3. In a right-angled triangle, if the square on one side be equal to the rectangle contained by the hypotenuse and the other side, the hypotenuse is cut in “extreme and mean ratio” by the perpendicular on it from the right angle. 4. If AB be cut in “extreme and mean ratio” at C, prove that (1) AB2 + BC2 = 3AC2. (2) (AB + BC)2 = 5AC2. 5. The three lines joining the pairs of points G, B; F, D; A, K, in the construction of Proposition xi., are parallel. 6. If CH intersect BE in O, AO is perpendicular to CH. 7. If CH be produced, it meets BF at right angles. 8. ABC is a right-angled triangle having AB = 2AC: if AH be made equal to the difference between BC and AC, AB is divided in “extreme and mean ratio” at H. PROP. XII.—Theorem. In an obtuse-angled triangle (ABC), the square on the side (AB) subtending the obtuse angle exceeds the sum of the squares on the sides (BC, CA) containing the obtuse angle, by twice the rectangle contained by either of them (BC), and its continuation (CD) to meet a perpendicular (AD) on it from the opposite angle. Dem.—Because BD is divided into two parts in C, we have BD2 = BC2 + CD2 + 2BC.CD [iv.] and AD2 = AD2. Hence, adding, since [I. xlvii.] BD2 + AD2 = AB2, and CD2 + AD2 = CA2, we get Therefore AB2 is greater than BC2 + CA2 by 2BC.CD. The foregoing proof differs from Euclid’s only in the use of symbols. I have found by experience that pupils more readily understand it than any other method. Or thus: By the First Book: Describe squares on the three sides. Draw AE, BF, CG perpendicular to the sides of the squares. Then it can be proved exactly as in the demonstration of [I. xlvii.], that the rectangle BG is equal to BE, AG to AF, and CE to CF. Hence the sum of the two squares on AC, CB is less than the square on AB by twice the rectangle CE; that is, by twice the rectangle BC.CD. Cor. 1.—If perpendiculars from A and B to the opposite sides meet them in H and D, the rectangle AC.CH is equal to the rectangle BC.CD. Exercises. 1. If the angle ACB of a triangle be equal to twice the angle of an equilateral triangle, AB2 = BC2 + CA2 + BC.CA. 2. ABCD is a quadrilateral whose opposite angles B and D are right, and AD, BC produced meet in E; prove AE.DE = BE.CE. 3. ABC is a right-angled triangle, and BD is a perpendicular on the hypotenuse AC; Prove AB.DC = BD.BC. 4. If a line AB be divided in C so that AC2 = 2CB2; prove that AB2 + BC2 = 2AB.AC. 5. If AB be the diameter of a semicircle, find a point C in AB such that, joining C to a fixed point D in the circumference, and erecting a perpendicular CE meeting the circumference in E, CE2 − CD2 may be equal to a given square.