5. Given that the area of $\triangle A B C$ is 4, the side lengths are $B C=a, C A=b, A B=c(c>b)$, and $A D$ is the angle bisector of $\angle A$. $C^{\prime}$ is the symmetric point of $C$ with respect to the line $A D$. If $\triangle C^{\prime} B D$ is similar to $\triangle A B C$, find the minimum perimeter of $\triangle A B C$.
$(2010$, New Knowledge Cup Shanghai Junior High School Mathematics Competition)
From the problem, we have
$$
B D=\frac{a c}{b+c}, C D=\frac{a b}{b+c}, B C^{\prime}=c-b .
$$

Since $\triangle C^{\prime} B D \backsim \triangle A B C$, we have
$$
\begin{array}{l}
\frac{B D}{A B}=\frac{B C^{\prime}}{B C} \Rightarrow \frac{\frac{a c}{b+c}}{c}=\frac{c-b}{a} \\
\Rightarrow c^{2}=a^{2}+b^{2} .
\end{array}
$$

Thus, $\triangle A B C$ is a right triangle,
$$
\begin{array}{l}
S=\frac{1}{2} a b=4, \\
a+b+c=(a+b)+\sqrt{a^{2}+b^{2}} \\
\geqslant 2 \sqrt{a b}+\sqrt{2 a b}=4 \sqrt{2}+4,
\end{array}
$$

with equality if and only if $a=b$, i.e., when $\triangle A B C$ is an isosceles right triangle.