Example 1 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 $O$, using 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 $S_{\triangle CAB}=S_{\triangle CDE}, S_{\triangle CAD}=S_{\triangle CBE}$
Therefore, $S_{\triangle ABCD}=S_{\triangle EBCD}$
Furthermore, quadrilateral $EBCD$ is an isosceles trapezoid with the upper base of 2, the lower base of 6, and the legs of 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 $2\sqrt{3}$.
Hence, $S_{\text{quadrilateral } ABCD}=S_{\text{quadrilateral } EBCD}=8\sqrt{3}$.