Complex numbers seem to work fine, but I present a normal solution. Claim. $EDB \sim FAB$ Obviously, $ADB\sim CAB$, thus $\frac{AB}{DB}=\frac{AC}{DA}=\frac{AF}{DE}$ and because $\angle EDB = \angle FAB$, the claim follows. $\quad \square$ Thus, $\angle FBA = \angle EBD$. Also, $\angle BXF=180^{\circ} - \angle BEF=90^{\circ} - \angle BED =\angle EBD=\angle FBA$ ($X$ is arbitrary point on opposite from $\overarc{BEF}$), therefore $AB$ is tangent to $(BEF)$. Hence, $AC \perp AB\perp BM\implies AC\parallel BM$. We are done.

It is clear that $\triangle ABC \sim \triangle DBA$ and therefore $\triangle CBF \sim \triangle ABE$. And since $CD||EF$ we have $\angle ABE=\angle CBF=\angle EFB$ So $AB$ is tangent with $(BEF)$ and we are done.