To PROPOSITION XXIX. PROBLEM IX. O a given straight line (A B), to apply a parallelogram (A G), equal to a given rectilineal figure (V), exceeding by a parallelogram (MI), fimilar to another given (T). Given. 1. The Araight line AB, & the pgr. T. II. The rectilineal figure V. THE Sought. The conftruction of a pgr. AG, applied to A B, equal to the rectilineal figure V, & baving for excefs a pgr. MI, fimilar to T. Refolution. 1. Divide A B into two equal parts in E. 3. Defcribe a pgr. X (or PS) = V +ED, fimilar & fimi- 4. Since X, (asV+E D), is > ED; the fide R Sis FD, P.10. B.I. P.18. B.1. pgr. F KG N, which will be equal & fimilar to the pgг. X. P.31. B.1, DEMONSTRATION. HE pgr. K N being equal and fimilar to the pgr. X, which is itfelf fimilar to the pgr. ED (Ref. 3). 1. The KN is fimilar to the pgr. pgr. ED. P.21. B.6. 2. Wherefore thofe two pgrs K N, ED are about the fame diagonal. P.26. B.6. Draw this diagonal F B G, & complete the description of the figure. Since X is to V + ED; & X=pgr. KN (Ref. 3. & 4). 3. The pgr. KNV+ED. Ax.1.B.I. Therefore taking away from both fides the common pgr. ED. 4. The remaining gnomon a b c is to the rectilineal figure V. But because A E≈ EB (Ref. 1). Ax.3. B.1. 5. The pgr. AK the pgr. E I. 6. Confequently, this pgr. A K is to the pgr. NB. 235 Therefore adding to both fides the common pgr. M K 7. There will refult the pgr. A G to the gnomon a bc. But the gnomon abc is to the rectilineal figure V (Arg. 4). 8. Confequently, the pgr. AG is to the rectilineal figure V. Since then this pgr. A C has for excefs the pgr. M I, fimilar to the pgr. ED (P. 24. B. 6.); & confequently fimilar to the pgr. T (Ref. 2. P. 21. B. 6). 9. There has been applied to A B, a pgr. A G to the rectilineal figure V, having for excefs a pgr. MT, fimilar to the pgr. T. Which was to be done. R E MARK. Ax.2. B.1. Ax.1. B.1. IF as in the foregoing cafe AB be made = a, the given fquare V_ (reduced to a pgr. equiangular to the pgr. T)=nl; the ratio of the fides QP, PR of the pgr. X (which is the fame as that of the fides of the pgr. T) m: n; & A M=x, conJequently, MB x — a. there will refult an equation of the fame kind. For fince the defect MI should be fimilar to the pgr. T or X, we will have as before the following proportion. QP: PR MB MG (D. 1. B. 6). m m: n =x1a: And because the pgr. AG (AM. MG) should be equal to the given Space V (nl), there refults the following equation, PROPOSITION XXX. PROBLEM X. To cut O cut a given ftraight line (A B) in extreme and mean ratio (in E). Given. The ftraight line A B. Refolution. Sought. BA: AE 1. Upon the ftraight line A B defcribe a fquare B C. AE; BE P.46. B.1. 2. Apply to the fide C A, a pgr. CD to the fquare B C, P.29. B.6. BECA DEMONSTRATION. ECAUSE BC=CD (Ref. 2); by taking away the common rgle. C E from each. 1. The remainder BFA D. But BF is alfo equiangular with AD (P. 15. B. 1). 2. Therefore their fides FE, E B, ED, A E about the equal angles, 3. Wherefore, But because BA is > AE (Ax. 8. B. 1). 4. The ftraight line A E is > E B. Ax. 3. B.1. P.14. B.6. D. 30. B.1. P.7.11.B.5. 5. Confequently, the straight line AB is cut in extreme & mean ratio in E. Otherwife. Divide B A in E, fo that the rect. A B. BE be to the of A E DEMONSTRATION. P.14. B.5. Which was to be done. P.11. B.2. BECAUSE AE: BE. P.17. B.6. (Ax. 8. B. 1). 3. Confequently, the straight line AB is cut in extreme & mean ratio in E. IN PROPOSITION XXXI. THEOREM XXI N every right angled triangle (A B C), the rectilineal figure (E) defcribed upon the hypothenule (A C) is equal to the fum of the fimilar and fimilarly defcribed figures (G & H), upon the fides (A B, BC) containing the right angle. Hypothefis. 1. A B C is a rgle. ▲ in B. II. The fig. E is defcribed upon the bypoth. A C of this A. III. And the figures G&H are fimilar to E, & fimilarly defcribed upon the two other fides AB, B C. BECAUS DEMONSTRATION. Thefis. fg. Efig. G+H ECAUSE the figures E, G, H are fimilar, & fimilarly described upon the homologous fides A C, A B, BC (Hvp. 3). I. And GE of A B□ of A C. Į HEO of B C of A C. J SP 20. B.6. Cor. 2. 2. Confequently, G+HE□ of A B+ of BC of A C. P.24. B.5. But because the AABC is rgle. in B (Hyp. 1). 3. The of AB+ of BC is to the of A C. 4. Therefore, the figure E is to the figures G + H. P.47. B.1. SP.16. B.5. Cor. Which was to be demonstrated. IF PROPOSITION XXXII. THEOREM XXII. F two triangles (A BC, CDE), which have two fides (A B, B C) of the one, proportional to two fides (CD, DE) of the other, be joined at one angle (C), so as to have their homologous fides (A B, CD, B C, D E) parallel to one another, the remaining fides (A C, CE) fhall be in a ftraight line. Hypothefis. Thefis. The remaining fides AC, CE of thofe A are in a fraight line A E. to DE. DEMONSTRATION. BECAUS ECAUSE the plles. A B, C D are cut by the ftraight line BC, & the plles. B C, DE by the ftraight line DC (Hyp. 2). 1. The B is to VBCD & V Ď is = to V BCD. 2. Confequently, V B is to VD. = And befides A B BCCD: DE (Hyp. 1). 3. The AABC, C D E are equiangular. 4. Therefore, VA is to VDCE, being oppofite to the = homologous fides B C, D E. Adding then to both fides VB, or its VBCD (Arg. 1), together with the common VB C A. 5. The VA+B+BCA will be to the VDCE+BCD+BCA. But the A+ B+ BCA are to 2 L (P. 32. B. 1). 6. Confequently the VDCE+BCD+BCA are alfo P.29. B.1. Ax.1. B. 1. P. 6. B.6. Ax.2. B.1. to 2 L. Ax.1. B.1. P.14. B.1. Which was to be demonftrated. 7. Wherefore the straight lines AC, CE are in the same straight line A E. |