67th Romanian Mathematical Olympiad

Question

Let ABC be a right triangle with legs AB and AC. The bisector of the angle ACB intersects AB in D and the perpendicular in B on BC in E. Denote F the reflection of E across B and P the intersection of the lines DF and BC. Prove that EP CF. Cătălin Cristea FIGURE_0

Accepted Answer

In BEC, m( CEB) = 180^ - m( EBC) - m( ECB) = 90^ - m( ECB). In ADC, m( ADC) = 180^ - m( DAC) - m( ACD) = 90^ - m( ACD). Since m( ACD) = m( ECB), it follows ADC CEB. Now ADC EDB. This yields [BD] = [BE]. Since [BE] = [BF], we infer that [BD] = [BE] = [BF], so DEF has the right angle D. In CEF, CB and FD are altitudes, hence P is the orthocenter. In conclusion, EP is an altitude, that is EP CF. FIGURE_0