As shown in Figure 8, extend $B A$ and $F D$ to intersect at point $K$, and draw $G H \parallel B D$ to intersect $C B$ and $F D$ at points $G$ and $H$ respectively. Then,
$$\angle A D B = \angle D A H.$$

Since $\angle C D F = \angle A D H$, it follows that $\angle D A H = \angle A D H$.
Thus, $A H = D H$.
Also, since $\angle C A K = 90^{\circ}$, we have
$\angle K A H = 90^{\circ} - \angle D A H$
$= 90^{\circ} - \angle A D H = \angle A K H$.
Therefore, $K H = A H = D H$.
Since $B D \parallel G H$, we have
$$\frac{A H}{B D} = \frac{K H}{K D} = \frac{1}{2}, \frac{B E}{E D} = \frac{A G}{A H}, \frac{A C}{D C} = \frac{A G}{B D}.$$

Hence, $\frac{B E}{E D} = \frac{2 A C}{D C}$.