Example 10 As shown in Figure 14, villages $A$, $B$, and $C$ are located along an east-west oriented highway, with $AB=2 \text{ km}$, $BC=3 \text{ km}$. To the due north of village $B$ is village $D$, and it is measured that $\angle ADC=45^{\circ}$. Now the $\triangle ADC$ area is planned to be a development zone, except for a pond of $4 \text{ km}^2$, the rest is designated for construction or green space. Try to find the area of the construction or green space in this development zone.
Solution: As shown in Figure 15, construct the axisymmetric figure of Rt $\triangle A D B$ with respect to the line containing $D A$, which is Rt $\triangle A D B_{1}$. It is easy to know that
Rt $\triangle A D B \cong$
Rt $\triangle A D B_{1}$.
Construct the axisymmetric figure of Rt $\triangle C D E$ with respect to the line containing $D C$, which is Rt $\triangle C D B_{2}$. It is easy to know that
Rt $\triangle C D B \cong \mathrm{Rt} \triangle C D B_{2}$.
Extend $B_{1} A$ and $B_{2} C$ to intersect at $E$, then $B_{1} D B_{2} E$ is a square.
Let $B D=x$, then
\[
\begin{array}{l}
B_{1} D=D B_{2}=B_{2} E=E B_{1}=x . \\
A B_{1}=A B=2, C B_{2}=C B=3, A C=5 .
\end{array}
\]

Therefore, $A E=x-2, C E=x-3$.
In Rt $\triangle A E C$, $A E^{2}+C E^{2}=A C^{2}$, i.e., $(x-2)^{2}+(x-3)^{2}=(2+3)^{2}$.
Rearranging and factoring, we get
\[
(x-6)(x+1)=0 \text{. }
\]

Since $x>0$, then $x+1>0$.
Therefore, $x=6$, i.e., $D B=6 \mathrm{~km}$.
Thus, $S_{\triangle A C D}=\frac{1}{2} \times 5 \times 6=15\left(\mathrm{~km}^{2}\right)$.
Given that there is a pond of $4 \mathrm{~km}^{2}$ in the development zone, the area for construction and greening in this development zone is
\[
15-4=11\left(\mathrm{~km}^{2}\right)
\]