Example 3 As shown in Figure 5, given that the radius of the circumcircle $\odot 0$ of quadrilateral $A B C D$ is
2, the diagonals $A C$ and $B D$
intersect at point $E, A E=E C$,
$A B=\sqrt{2} A E$, and $B D=$
$2 \sqrt{3}$. Find the area of quadrilateral $A B C D$.
(2000, National Junior High School Mathematics Competition)
【Analysis】This problem essentially involves the combination of figures 6 and 7.
Solution From figure 6 and the problem statement, we have
$$
\begin{array}{l}
A B^{2}=A E \cdot A C \Rightarrow \triangle A B E \backsim \triangle A C B \\
\Rightarrow \angle A B E=\angle A C B \Rightarrow A B=A D .
\end{array}
$$

From figure 7, since $A B=A D$, it follows that $A O \perp B D$. Solving this, we get $O F=1$.
Therefore, the area of $\triangle A B D$ is $\sqrt{3}$.