Prove that $S_{A B C D}=\frac{1}{2} S_{A B C D E F}=S_{B C D E}$,
and $S_{A B C D}=S_{\triangle A B D}+S_{\triangle D B C}$,
and $S_{B C D E}=S_{\triangle E B D}+S_{\triangle D B C}$,
thus $S_{\triangle A B D}=S_{\triangle E B D}$.
Since they share the same base, then $A E / / B D$.
Similarly, we can prove that $A C / / D F, C E / / B F$.
Assume $A D, B E$ intersect at $M$. Consider the homothety centered at $M$ that maps $A$ to $D$. Since $A E / / B D$, $B$ is the image of $E$ under this transformation.

The image of $A C$ is a line through $D$ and parallel to it, and the image of $E C$ is a line through $B$ and parallel to it, so $F$ is the common point of these two image lines, thus $F$ is the image of point $C$.
Therefore, $C F$ passes through the homothety center $M$.