Example 1 In $\triangle A B C$, $A C=B C, \angle A C B=$ $90^{\circ}, D, E$ are two points on side $A B$, $A D=3, B E=4$, $\angle D C E=45^{\circ}$. Then the area of $\triangle A B C$ is $\qquad$
(2006, Beijing Middle School Mathematics Competition (Grade 8))
Solve as shown in Figure 1, construct square $C A H B$, and extend $C D$, $C E$ to intersect $A H$, $B H$ at points $G$, $F$ respectively.
Let $D E=x$.
Since $A C / / B F$
$$
\Rightarrow \frac{3+x}{4}=\frac{A C}{B F} \text {, }
$$
$B C / / A G$
$$
\begin{array}{c}
\Rightarrow \frac{4+x}{3}=\frac{B C}{A G} . \\
\text { Then } \frac{(3+x)(4+x)}{12}=\frac{A C^{2}}{B F \cdot A G} .
\end{array}
$$

Also, $\angle D C E=\angle F B E=\angle G A D=45^{\circ}$, hence $\triangle A D G \backsim \triangle C D E \backsim \triangle B F E$
$$
\Rightarrow \frac{B F}{A D}=\frac{B E}{A G} \Rightarrow B F \cdot A G=12 \text {. }
$$

Notice that,
$$
A C^{2}=B C^{2}=\frac{1}{2} A B^{2}=\frac{1}{2}(3+x+4)^{2} \text {. }
$$

Substitute equations (2) and (3) into equation (1), simplify and rearrange to get
$$
x^{2}=3^{2}+4^{2}=25 \Rightarrow x=5 \text {. }
$$

Therefore, $S_{\triangle A B C}=\frac{1}{4} A B^{2}=\frac{1}{4}(3+5+4)^{2}=36$.