2004 IMO Problems/Problem 5

Problem

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 $AP = CP.$

Solution

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Let $K$ be the intersection of $AC$ and $BE$ , let $L$ be the intersection of $AC$ and $DF$ ,

2004 IMO (Problems) • Resources
Preceded by Problem 4	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 6
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2004 IMO Problems/Problem 5

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