2004 IMO Problems/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 $AP = CP.$


This problem needs a solution. If you have a solution for it, please help us out by adding it.

See Also

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