PROP. XIII. THEOR. Book II. IN N every triangle, the fquare of the fide fubtending any of the acute angles, is lefs than the fquares of the fides containing' that angle, by twice the rectangle contained by either of these fides, and the ftraight line intercepted between the perpendicular, let fall upon it from the oppofite angle, and the acute angle. Let ABC be any triangle, and the angle at B one of its acute angles, and upon BC, one of the fides containing it, let fall the perpendicular 2 AD from the oppofite angle: The a 12.1. fquare of AC, oppofite to the angle B, is lefs than the squares a of CB, BA by twice the rectangle CB.BD. Firft, Let AD fall within the triangle ABC; and be cause the straight line CB is di vided into two parts in the point Db, BC2+BD2 = 2BC.BD+CD2. Add to each AD2; then BC2 + BD2 + AD2 = 2BC.BD + CD2 + AD2. But BD2+ AD2= AB2, and CD2+DA2 AC2c; therefore BC2+AB2BC,BD+AC2; that is, AC is lefs than BC2+ B. D AB2 by 2BC.BD. Secondly, Let AD fall without the triangle ABC*: Then because the angle at D is a right angle, the angle ACB is great b 7.2. 1 € 47. I. erd than a right angle, and AB2e AC2+BC2+2BC.CD. d. 16. 1. Add BC to each; then AB2+BC2 AC2+2BC2+2BC.CD. But e 12. 2. See figure of the last Froposition. Book II. £ 3 2. But because BD is divided into two parts in C, BC2 + Laftly, Let the fide AC be perpendicular to BC; then is BC the ftraight line between the perpendicular and the acute angle at B; and it is manifest that 847.1. g AB2 + BC2 = AC2 + 2BC2 = AC2 + Therefore in every triangle, 2BC.BC &c. Q. E. D. A PROP. XIV. PRO B. B a 45. I. b 5.2. defcribe a fquare that fhall be equal to a T given rectilineal figure. Let A be the given rectilineal figure; it is required to defcribe a fquare that fhall be equal to A. Describe a the rectangular parallelogram BCDE equal to the rectilineal figure A. If then the fides of it, BE, ED are equal to one another, it is a square, and what was required is done; but if they are not equal, produce one of them BE to F, and make EF equal to ED, and bisect BF in G: and from the centre G, at the distance GB, or GF, defcribe the femicircle BHF, and produce DE to H, and join GH. Therefore, because the straight line BF is divided into two equal parts in the point G, and into two unequal in the point E, the rectangle BE.EF, together with the fquare of EG, is equal to the square of GF: but GF is equal to GH; there b fore fore the rectangle BE.EF, together with the fquare of EG, Book II. is equal to the fquare of GH: But the fquares of HE and EG are equal to the fquare of GH : Therefore also the rectangle BE.EF together with the fquare of EG, is equal to the fquares of HE and EG. Take away the fquare of EG, which is common to both, and the remaining rectangle BE.EF is equal to the fquare of EH: But BD is the rectangle contained by BE and EF, because EF is equal to ED; therefore BD is equal to the fquare of EH; and BD is also equal to the rectilineal figure A; therefore the rectilineal figure A is equal to the fquare of EH: Wherefore a square has been made equal to the given rectilineal figure A, viz. the square defcribed upon EH. Which was to be done. PROP. A. THEOR. If one of F one fide of a triangle be bifected, the fum of the See N. fquare of half the fide bifected, and of the fquare of the line drawn from the point of bifection to the oppofite angle of the triangle. Let ABC be a triangle, of which the fide BC is bifected in D, and DA drawn to the oppofite angle; the fquares of BA and AC are together double of the fquares of BD and DA. Book II. a 47. I. From A draw AE perpendicular to BC, and becaufe BEA is a right angle, AB a BE2+AE2, and ACCE2+ AE; wherefore AB+AC BE2+CE2+2AE2. But because the line BC is cut equally in D, b9 2. andunequally in E, BE2+CE 2BD +2DE; therefore AB2+ D 2DE+2AE22AD2; wherefore AB2+ AC2 2 BD2+ 2AD. Therefore, &c. Q. E. D. a 15. 1. b 29. 1. PROP. B. THEOR. HE fum of the fquares of the diameters of any T'parallelogram is equal to the fum of the fquares of the fides of the parallelogram. Let ABCD be a parallelogram, of which the diameters are AC and BD; the fum of the fquares of AC and BD is equal to the fum of the fquares of AB, BC, CD, DA. Let AC and BD interfect one another in E: and because the vertical angles AED, CEB are equal a, and also the alternate angles EAD; ECB, the triangles ADE, CEB have two angles in the one equal to two angles in the other, each to each: but the fides AD and BC, which are oppofite to equal angles in these triangles, 34. 1. are alfo equal; therefore the other fides which are oppofite to the equal d 26. 1. angles are alfo equal, viz. AE to EC, and ED to EB. ė A. 2. Since, therefore, BD is bifected in E, AB2 + B AD2e2BE2+2AE2; and for the fame reafon, CD2+ A D E BC22BE2 + 2EC2 = 2BE2 + 2AE2, becaufe ECAE. 4BE2= |