Let $AE$ intersect $DC$ at point $H$,

and $AF$ intersect $BD$ at point $G$. Join $AB$, $BC$, $BH$, $BE$, $CE$ and $GH$. By symmetry, $CE$ is also a tangent line of the circle $O_1$ and $H$ is the midpoint of $AE$.

Since $AF \perp DE$, we have
$$\angle AGB = \frac{\pi}{2} - \angle BDE.$$

By $AE \perp DC$,
$$\angle AHB = \frac{\pi}{2} - \angle BHC.$$

By ①, ② and ③, $\angle AGB = \angle AHB$. Therefore $A$, $G$, $H$, $B$ lie on the same circle, and $\angle AHG = \angle ABG = \angle AED$. Thus $GH \parallel DE$. Since $H$ is the midpoint of $AE$, $G$ is the midpoint of $AF$.