Problem

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Tags: geometry, circumcircle, trigonometry, parallelogram, geometry unsolved



Let $ABCD$ be a convex quadrilateral. Assume line $AB$ and $CD$ intersect at $E$, and $B$ lies between $A$ and $E$. Assume line $AD$ and $BC$ intersect at $F$, and $D$ lies between $A$ and $F$. Assume the circumcircles of $\triangle BEC$ and $\triangle CFD$ intersect at $C$ and $P$. Prove that $\angle BAP=\angle CAD$ if and only if $BD\parallel EF$.