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.
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.