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Prove that <APQ = 2<CAP if AQ/QB=DP/DE

Source: Serbia NMO 2010 problem 4

March 11, 2011
geometrycircumcirclegeometric transformationreflectiontrigonometryincentergeometry unsolved

Problem Statement

Let OO be the circumcenter of triangle ABCABC. A line through OO intersects the sides CACA and CBCB at points DD and EE respectively, and meets the circumcircle of ABOABO again at point P≠OP \neq O inside the triangle. A point QQ on side ABAB is such that AQQB=DPPE\frac{AQ}{QB}=\frac{DP}{PE}. Prove that ∠APQ=2∠CAP\angle APQ = 2\angle CAP.
Proposed by Dusan Djukic
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