Three. (25 points) As shown in Figure 3, in isosceles $\triangle ABC$, $AB = AC = \sqrt{5}$, $D$ is a point on side $BC$ other than the midpoint, the symmetric point of $C$ with respect to line $AD$ is $E$, and the extension of $EB$ intersects the extension of $AD$ at point $F$. Find the value of $AD \cdot AF$.
Three, connect $A E$, $E D$, $C F$. From the given conditions, we know
$$
\angle A B C=\angle A C B=\angle A E D \text {. }
$$

Then $A$, $E$, $B$, $D$ are concyclic
$$
\Rightarrow \angle B E D=\angle B A D \text {. }
$$

Since points $C$, $E$ are symmetric with respect to line $A D$
$$
\begin{array}{l}
\Rightarrow \angle B E D=\angle B C F \\
\Rightarrow \angle B A D=\angle B C F \\
\Rightarrow A, B, F, C \text { are concyclic. } \\
\text { Also } A B=A C \\
\Rightarrow \angle A B D=\angle A C B=\angle A F B \\
\Rightarrow \triangle A B D \backsim \triangle A F B \Rightarrow \frac{A B}{A F}=\frac{A D}{A B} \\
\Rightarrow A D \cdot A F=A B^{2}=5 .
\end{array}
$$