Note that $\triangle ABD\cong\triangle EDB$ since $AB=DE$, $AD=BE$ and $BD=DB$. Thus, $ABDE$ is a symmetrical trapezoid, i.e. the perpendicular bisector of $BD$ is the perpendicular bisector of $EA$. Since $BC=CD$, $C$ lies on this perpendicular bisector. Similarly, $F$ lies on this perpendicular bisector since $AF=EF$. Thus, reflection across $CF$ maps $ABD$ to $EDB$, so in particular, $BE\cap AD$ lies on $CF$.