz. A H), and from the remainder (H B) more than its half (viz. H K), and so on : there shall at length remain a magtitude (K B), less than the least (C), of the proposed magnitudes. Preparation. Pol.1. B. five remainders, until the number of times, be equal to the DEMONSTRATION. If there be taken from it a magnitude GI=C. 1. The remainder E G will be the half of El. But E I is > AB (Prep. 1). 2. Consequently, the half of EI is > the half of A B. 2.19. B. 5 3. Therefore, GE will be much > the half of A B. But H B is < the half of AB (Prep. 2). 4. Much more then G E is > HB. 5. Therefore, EF, the half of E G, is > the half of H B. And KB is the half of H B (Prep. 3). 6. Consequently, E F is > K B. And as the same reasoning may be continued until a part (E F) of the multiple of the magnitude C be attained, which will be equal to C (Prep. 4). > It follows, that the magnitude C will be the remaining part (KB) of the greater A B. Which was to be demonstrated. BECAUSE CIRCLE PROPOSITION II. THEOREM II. IRCLES (A F D & IL P), are to one another as the squares of their diameters (A E & IN). Hypothesis. Thesis. in the circles AFDES I L P there has O AFD: OILP= AE : IN2. been drawn the diameters A E ES I N. DEMONSTRATION. A E2 is to I N2 as the O AF D is to a space T (which P. 6. B. 4. 1. Supposition. 1. Preparation. P.30. B. 3: Pof.1. B. 1. 4. Thro the point K, draw SR plle. to LI. P.31. B. 1. 5. Produce N'L & P I to R&S; which will form the rgle. SRIL. the OIL P. Becaus ECAUSE the described about the OIL P is > the 0 itself. Ax.8. B. 1. 1. The half of this will be > the half of the O IL P. P.19. B. 5 But the inscribed O ILNP is = to half of the circumscribed o (the side of the circumscribed being to the diameter, & the of the diameter = OLI+OLN=2OLI). P.47. B. 1. 2. Therefore, the LIPN is > the half of the O IL P. Axi. B. 1. The rgle. Siis > the segment LKI (Prep. 5. & Ax. 8. B. 1). 3. Consequently, the half of the rgle. S I is the half of the segment LKI. P.19. B. 5 The LKI is = to half of the rgle. S I. P 41. B. i. 4. Therefore, the ALKI is > the half of the segment L KI. P.i9. B. 5. 5 5. It may be proved after the same manner, that all the ALMN, NOP, &c. are each > the half of the segment in which it is placed. half of all those segments. It will be proved after the same manner. ments, are together > the half of the segments in which thote be taken more than the half, & so on. Lem. B.12. But the O ILP is = T + V (1. Sup.). Therefore, taking those segments L KI, &c. from the OIL P. And the space V, from T +V (which is > those segments). 9. The remainder, viz. the polygon I KLMNOPQ will be > T. Ax.5. B. 1. But the polyg. ADFK: polyg. I LOQ=of AE: of IN. P. 1. B.12. <V. |