5th Iberoamerican 1990

I is the incenter of the triangle ABC and the incircle touches BC, CA, AB at D, E, F respectively. AD meets the incircle again at P. M is the midpoint of EF. Show that PMID is cyclic (or the points are collinear).

Solution

∠AEI = ∠AME = 90^o, so AEI and AME are similar. Hence AM/AE = AE/AI or AM·AI = AE². AE is tangent to the incircle, so AE2 = AP·AD. Hence AM·AI = AP·AD, so if P,M,I,D are not collinear, then they are cyclic.

Thanks to Johann Peter Gustav Lejeune Dirichlet

5th Ibero 1990

© John Scholes
jscholes@kalva.demon.co.uk
1 July 2002