The incircle of the triangle ABC touches BC, CA, AB at D, E, F respectively. AD meets the circle again at Q. Show that the line EQ passes through the midpoint of AF iff AC = BC.
Solution
∠AQM = ∠EQD (opposite angle) = ∠EDC (CD tangent to circle EQD) = (180o - ∠C)/2 = ∠A/2 + ∠B/2 (*).
MF2 = MQ.ME (MF tangent to circle FQE). So AM = AF is equivalent to AM2 = MQ.ME or AM/MQ = ME/AM. But since triangles AMQ and EMA have a common angle M, AM/MQ = ME/AM iff they are similar, and hence iff ∠AQM = ∠A. Using (*) AM = AF iff ∠A = ∠B.
© John Scholes
jscholes@kalva.demon.co.uk
24 Oct 2002