Example 1. As shown in the figure, in $\triangle A B C$, $A D \perp B C$ at $D, \angle C A B=45^{\circ}, B D=3, C D=2$. Find the area of $\triangle A B C$.
Solve: Construct $\triangle A E B$ and $\triangle A C F$ symmetric to $\triangle A B D$ and $\triangle A D C$ with respect to $A B$ and $A C$ respectively. Extend $E B$ and $F C$ to intersect at point $G$. In quadrilateral $A F G E$, according to the transformation, we have
$$
A F=A D=A E,
$$
$$
\begin{array}{l}
\angle F A E=\angle A F G=\angle A E G=\angle F G E \\
=90^{\circ} .
\end{array}
$$

Therefore, quadrilateral $A F G E$ is a square.
Let $A F=A D=x$, then
$$
\begin{array}{l}
B G=x-3, C G=x-2 . \\
(x-3)^{2}+(x-2)^{2}=(2+3)^{2} .
\end{array}
$$

Solving for $x$ gives $x=6$ (negative value is discarded).
Thus, the area of $\triangle A B C$ is $=\frac{1}{2} \times 5 \times 6=15$.