We have $ \angle DBF = \angle DBC$ because $ DB$ is the bisector of $ \angle FBC$. We also have $ \angle DFB = \angle DEA = \angle BCD$ (because $ DEFA$ and $ BCEA$ are cyclic quadrilaterals). Hence, triangles $ DBF$ and $ DBC$ are similar. But, since they share the common side $ BD$, they are congruent.

easy,$\angle DFA+\angle DEA=180$,so $\angle DFB=\angle C$($\angle AEB=\angle C$),$BD$ is the angle bisector,hence done.