## Τρίτη, 17 Οκτωβρίου 2017

### 2004 IMO Shortlist G4

In a convex quadrilateral $ABCD$ the diagonal $BD$ does not bisect the angles $ABC$ and $CDA$.  The point $P$  lies inside $ABCD$ and satisfies $\angle PBC = \angle DBA$   and   $\angle PDC = \angle BDA$. Prove that $ABCD$ is a cyclic quadrilateral if and only if $AP = CP$ .

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