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APF = BPD in equilateral triangle

Source: Mexican Math Olympiad Regional Contest 2010 Problem 6

July 21, 2015
geometrytangent2010

Problem Statement

Let ABCABC be an equilateral triangle and DD the midpoint of BCBC. Let EE and FF be points on ACAC and ABAB respectively such that AF=CEAF=CE. P=BEP=BE ∩\cap CFCF. Show that ∠\angleAPF=APF= ∠\angleBPDBPD
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