Let $ABCD$ be cyclic quadrilateral and $E$ the midpoint of $AC$. The circumcircle of $\triangle CDE$ intersect the side $BC$ at $F$, which is different from $C$. If $B'$ is the reflection of $B$ across $F$, prove that $EF$ is tangent to the circumcircle of $\triangle B'DF$. Proposed by Nikola Velov