【Example 4.24】In a cyclic quadrilateral $ABCD$ (Figure 4.16), $AB=2$, $BC=6$, $CD=DA=4$. Find the area of quadrilateral $ABCD$.
Connect $B D$, we have $B D^{2}=B D^{2}$, that is,
$$
\begin{aligned}
2^{2}+4^{2}-2 \cdot 2 \cdot 4 \cos A= & 6^{2}+4^{2}-2 \cdot 6 \cdot 4 \cos C \\
& =6^{2}+4^{2}+2 \cdot 6 \cdot 4 \cos A
\end{aligned}
$$

One equation with one unknown $(\cos A)$, solving it yields $A=120^{\circ}, C=60^{\circ}$.
$$
S_{A B C D}=\frac{1}{2} \cdot 2 \cdot 4 \sin 120^{\circ}+\frac{1}{2} \cdot 6 \cdot 4 \sin 60^{\circ}=8 \sqrt{3} \text {. }
$$

Of course, by noting the auxiliary point $E$, we know that $\triangle B E A$ and $\triangle D E A$ are isosceles triangles, and $\triangle D E C$ is an equilateral triangle. Thus, $S_{A B C D}=2 \cdot \frac{1}{2} \cdot 2 \cdot 4 \sin 120^{\circ}+\frac{1}{2}$ - $4 \cdot 4 \sin 60^{\circ}=8 \sqrt{3}$.
This example is from the 2001 National Liberal Arts College Entrance Examination.