Let $\angle CBD = \angle CDB = \alpha$ Here, $A$ is the circumcenter of $\triangle EBF$ $\angle BAE = 180^{\circ} - 2\alpha$ so $\angle EFB = \alpha$ Then $\angle EFB = \angle DBC$ so $EF \parallel BD$

We let $\angle DBC=x$ and so $\angle BDC=x$ and $A$ is the circumcenter of triangle $EBF$ so we have $\angle EAB=180-2x \implies \angle EFB=x$ so $\angle EFB=\angle CBD$ so $EF \parallel BD.$ Hence, proved.

A is the center of BEF, and that's literally it.