Proof: Let the perpendicular bisectors of $BD$ and $CE$ intersect $BD$ and $CE$ at points $M$ and $N$, respectively, and intersect the circle $\Gamma$ at the second points $J$ and $I$. Extend $JD$ and $IE$ to intersect the circle at points $X$ and $Y$. Draw the auxiliary lines as shown in Figure 5.
$$\begin{array}{l} \text { By } B F=F D, B J=J D, \\ \angle F X D=180^{\circ}-\angle F B J \\ =180^{\circ}-\angle F D J=\angle F D X, \end{array}$$

we get $F D=F X$.
Since $\triangle B D J$ is an isosceles triangle, then $\triangle A X D \backsim \triangle J B D$.
Thus, $A D=A X$.
Similarly, $A E=A Y$.
Therefore, points $X$, $E$, $D$, and $Y$ lie on a circle with $A$ as the center and $AD$ as the radius, so
$$\begin{array}{l} \angle E D Y=\frac{1}{2} \angle E A Y=\angle Y J G . \\ \text { Then } \angle D E Y=180^{\circ}-\angle D E I \\ =180^{\circ}-\angle D X Y=\angle Y G J . \end{array}$$

Hence, $\triangle G E Y \backsim \triangle J G Y$
$$\begin{aligned} \Rightarrow & \angle J F G=\angle J Y G=\angle D Y E \\ & =\frac{1}{2} \angle D A E=\angle D A Z \\ \Rightarrow & F, M, Z, A \text { are concyclic } \\ \Rightarrow & F G \perp A Z . \end{aligned}$$

Since $A Z \perp D E$, we have $F G \parallel D E$.