5. As shown in Figure 1, given a cyclic quadrilateral $A B E C, A E \perp$ $B C$ at point $D$, and $B D=2$, $C D=3, \angle B E C=135^{\circ}$. Then the area of quadrilateral $A B E C$ is ( ).
(A) 12
(B) 18
(C) $\frac{35}{2}$
(D) $\frac{25}{2}$
5. C.

Notice that, $\angle B A C=45^{\circ}$.
Let $A \dot{D}=x$.
Since $A D \perp B C$
$$
\begin{array}{l}
\Rightarrow A B^{2}=x^{2}+4, A C^{2}=x^{2}+9 \\
\Rightarrow S_{\triangle A B C}^{2}=\left(\frac{1}{2} A B \cdot A C \sin 45^{\circ}\right)^{2} . \\
\text { Also } S_{\triangle B C}^{2}=\left(\frac{1}{2} B C \cdot A D\right)^{2} \\
\Rightarrow\left(x^{2}+4\right)\left(x^{2}+9\right) \sin ^{2} 45^{\circ}=[(2+3) x]^{2} \\
\Rightarrow\left(x^{2}\right)^{2}-37 x^{2}+36=0 \\
\Rightarrow x^{2}=36 \text { or } 1 \\
\Rightarrow A D=6, D E=1 \\
\Rightarrow S_{\text {quadrilateral } A B E C}=\frac{1}{2} A E \cdot B C=\frac{35}{2} .
\end{array}
$$