olymmath

PL-OM 1997 Problem 5

    Dany jest pięciokąt wypukły ABCDEABCDE, w którym DC=DEDC = DE oraz ∠DCB=∠DEA=90∘\angle DCB = \angle DEA = 90^\circ. Niech FF będzie punktem boku ABAB wyznaczonym przez warunek AF:BF=AE:BCAF : BF = AE : BC. Udowodnić, że ∠FCE=∠ADE\angle FCE = \angle ADE oraz ∠FEC=∠BDC\angle FEC = \angle BDC.

    Official text Polish Mathematical Olympiad 1997 official paper, page 1. Quoted with source.

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