11. Given that the area of $\triangle A B C$ is 4, with side lengths $B C=a, C A=b, A B=c(c>b)$, and $A D$ is the angle bisector of $\angle A$, $C^{\prime}$ is the reflection of point $C$ about the line $A D$. If $\triangle C^{\prime} B D$ is similar to $\triangle A B C$, find the minimum perimeter of $\triangle A B C$.
II. 11. As shown in Figure 7, it is clear that point $C^{\prime}$ lies on side $A B$, and
$B C^{\prime}=c-b$.
By the property of internal angle bisectors, we have $B D=\frac{c a}{b+c}$.

Since $\triangle C^{\prime} B D \backsim \triangle A B C$, and $\angle B$ is a common angle, we have
$$
\frac{c-b}{c}=\frac{\frac{c a}{b+c}}{a} \text { or } \frac{c-b}{a}=\frac{\frac{c a}{b+c}}{c} \text {. }
$$

From the former, we get $c^{2}-b^{2}=c^{2} \Rightarrow b^{2}=0$, which is impossible.
From the latter, we get $c^{2}-b^{2}=a^{2}$, i.e., $c^{2}=a^{2}+b^{2}$.
Thus, $\angle C=90^{\circ}$.
Given $S_{\triangle M B C}=4$, we have $a b=8$. Therefore,
the perimeter of $\triangle A B C$ is
$$
\begin{array}{l}
=a+b+c=a+b+\sqrt{a^{2}+b^{2}} \\
\geqslant 2 \sqrt{a b}+\sqrt{2 a b}=4 \sqrt{2}+4 .
\end{array}
$$

The equality holds when $a=b=2 \sqrt{2}, c=4$.
Therefore, the minimum perimeter of $\triangle A B C$ is $4 \sqrt{2}+4$.