Take the perpendicular from $B$ to $AC$. Now $EF.DC=BD.DE$ and $ED.EF=EX.EA$ by pop on the circle with diameter $AB$, and substituting implies to prove $ED^2.BD=EX.DC$ but $ED^2=EA.EC$. Thus we just need $EX.DC=EC.BD$, which is true by parallelism.

First, it's clear that $\triangle EDA \sim \triangle DCA$, then $\frac{ED}{DC} = \frac{AE}{AD}$. Therefore, $AF \perp BF \Leftrightarrow AFDB$ is cyclic $\Leftrightarrow \angle ABD = \angle AFE \Leftrightarrow \triangle AEF \sim \triangle ADB$ (since $\angle ADB = \angle AEF = 90^{\circ}$) $\Leftrightarrow \frac{EF}{BD} = \frac{AE}{AD} = \frac{ED}{DC} \Leftrightarrow EF \cdot DC = BD \cdot DE$ $\blacksquare$