2. (25 points) As shown in Figure 1, given that $E$ is a point on side $A B$ of square $A B C D$, and the symmetric point of $A$ with respect to $D E$ is $F, \angle B F C = 90^{\circ}$. Find the value of $\frac{A B}{A E}$.
2. As shown in Figure 4, extend $E F$ to intersect $B C$ at point $M$, connect $D M$, and let it intersect $C F$ at point $G$.
Then, Rt $\triangle D F M \cong$ Rt $\triangle D C M$. Therefore, $\angle F D M = \angle M D C$, and $F M = C M$.
Thus, $M$ is the midpoint of $B C$.
$$
\begin{array}{l}
\text { Also, } \angle E D M = \angle E D F + \angle F D M \\
= \frac{1}{2} \angle A D F + \frac{1}{2} \angle F D C = 45^{\circ} .
\end{array}
$$

Rotate $\triangle M D C$ around point $D$ by $90^{\circ}$ counterclockwise to get $\triangle H D A$.
Then, $\triangle M D E \cong \triangle H D E$.
Therefore, $E M = H E = A E + M C$.
Let the side length of the square be $1$, and $A E = x$.
By $E B^{2} + B M^{2} = E M^{2}$, we have
$$
\begin{array}{l}
(1-x)^{2} + \left(\frac{1}{2}\right)^{2} = \left(x + \frac{1}{2}\right)^{2} \\
\Rightarrow x = \frac{1}{3} \Rightarrow \frac{A B}{A E} = 3 .
\end{array}
$$