ABC is a triangle. The incircle touches BC, CA, AB at D, E, F respectively. X is a point inside the triangle such that the incircle of XBC touches BC at D. It touches CX at Y and XB at Z. Show that EFZY is cyclic.
Solution
If EF is parallel to BC, then AB = AC, D is the midpoint of BC, and AD is perpendicular to BC. Hence X lies on AD and YZ is also parallel to BC. So EF is parallel to YZ and AD is an axis of symmetry. Hence EFZY is cyclic.
So we may assume EF meets BC at P. Hence, by Menelaus' theorem, (AF/FB) (BP/CP) (CE/EA) = 1. But AF = AE (equal tangents to incircle), so (1/FB) (BP/CP) CE = 1. But XY = XZ (equal tangents), and BF = BD (equal tangents) = BZ (equal tangents), and CE = CD (equal tangents) = CY (equal tangents). So (XZ/BZ) (BP/CP) (CY/XY) = 1. Hence applying Menelaus to the points Z, P, Y on the sides of the triangle XBC, we conclude that Z, P, Y are collinear.
Hence PE.PF = PD2 and PY.PZ = PD2. So PE.PF = PY.PZ. Hence EFZY is cyclic.
© John Scholes
jscholes@kalva.demon.co.uk
20 Sep 2002