# Prove that $ABCD$ is a cyclic quadrilateral

Points $E$ and $F$ are on side $BC$ of convex quadrilateral $ABCD$(with $E$ closer than $F$ to $B$). It is known that $\angle {BAE}=\angle {CDF}$ and $\angle {EAF}=\angle {FDE}$.Prove that $\angle {FAC}=\angle{EDB}$.