10. As shown in Figure 4, in $\triangle A B C$, $\angle B A C=45^{\circ}$, $A D \perp B C$ at point $D$. If $B D=3, C D=2$, then $S_{\triangle A B C}$ $=$ $\qquad$ .
10.15 .

Solution 1: Let $A B=c, A C=b$. Then
$$
\begin{aligned}
A D^{2} & =c^{2}-9=b^{2}-4, b=\sqrt{c^{2}-5}, \\
A D & =\sqrt{c^{2}-9} .
\end{aligned}
$$

As shown in Figure 11, draw $C E \perp A B$ at point $E$, $\triangle A E C$ is an isosceles right triangle. Thus,
$$
\begin{array}{l}
E C=\frac{\sqrt{2}}{2} A C=\frac{\sqrt{2}}{2} b \\
=\frac{\sqrt{2}}{2} \sqrt{c^{2}-5}, \\
2 S_{\triangle A B C}=\frac{\sqrt{2}}{2} c \cdot \sqrt{c^{2}-5}=5 \sqrt{c^{2}-9} .
\end{array}
$$

Rearranging gives $\left(c^{2}-10\right)\left(c^{2}-45\right)=0$.
Solving yields $c_{1}=\sqrt{10}$, so $h=1$ or $h=6$.
When $h=1$, $\angle B A D>\angle A B C, \angle C A D>$ $\angle A C B$, so, $\angle B A C>\angle A B C+\angle A C B$.

Thus, $\angle B A C>\frac{1}{2} \times 180^{\circ}=90^{\circ}$, which contradicts $\angle B A C$ $=45^{\circ}$.
Therefore, $A D=6$.
So, $S_{\triangle A B C}=\frac{1}{2} \times 5 \times 6=15$.