MathDB

7

Part of 2013 JBMO TST - Turkey

Problems(1)

Find angle APD

Source: Turkey JBMO Team Selection Test 2013, P7

5/31/2013
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.
geometrycircumcirclegeometry proposed