THEOREM . EQUAL parallelograms (A B, BC), which have one angle of the one (F B D) equal to one angle of the other (GBE), have their fides (FB, BD & GB, B E), 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 (G B E) and the sides (F B, BD & G B, B E), about the equal angles reciprocally proportional, are equal. Hypothesis. Thesis. 1. The pgr. A B is = 10 the pgr. B C. FB : BE = GB : BD. II. V F B D is = to Y G B E. Preparation. may be in a straight line F E. Pof.2.B.1. I DEMONSTRATION. ECAUSE the y FB D, G Be are equal (Hyp. 2); & FB, B E are in a straight line FE (Prep. 1). 1. Therefore, GB, B D are in a ftraight line G D. P.14. B. 1. But the pgr. A B being = to the pgr. B C (Hyp. 1). 2. The Pgr. P. 7. B. 5. But the pgrs. AB, DE also BC, DE have the same altitude (D.4.B.6). 3: 'Hence Pgr. DE=FB:BE & k pgr. BC: pgr. DE=GB: BD. P. 1. B.6. 4. Consequently, FB:BE = GB : B D ( Arg. 2). P.11.B.5. Thesis. 1. FB; BE=GB: B.D. The pgr. A B is = to the pgr. B C. II. V FB Dis = to GBĖ. II. DEMONSTRATION. T may be demonstrated as before, that GB, BD are in the line GD. But the pgrs. AB, DE, & BC, DE, have the same altitude (D.4.B.6). 2. Hence, pgr. AB : pgr.DE=FB : BE, & pgr. BC: pgr.DE=GB: BD. P. 1. B.6. But FB:BE=GB: BD (Hyp.). BC. B C : pgr. Pgr. AB : 1.IT P.U. B.5. P. 9. B.5. PROPOSITION XV. THEOREM X. EQUAL triangles (AC B, E C D) which have one angle of the one (m) equal to one angle of the other (n): have their sides (A C, C B, & 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 sides (A C, CB, & EC, CD), about the equal angles reciprocally proportional, are equal to one another. CASE I. Thelis. 1. The A ACB is = to AECD. The fides AC, CB & EC, CD, II. Vm is = to y n. are reciprocally proportional, or AC: CD = EC: CB. Preparation. may be in the same straight line A D. Pof.1. B.1. DEMONSTRATION. BECAUS 1. P.14. B.I. 2. ECAUSE Vm= yn (Hyp. 2.), & the straight lines AC, CD P. 7. B.5. But the AACB, CB D also E CD, CB D have the same altitude (Prep. 2. Arg. 1. & D. 4. Rem. B. 6). 3. Wherefore the A ACB : ACBD = AC:CD. P. 1. B.6. & the AECD: A CBD = EC:CB. 4. Consequently, AC: CD= EC: CB (Arg.2.&P.11.B.5). Which was to be demonstrated. CASE II. Thesis. 1. AC: CD= EC:CB. The AACB, is in the AECD. II. & Vm = V n. Preparation. may be in the fame straight line A D. DEMONSTRATION. 1. 2. The T may be demonstrated, as in the first Case, that EC, CB are in P. 1. B.6. AC: CD=EC: CB. (Hyp.1). 3. Wherefore A ABC: ACBD = A ECD: ACBD 4. Consequently, the AABC is = to the AECD P. 9. B. 5. Which was to be demonstrated. P.U. B.5. 2. Therefore the sides of the rgles E B, F D about the equal VA&7. 01. B.5. N M C H D Thesis. Rgle. A B. N = Rgle. C D. M. Preparation. 1. At the extremities A & C, of AB,CD, erea the I AE,CF. P.11. B.5. 2. Make A E=N, & CFM. P. 3. B.I. 3. Complete the rgles. E B, F D. P.31. B.1. , D. 2. B.6. 3. Consequently, the rgle. EB = rgle. FD, or the rgle under AB.AE SP.14.B.6. = the rgle. under CD. CF. {D. IB.1. Consequently, A E being =N&CF=M (Prep. 2). 4. The rgle. under A B. N is also = to the rgle. under CD. M. Ax.2. B.2. Which was to be demonstrated. Thesis. AB:CD=M:N. rgle. N AE=N, & CF=M (Prep. 2). 1. The igle. under A B. A E is = to the rgle under CD.CF. Ax.2. B.s. But these fides being about the equal ÞEAB, FCD (Prep.1. & AX.10.B.1). AB; CD= CĖ: A E. P.14. B.6. CF being = M & A ESN (Prep. 2). P.7. && 11. B.5, Because P.7 3. (F three straight lines (AB, CD, M) be proportionals, the re&angle (AB.M) contained by the extremes is equal to the square of the mean (CD): And if the re&tangle 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. Thesis. AB :CD=CD: M. The rgle. AB.M is = to the of CD Preparation 1. At the extremities B & D of AB, CD erect the IBE, DF. Pu. B.s. 2. Make B E = M & DF =DC. P. 3. B.1. 3. Complete the rgles. E A, FC. P.zi. B.1. 1. DEMONSTRATION. ECAUSE AB: CD=M (Hyp.), & CD=DF & M=BE (Prep. 2) AB:CD=DF:BE. P.7. &U.B.S, herefore the sides of the rgles. E A, FC about the equal V B & D (Prep. 1. & Ax. 10. B. 1) are reciprocal. D. 2. B.6 2. Consequently, the rgle. E A is = to the rgle. FC, or the rgle. under A B. BE= the rgle Ç D. DF. SP.14. B.1. 3. Wherefore, B e being = M&DF=CD (Prep. 2), the rgle. D. 1.B.6. A B. M is also = to the O of C D. Ax. Thesis. AB:CD=CD:M II. DeMONSTRATION. BECAUS: ECAUSE 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. B E is = to the rgle, under CD. DF. Ax. 2.B.Z But those lides are about the equal y EBA, FDC (Ax. 10. B. 1. & Prep. 1). 2. Therefore, AB:CD=DF:BE. And fince DF=CD&B E=M (Prep. 2). 3. AB:CDCD: M. P.7. &11.B.S. Which was to be demonstrated. B.2. P.14. B.6. |