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Turkey TST 1996 Problem 2, P,E,F collinear

Source: Turkey TST 1996 Problem 2

September 28, 2011
geometryparallelogramtrigonometrygeometry proposed

Problem Statement

In a parallelogram ABCDABCD with ∠A<90\angle A < 90, the circle with diameter ACAC intersects the lines CBCB and CDCD again at EE and FF , and the tangent to this circle at AA meets the line BDBD at PP . Prove that the points PP, EE, FF are collinear.
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