In like manner, we have, BC2 : AC2 :: BC : DC. Cor. 3. The squares of the sides about the right angle are to cach other as the adjacent segments of the hypothenuse. For, by combining the propor tions of the preceding corollary (B. II., P. IV., C.), we have, B AB2 : 40 :: BD : DC. Α Cor. 4. The square described on the diagonal of a square is double the given square. For, the square of the diagonal is equal to the sum of the squares of the two sides ; but the square of each side is equal to the given square: hence, - AC2 ÁÇ2 = 2ĀB2 ; or, AC2 = 2BC2. A Cor. 5. From the last corollary, we have, AC2 : AB2 :: 2 : 1 ; H D G hence, by extracting the square root of each term, we have, AC : AB :: √2 : 1; that is, the diagonal of a square is to the side, as the square root of two to one; consequently, the diagonal and the side of a square are incommensurable. PROPOSITION XII. THEOREM. In any triangle, the square of a side opposite an acute angle, is equal to the sum of the squares of the base and the other side, diminished by twice the rectangle of the base and the distance from the vertex of the acute angle to the foot of the perpendicular drawn from the vertex of the opposite angle to the base, or to the base produced. Let ABC be a triangle, C one of its acute angles, BC its base, and AD the perpendicular drawn from A to BC, or BC produced; then will A AB2 = BC2 + AC2 – 2BC × CD. · D For, whether the perpendicular meets the base, or the base produced, we have BD equal to the difference of BC and CD: hence (P. IX.), In any obtuse-angled triangle, the square of the side opposite the obtuse angle is equal to the sum of the squares of the base and the other side, increased by twice the rectangle of the base and the distance from the vertex of the obtuse angle to the foot of the perpendicular drawn from the vertex of the opposite angle to the base produced. Let ABC be an obtuse-angled triangle, B its obtuse angle, BC its base, and AD the perpendicular drawn from A to BC produced; then will AC2 = BC2 + AB2 + 2BC × BD; which was to be proved. Scholium. The right-angled triangle is the only one in which the sum of the squares described on two sides is equal to the square described on the third side. PROPOSITION XIV. THEOREM. In any triangle, the sum of the squares described on two sides is equal to twice the square of half the third side increased by twice the square of the line drawn from the middle point of that side to the vertex of the opposie angle. Let ABC be any triangle, and EA a line drawn from the middle of the base BC to the vertex A: then will Draw AD perpendicular to BC; then, from Proposition XII., we have, Adding these equations, member to member (A. 2), recollect. ing that BE is equal to EC, we have, AB2 + A02 2BE2 + 2 EA2 ; which was to be proved. Cor. Let ABCD be a parallelogram, and BD, AC, its diagonals. Then, since the diagonals mutually bisect each other (B. I., P. XXXI.), we shall have, B E and, A ; CD2 + DA2 = 2 CE2 + 2DE whence, by addition, recollecting that AE is equal to CE, and BE to DE, we have, AB2 + BC2 + CD2 + DA = 4CE2 + 4DE2; but, 4 CE is equal to AC2, and 4DE2 to BD2 (P. VIII, C.): hence, AB2 + BC2 + CD2 + DA2 = AC2 + BD'. That is, the sum of the squares of the sides of a parallelogram, is equal to the sum of the squares of its diagonals. PROPOSITION XV. THEOREM. In any triangle, a line drawn parallel to the base divides the other sides proportionally. Let ABC be a triangle, and DE a line parallel to the base BC: then AD : ᎠᏴ :: AE Draw EB and DC. Then, because the triangles AED and DEB have their bases in the same line AB, and their they will vertices at the same point E, CE. E C.) AED : DEB :: AD : DB. B The triangles AED and EDC, have their bases in the same line AC, and their vertices at the same point D; they have, therefore, a common altitude; hence, But the triangles DEB and EDC have a common base DE, and their vertices in the line BC, parallel to DE; they are, therefore, equal: hence, the two preceding proportions have a couplet in each equal; and consequently, the remaining terms are proportional (B. II., P. IV.), hence, AD : ᎠᏴ :: AE : EC; which was to be proved. Cor. 1. We have, by composition (B. II., P. VI.), AD + DB : AD :: AE EC : AE; |