MathDB

6

Part of 2010 Regional Olympiad of Mexico Center Zone

Problems(1)

APF = BPD in equilateral triangle

Source: Mexican Math Olympiad Regional Contest 2010 Problem 6

7/21/2015
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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