14. As shown in Figure 4, given that the radius of the circumcircle $\odot O$ of quadrilateral $ABCD$ is 2, the intersection point of diagonals $AC$ and $BD$ is $E, AE=$ $EC, AB=\sqrt{2} AE$, and $BD=$ $2 \sqrt{3}$. Find the area of quadrilateral $ABCD$.
14. From the given, we have $A B^{2}=2 A E^{2}=A E \cdot A C$. Therefore,
$$
\frac{A B}{A C}=\frac{A E}{A B} \text {. }
$$

Also, since $\angle E A B=\angle B A C$, it follows that $\triangle A B E \sim \triangle A C B$.
Thus, $\angle A B E=\angle A C B$, which implies $A B=A D$.
As shown in Figure 5, connect $A O$, intersecting
$B D$ at $H$, then $B H=H D=$
$\sqrt{3}$. Therefore,
$$
\begin{aligned}
O H & =\sqrt{O B^{2}-B H^{2}} \\
& =1, \\
A H & =O A-O H=1 .
\end{aligned}
$$

Hence, $S_{\triangle A K D}$
$$
=\frac{1}{2} B D \cdot A H=\sqrt{3} \text {. }
$$

Since $E$ is the midpoint of $A C$, we have
$$
S_{\triangle A R E}=S_{\triangle B C E}, S_{\triangle A D E}=S_{\triangle C D E} \text {. }
$$

Thus, $S_{\triangle A B D}=S_{\triangle H C D}$.
Therefore, $S_{\text {quadrilateral } A B C D}=2 S_{\triangle A B D}=2 \sqrt{3}$.