IMO 1990 SL 12

Origin: IRE

Problem

Let ABC be a triangle and L the line through C parallel to the side AB. Let the internal bisector of the angle at A meet the side BC at D and the line L at E and let the internal bisector of the angle at B meet the side AC at F and the line L at G. If GF = DE, prove that AC = BC.

Solution

Let d(X, l) denote the distance of a point X from a line l. Using the elementary facts that AF : FC = c : a and BD : DC = c : b, we obtain d(F, L) = a a+chc and d(D, L) = b b+chc, where ha is the altitude of \triangleABC from A. We also have \angleFGC = \beta/2, \angleDEC = \alpha/2. It follows that DE = d(D, L) sin(\alpha/2) and FG = d(F, L) sin(\beta/2). (1) Now suppose that a > b. Since the function f(x) = x x+c is strictly increas- ing, we deduce d(F, L) > d(D, L). Furthermore, sin(\alpha/2) > sin(\beta/2), so we get from (1) that FG > DE. Similarly, a < b implies FG < DE. Hence we must have a = b, i.e., AC = BC.