33rd CanMO 2001 Problem 3 The triangle ABC has AB and AC unequal. The angle bisector of A meets the perpendicular bisector of BC at X. The line joining the feet of the perpendiculars from X to AB and AC meets BC at D. Find BD/DC.
Let the perpendiculars from X to the lines AB, AC meet them at Z, Y respectively. Triangles XBZ, XYC are congruent because XB = XC (X lies on angle bisector), XZ = XY (X lies on perpendicular bisector) and ∠BZX = ∠CYX = 90 o . Hence BZ = CY. Also AZ = AY. By Ceva's theorem, (AZ/ZB) (BD/DC) (CY/YA) = 1. Hence BD/DC = 1. 33rd CanMO 2001 © John Scholes jscholes@kalva.demon.co.uk 10 June 2002