Example 1 In quadrilateral $A B C D$, $A B=30, A D$ $=48, B C=14, C D=40, \angle A B D+\angle B D C=$ $90^{\circ}$. Find the area of quadrilateral $A B C D$.
Solution: As shown in Figure 2, we have
$$
\begin{array}{l}
S_{\triangle A B D}=S_{\triangle A_{1} D B}, \\
A_{1} D=A B=30, \\
A_{1} B=A D=48, \\
\angle A_{1} D B=\angle A B D .
\end{array}
$$

Thus, we have
$$
\begin{array}{l}
\angle A_{1} D C=\angle A_{1} D B+\angle B D C \\
=\angle A B D+\angle B D C=90^{\circ} .
\end{array}
$$

Therefore, $\triangle A_{1} D C$ is a right triangle.
So $A_{1} C=\sqrt{30^{2}+40^{2}}=50$.
In $\triangle A_{1} B C$,
$$
\begin{array}{l}
B C^{2}+A_{1} B^{2}=14^{2}+48^{2} \\
=2500=50^{2}=A_{1} C^{2} .
\end{array}
$$

By the converse of the Pythagorean theorem, $\angle A_{1} B C=90^{\circ}$. Therefore,
$$
\begin{array}{l}
S_{\text {quadrilateral } B C C D}=S_{\text {quadrilateral } A_{1} B C D} \\
=S_{\triangle A_{1} B C}+S_{\triangle A_{1} D C} \\
=\frac{1}{2} A_{1} B \cdot B C+\frac{1}{2} A_{1} D \cdot C D \\
=\frac{1}{2} \times 48 \times 14+\frac{1}{2} \times 30 \times 40 \\
=336+600=936 .
\end{array}
$$

By using the axis of symmetry, we cleverly calculated the area of quadrilateral $A B C D$.