Example 5 In $\triangle A B C$, it is known that $\angle B A C=45^{\circ}$, $A D \perp B C$ at point $D$. If $B D=2, C D=3$, then $S_{\triangle A B C}$ $=$ $\qquad$
(2007, Shandong Province Junior High School Mathematics Competition)
Solve as shown in Figure 5, with $AB$ as the axis of symmetry, construct the symmetric figure of $\triangle ADB$ as $\triangle AGB$, and with $AC$ as the axis of symmetry, construct the symmetric figure of $\triangle ADC$ as $\triangle AFC$, and extend $GB$ and $FC$ to intersect at point $E$. Then it is easy to know that quadrilateral $AGEF$ is a square.
Assume $AD=h$. Then
$$
\begin{array}{l}
BE=h-2, CE=h-3 . \\
\text { By } BC^{2}=BE^{2}+CE^{2} \\
\Rightarrow(h-2)^{2}+(h-3)^{2}=5^{2} \\
\Rightarrow h^{2}-5 h-6=0 \\
\Rightarrow h=6 \\
\Rightarrow S_{\triangle ABC}=\frac{1}{2} BC \cdot AD=\frac{1}{2} \times 5 \times 6=15 .
\end{array}
$$