20th Brazil 1998 Problem A2 ABC is a triangle. D is the midpoint of AB, E is a point on the side BC such that BE = 2 EC and ∠ADC = ∠BAE. Find ∠BAC.
Let F be the midpoint of BE. Reflect BC in the line BD to get BH. Let G be the reflection of F. Then ∠GDB = ∠FDB. But D is the midpoint of AB and F is the midpoint of BD, so AE is parallel to DF and hence ∠FDB = ∠EAB. Hence ∠GDB = ∠ADC, so D lies on the line CG. Let CH meet the line BD at A'. Evidently ∠CA'B = 90 o . We want to show that A and A' coincide. Consider the triangle A'HB and the line CG. We have (CA'/CH)(DB/DA')(GH/GB) = 1. But CA'/CH = 1/2 and GH/GB = 2, so DB/DA' = 1. Hence AB = A'B = 2 DB and so A and A' do coincide. 20th Brazil 1998 © John Scholes jscholes@kalva.demon.co.uk 22 Oct 2003 Last corrected/updated 22 Oct 03