EQ PROPOSITION XIV. THEOREM IX. QUAL parallelograms (A B, BC), which have one angle of the one (FBD) equal to one angle of the other (G BE), have their fides (FB, B D & G B, BE), about the equal angles reciprocally proportional, (that is, FB BE GB: BD). And parallelograms that have one angle of the one (FBD) equal to one angle of the other (GBE) and the fides (F B, BD & GB, B E), about the equal angles reciprocally proportional, are equal, Thefis. Hypothefis. 1. The pgr. A B is to the pgr. B C. II. VFBD is to YG BE. Preparation. FB: BE GB: BD. 1. Place the two pgrs. A B, BC fo as the fides FB, BE DE. 2. Complete the pgr. D E. BECAUSE I. DEMONSTRATION. ECAUSE the VFBD, GBE are equal (Hyp. 2) ; & F B, 1. Therefore, GB, B D are in a straight line G D. But the pgr. A B being to the pgr. B C (Hyp. 1). 2. The pgr. A B: pgr. DE = pgr. BC: pgr. DE. But the pgrs. AB, DE alfo BC, DE have the fame altitude (D.4.B.6). 3. Hence pgr. AB: pgr. DE FB: BE & pgr. BC: pgr. DE 4. Confequently, FB: BE GB BD (Arg. 2). 1.IT may = Pof.2. B.1. P.14. B. 1. P. 7. B. 5. GB: BD. P. 1. B. 6. Which was to be demonftrated. II. DEMONSTRATION. be demonstrated as before, that GB, BD are in the line GD. But the pgrs. AB, DE, & BC, DE, have the fame altitude (D.4 B.6). 2. Hence, pgr. AB: pgr.L .DE=FB: BE, & pgr. BC: pgr.l .DE=GB": : BD. P. 1. B.6. PROPOSITION XV. E THEOREM X. EQUAL triangles (ACB, ECD) which have one angle of the one (m) equal to one angle of the other (n): have their fides (A C, CB, & EC, CD), about the equal angles, reciprocally proportional; & the triangles (ACB, ECD) which have one angle in the one (m) equal to one angle in the other (n), and their fides (A C, CB, & E C, CD), about the equal angles reciprocally proportional, are equal to one another. Hypothefis. CASE I. 1. The AAC B is to AEC D. Thefis. The fides AC, CB& EC, CD, are reciprocally proportional, er AC: CDEC: CB. 1. Place the AACB, ECD fo that the fides AC, CD may be in the fame ftraight line A D. 2. Draw the straight line B D. BECAUS DEMONSTRATION. ECAUSE Vmn (Hyp. 2.), & the ftraight lines AC, CD are in the fame ftraight line A D (Prep. 1). 1. The lines E C, C B are also in a straight line E B. But the AAC B being to the AECD (Hyp. 1). 2. The AACB: ACBD-AECD: ACBD. But the AAC B, C B D alfo E CD, CBD have the fame altitude (Prep. 2. Arg. 1. & D. 4. Rem. B. 6). 3. Wherefore the AACB: ACBD AC: CD. & the AECD: ACBD = EC: CB. Pof.1. B.1. P.14. B.I. P. 7. B.5. }P. 1. P. 1. B.6. Which was to be demonftrated. 4. Confequently, AC: CD EC: CB (Arg.2.&P.11.B.5). 1. Place the two AA CB, E CD fo that the fides A C, CD, 2. Draw the straight line B D. DEMONSTRATION. 1.IT may be demonftrated, as in the firft Cafe, that EC, CB are in the fame ftraight line EB. And because the ▲ A CB, CBD, alfo the AECD, CBD have the fame altitude (Prep. 2. Arg. 1. & D. 4 Rem. B. 6). AACB ACBD=AC: CD. : Likewife A ECD: ACBD 2. The EC: CB. EC: CB. (Hyp.1). 4. Confequently, the AABC is to the AECD. Which was to be demonstrated. } P. 1. P. 1. B.6. P.11. B.5. PROPOSITION XVI. THEOREM XI. F four ftraight lines (A B, C D, M, N) be proportionals, the rectangle contained by the extremes (A B. N) is equal to that of the means (C D. M). And if the rectangle contained by the extreames (A B. N) be equal to the rectangle contained by the means (C D. M), the four straight lines (AB, CD, M, N) are proportionals. Hypothefis. AB: CDM: N. Thefis. Preparation. 1. At the extremities A & C, of AB,CD, erect the L AE,CF. P.11. B.5. 2. Make A E=N, & CF = M. 3. Complete the rgles. E B, F D. I. DEMONSTRATION. BECAUSE AB: CD=M: N (Hyp.): & M-CF & N=AE (Prep.2). 1. AB: CDCF: AE. P. 3. B.i. P.7.& 11. B.5. 2. Therefore the fides of the rgles E B, F D about the equal VA & Ć, (Prep. 1. & Ax. 10. B. 1.) are reciprocal. 3. Confequently, the rgle. EB 1 the rgle. under CD. CF. Confequently, A E being = N 4. The rgle. under A B. N is also Hypothefis. D. 2. B.6. rgle. FD, or the rgle under AB.AE SP.14 B.6. & CFM (Prep. 2). The rgle. AB. N is to the rgle. CD. M. II. DEMONSTRATION. {D. B.I. Ax.2. B.2. Thefis. AB: CDM: N, BECAUSE the rgle. AB. N is to the rgle CD. M (Hyp.) : & AEN, & CFM (Prep. 2). 1. The rgle. under A B. A E is to the rgle under C D. C F. But these fides being about the equal VEAB, FCD (Prep.1. &Ax.10.B.1). IF PROPOSITION XVII. THEOREM XII. F three ftraight lines (AB, CD, M) be proportionals, the rectangle (AB.M) contained by the extremes is equal to the fquare of the mean (CD): And if the rectangle contained by the extreams (AB.M) be equal to the square of the mean (CD), the three straight lines (AB, CD, M) are proportionals. Hypothefis. AB: CD CD: M. Thefis. The rgle. AB.M is to the of CD. Preparation. 1. At the extremities B & D of AB, CD erect the LBE, DF. P.11. B.1. 2. Make BEM & DFDC. 3. Complete the rgles. E A, FC. BECAUS 1. I. DEMONSTRATION. ECAUSE AB: CD=M (Hyp.), & CD=DF & M = BE (Prep. 2). AB: CD DF: BE. P. 3. B.1. P.7. &11.B.5. Therefore the fides of the rgles. E A, FC about the equal V B & D (Prep. 1. & Ax. 10. B. 1) are reciprocal. 2. Confequently, the rgle. E A is to the rgle. FC, or the rgle. under A B. BE the rgle C D. DF. D. 2. B.6. 3. Wherefore, BE being M & DF CD (Prep. 2), the rgle. D. 1. B.6. A B. M is also to the of CD. Ax. 2. B.2. AB: CD CD: M BECAUSE the rgle. under AB.M is to the of CD (Hyp.), & that B E is M&DF = CD (Prep. 2). 1. The rgle. under A B. BE is to the rgle. under C D. DF. |
