II. (25 points) Let $D$ be the midpoint of the base $BC$ of isosceles $\triangle ABC$, and $E$, $F$ be points on $AC$ and its extension, respectively. Given that $\angle EDF=90^{\circ}$, $DE=a$, $DF=b$, $AD=c$. Try to express the length of $BC$ in terms of $a$, $b$, and $c$.

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II. As shown in Figure 8, draw $D G \perp A C$ at $G$. By the properties of isosceles triangles, we know that $A D \perp B C$ and $A D$ bisects $\angle B A C$.

In the right triangle $\triangle E D F$, by the Pythagorean theorem, we have
$$
\begin{array}{l}
E F=\sqrt{D E^{2}+D F^{2}} \\
=\sqrt{a^{2}+b^{2}} .
\end{array}
$$

By $D E \cdot D F=E F \cdot D G$, we get
$$
D G=\frac{D E \cdot D F}{E F}=\frac{a b}{\sqrt{a^{2}+b^{2}}} \text {. }
$$

In the right triangle $\triangle A D G$,
$$
A G=\sqrt{A D^{2}-D G^{2}}=\frac{\sqrt{a^{2} c^{2}+b^{2} c^{2}-a^{2} b^{2}}}{\sqrt{a^{2}+b^{2}}} \text {. }
$$

By $\triangle A B D \backsim \triangle A D G$, we get $\frac{B D}{D G}=\frac{A D}{A G}$.
Thus, $B D=\frac{A D \cdot D G}{A G}=\frac{a b c}{\sqrt{a^{2} c^{2}+b^{2} c^{2}-a^{2} b^{2}}}$.
Therefore, $B C=2 B D=\frac{2 a b c}{\sqrt{a^{2} c^{2}+b^{2} c^{2}-a^{2} b^{2}}}$.