28th CanMO 1996 Problem 4 In the triangle ABC, AB = AC and the bisector of angle B meets AC at E. If BC = BE + EA find angle A.
Take X on BC with BX = BE. Then CX = AE and so CX/CE = AE/CE = AB/BC = AC/BC. The the triangles BAC and CXE have a common angle C and the ratio of the two sides containing the angle the same. Hence they are similar. So XC = XE. Hence ∠BXE = 2 ∠C. But ∠C = 90 o - A/2 and ∠BXE = 90 o - B/4 (since BEX is isosceles) = 90 o - (90 o - A/2)/4. Solving gives A = 100 o . 28th CanMO 1996 © John Scholes jscholes@kalva.demon.co.uk 15 June 2002 Last corrected/updated 28 Nov 03