APMO

From the conditions, we have $$\begin{array}{rl}\angle CBA& =180^{\circ }-\angle EDB=180^{\circ }-\angle EFB\\ & =180^{\circ }-\angle EFA-\angle AFB\\ & =180^{\circ }-\angle CBA-\angle ACB=\angle BAC.\end{array}$$ Let $P$ be the intersection of $AC$ and $BF$. Then we have $$\angle PAE=\angle CBA=\angle BAC=\angle BFC.$$ This implies $A,P,F,E$ are concyclic. It follows that $$\angle FPE=\angle FAE=\angle FBA,$$ and hence $AB$ and $EP$ are parallel. So $E,P,D$ are collinear, and the result follows.

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Alternative solution.

Let $E^{\prime }$ be any point on the extension of $EA$. From $\angle AED=\angle E^{\prime }AB=\angle ACD$, points $A,D,C,E$ are concyclic. Let $P$ be the intersection of $BF$ and $DE$. From $\angle AFP=\angle ACB=\angle AEP$, the points $A,P,F,E$ are concyclic. In addition, from $\angle EPA=\angle EFA=\angle DBA$, points $A,B,D,P$ are concyclic. By considering the radical centre of $(BDFE),(APFE)$ and $(BDPA)$, we find that the lines $BD,AP,EF$ are concurrent at $C$. The result follows.