Example I Given a cyclic quadrilateral $ABCD$ with side lengths $AB=2, BC=6, CD=DA=4$, find the area of quadrilateral $ABCD$.
Analysis: Let the circumcenter of quadrilateral $ABCD$ be $O$. Utilize the unique property of a circle: rotational invariance. By swapping the positions of triangle $OAB$ and triangle $OAD$, we obtain an isosceles trapezoid with an area equal to that of quadrilateral $ABCD$.

Solution: Let the circumcenter of quadrilateral $ABCD$ be $O$. Take a point $E$ on $\overparen{AD}$ such that $\overparen{DE} = \overparen{AB}$, then $\overparen{BE} = \overparen{AD}$.
And $S_{\triangle OAB} = S_{\triangle ODE}, S_{\triangle CDE} = S_{\triangle BDE}$
Therefore, $S_{\triangle ABCD} = S_{\triangle EBCD}$
Furthermore, quadrilateral $EBCD$ is an isosceles trapezoid with the upper base being 2, the lower base being 6, and the legs being 4. Draw $DF \parallel EB$ intersecting $BC$ at $F$, resulting in an equilateral triangle $CDF$ with side length 4, thus the height of the trapezoid is $2\sqrt{3}$.
$$
\text{Hence } S_{\text{quadrilateral } ABCD} = S_{\text{trapezoid } EBCD} = 8\sqrt{3} \text{.}
$$