Example 5 As shown in Figure 5, in $\triangle A B C$, $\angle C=90^{\circ}$, $I$ is the intersection of the angle bisectors $A D$ and $B E$ of $\angle A$ and $\angle B$. Given that the area of $\triangle A B I$ is 12. Then the area of quadrilateral $A B D E$ is $\qquad$
(2004, Beijing Middle School Mathematics Competition (Grade 8))
Solve As shown in Figure 5, construct the symmetric points $F, G$ of points $E, D$ with respect to $AD, BE$ respectively. Then $F, G$ lie on $AB$. Connect $IF, IG$. It is easy to know that
$$
\angle AIB=90^{\circ}+\frac{1}{2} \angle C=135^{\circ}.
$$

By the properties of axial symmetry, we have
$$
\begin{array}{l}
IF=IE, ID=IG, \\
\angle AIE=\angle AIF=\angle BID=\angle BIG=45^{\circ}. \\
\text{Therefore, } \angle FIG=\angle AIB-\angle AIF-\angle BIG \\
=135^{\circ}-45^{\circ}-45^{\circ}=45^{\circ}=\angle BID.
\end{array}
$$

Construct $DH \perp BE$ at point $H$, and $GK \perp IF$ at point $K$.
It is easy to prove $\triangle IDH \cong \triangle IGK$.
Thus, $GK=DH$.
Therefore, $\frac{1}{2} IE \cdot DH=\frac{1}{2} IF \cdot GK$, which means
$S_{\triangle DDE}=S_{\triangle ICF}$.
Hence, $S_{\text{quadrilateral } ABDE}=2 S_{\triangle AIB}=24$.