IMO 2004 Problem 5

In a convex quadrilateral $ABCD$ , the diagonal $BD$ bisects neither the angle $ABC$ nor the angle $CDA$ . The point $P$ lies inside $ABCD$ and satisfies $$\angle PBC=\angle DBA \text{and} \angle PDC=\angle BDA.$$ Prove that $ABCD$ is a cyclic quadrilateral if and only if $AP=CP$ .

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