5. As shown in Figure 3, in $\triangle A B C$, $A B$ $=A C, \angle A=36^{\circ}, C D$ is the angle bisector. Then the ratio of the area of $\triangle D B C$ to the area of $\triangle A B C$ is ( ).
(A) $\frac{\sqrt{5}-2}{2}$
(B) $\frac{\sqrt{5}-2}{3}$
(C) $\frac{3-\sqrt{5}}{2}$
(D) $\frac{3-\sqrt{5}}{3}$
5.C.
$$
\begin{array}{l}
\text { It is easy to prove } A D=D C=B C \Rightarrow \triangle C B D \subset \triangle A B C \\
\Rightarrow \frac{B C}{A B}=\frac{B D}{B C} \Rightarrow \frac{B C}{A B}=\frac{A B-B C}{B C}=\frac{A B}{B C}-1 \\
\Rightarrow \frac{B C}{A B}=\frac{\sqrt{5}-1}{2} \\
\Rightarrow \frac{S_{\triangle D B C}}{S_{\triangle A B C}}=\frac{B D}{A B}=\frac{A B-A D}{A B}=1-\frac{B C}{A B}=\frac{3-\sqrt{5}}{2} .
\end{array}
$$