16th Iberoamerican 2001

ABC is a triangle. The incircle has center I and touches the sides BC, CA, AB at D, E, F respectively. The rays BI and CI meet the line EF at P and Q respectively. Show that if DPQ is isosceles, then ABC is isosceles.

Solution

AF = AE, so ∠AFE = 90^o - A/2. Hence ∠BFP = 90^o + A/2. But ∠FBP = B/2, so ∠FPB = C/2. But BFP and BDP are congruent (BF = BD, BP common, ∠FBP = ∠FDP), so ∠DPB = C/2 and ∠DPQ = C. Similarly, ∠DQP = B. Hence ∠PDQ = A. So DQP and ABC are similar. So if one is isosceles, so is the other.

Thanks to Johann Peter Gustav Lejeune Dirichlet

16th Ibero 2001

© John Scholes
jscholes@kalva.demon.co.uk
16 January 2004
Last corrected/updated 16 Jan 04