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Find angle APD

Source: Turkey JBMO Team Selection Test 2013, P7

May 31, 2013
geometrycircumcirclegeometry proposed

Problem Statement

In a convex quadrilateral ABCDABCD diagonals intersect at EE and BE=2⋅ED, ∠BEC=45∘.BE = \sqrt{2}\cdot ED, \: \angle BEC = 45^{\circ}. Let FF be the foot of the perpendicular from AA to BCBC and PP be the second intersection point of the circumcircle of triangle BFDBFD and line segment DCDC. Find ∠APD\angle APD.
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