Example 14 As shown in Figure 1.4.16, in $\triangle A B C$, $A D \perp B C$ at $D$, $\angle C A B=45^{\circ}, B C=3, C D=2$, find $S_{\triangle A B C}$.
Given $\angle E A B=\angle B A D, \angle F A C=\angle C A D$, then draw $G E / / A F$ through $B$, and draw $F G$ // $A E$ through $C$ to form parallelogram $A F G E$.
It can be proven that $\mathrm{Rt} \triangle A E B \cong \mathrm{Rt} \triangle A D B, \mathrm{Rt} \triangle A F C \cong \mathrm{Rt} \triangle A D C$, thus $A F G E$ is a square.
Let $A F=A D=x$, then $B G=x-3, C G=x-2$, by
$$
(x-3)^{2}+(x-2)^{2}=(2+3)^{2}
$$

we get $x=6$ (discard the negative value), then $S_{\triangle A B C}=15$ is the required area.