Problem

[107]

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

Solution

Attributes Олімпіадна
Source International Mathematical Olympiad
Year 2004
Number 5
Difficulty 10.0
Themes Геометрія