Example 3 (2001 National College Entrance Examination) Given a cyclic quadrilateral $ABCD$ with side lengths $AB=2$, $BC=6$, $CD=DA=4$, find the area of quadrilateral $ABCD$.


Analysis: Connect the diagonal, and convert the area of the quadrilateral into the sum of the areas of two triangles.
Solution: Connect $B D$. Then the area of quadrilateral $A B C D$ is:
$$
\begin{aligned}
S= & S_{\triangle A B D}+S_{\triangle C D B} \\
& =\frac{1}{2} A B \cdot A D \cdot \sin A+\frac{1}{2} B C \cdot C D \cdot \sin C
\end{aligned}
$$

Since $A+C=180^{\circ}$
Therefore, $\sin A=\sin C$
Thus, $S=\frac{1}{2}(A B \cdot A D+B C \cdot C D) \sin A$
$$
=\frac{1}{2}(2 \times 4+6 \times 4) \sin A=16 \sin A
$$

By the cosine rule, in $\triangle A B D$:
$$
B D^{2}=A B^{2}+A D^{2}-2 A B \cdot A D \cos A=2^{2}+4^{2}-2 \times 2 \times 4 \cos A=20-16 \cos A
$$

In $\triangle C D B$:
$$
B D^{2}=C B^{2}+C D^{2}-2 C B \cdot C D \cos C=6^{2}+4^{2}-2 \times 6 \times 4 \cos C=52-48 \cos C
$$

Therefore, $20-16 \cos A=52-48 \cos C$
Since $\cos C=-\cos A$
Thus, $64 \cos A=-32$
$$
\cos A=-\frac{1}{2}
$$

Therefore, $A=120^{\circ}$
Thus, $S=16 \sin 120^{\circ}=8 \sqrt{3}$