From the conditions, we have

$$\begin{aligned} \angle C B A & =180^\circ-\angle E D B=180^\circ-\angle E F B \\ & =180^\circ-\angle E F A-\angle A F B \\ & =180^\circ-\angle C B A-\angle A C B=\angle B A C . \end{aligned}$$
Let $P$ be the intersection of $A C$ and $B F$. Then we have
$$\angle P A E=\angle C B A=\angle B A C=\angle B F C .$$
This implies $A, P, F, E$ are concyclic. It follows that
$$\angle F P E=\angle F A E=\angle F B A,$$
and hence $A B$ and $E P$ are parallel. So $E, P, D$ are collinear, and the result follows.

Let $E^{\prime}$ be any point on the extension of $E A$. From $\angle A E D=\angle E^{\prime} A B=\angle A C D$, points $A, D, C, E$ are concyclic.

Let $P$ be the intersection of $B F$ and $D E$. From $\angle A F P=\angle A C B=\angle A E P$, the points $A, P, F, E$ are concyclic. In addition, from $\angle E P A=\angle E F A=\angle D B A$, points $A, B, D, P$ are concyclic.
By considering the radical centre of $(B D F E),(A P F E)$ and $(B D P A)$, we find that the lines $B D, A P, E F$ are concurrent at $C$. The result follows.