6. If the square of a line CD, drawn from the angle C of an equilateral triangle ABC to a point D in the side AB produced, be equal to 2AB2; prove that AD is cut in “extreme and mean ratio” at B. PROP. XIII.—Theorem. In any triangle (ABC), the square on any side subtending an acute angle (C) is less than the sum of the squares on the sides containing that angle, by twice the rectangle (BC, CD) contained by either of them (BC) and the intercept (CD) between the acute angle and the foot of the perpendicular on it from the opposite angle. Dem.—Because BC is divided into two segments in D, and AD2 = AD2. Hence, adding, since CD2 + AD2 = AC2 [I. xlvii.], and BD2 + AD2 = AB2, we get BC2 + AC2 = AB2 + 2BC.CD. Therefore AB2 is less than BC2 + AC2 by 2BC.CD. Or thus: Describe squares on the sides. Draw AE, BF, CG perpendicular to the sides; then, as in the demonstration of [I. xlvii.], the rectangle BG is equal to BE; AG to AF, and CE to CF. Hence the sum of the squares on AC, CB exceeds the square on AB by twice CE—that is, by 2BC.CD. Observation.—By comparing the proofs of the pairs of Props. iv. and vii.; v. and vi.; ix. and x.; xii. and xiii., it will be seen that they are virtually identical. In order to render this identity more apparent, we have made some slight alterations in the usual proofs. The pairs of Propositions thus grouped are considered in Modern Geometry not as distinct, but each pair is regarded as one Proposition. Exercises. 1. If the angle C of the △ ACB be equal to an angle of an equilateral △, AB2 = AC2 + BC2 − AC.BC. 2. The sum of the squares on the diagonals of a quadrilateral, together with four times the square on the line joining their middle points, is equal to the sum of the squares on its sides. 3. Find a point C in a given line AB produced, so that AC2 + BC2 = 2AC.BC. PROP. XIV.—Problem. To construct a square equal to a given rectilineal figure (X). Sol.—Construct [I. xlv.] the rectangle AC equal to X. Then, if the adjacent sides AB, BC be equal, AC is a square, and the problem is solved; if not, produce AB to E, and make BE equal to BC; bisect AE in F; with F as centre and FE as radius, describe the semicircle AGE; produce CB to meet it in G. The square described on BG will be equal to X. Dem.—Join FG. Then because AE is divided equally in F and unequally in B, the rectangle AB.BE, together with FB2 is equal to FE2 [v.], that is, to FG2; but FG2 is equal to FB2 + BG2 [I. xlvii.]. Therefore the rectangle AB.BE + FB2 is equal to FB2 + BG2. Reject FB2, which is common, and we have the rectangle AB.BE = BG2; but since BE is equal to BC, the rectangle AB.BE is equal to the figure AC. Therefore BG2 is equal to the figure AC, and therefore equal to the given rectilineal figure (X).