3. As shown in Figure 4, the diagonals $AC$ and $BD$ of quadrilateral $ABCD$ intersect at point $O$, and $E$, $F$, and $G$ are the midpoints of $AB$, $OC$, and $OD$ respectively. Given that $OA = AD$, $OB = BC$, and $CD = \sqrt{3} AB$. Then the degree measure of $\angle FEG$ is $\qquad$
3. $120^{\circ}$.

As shown in Figure 13, connect $A G$, $B F$, and $F G$, and draw $E P \perp F G$ at point $P$.
Let $A B=2 a$, then $C D=\sqrt{3} A B=2 \sqrt{3} a$.
Since $O A=A D$ and $G$ is the midpoint of $O D$, we have
$A G \perp O D$.
Therefore, $\angle A G B=90^{\circ}$.
Similarly, $\angle A F B=90^{\circ}$.
Thus, points $A$, $B$, $F$, and $G$ are concyclic, with diameter $A B$ and center $E$.
Since $F$ and $G$ are the midpoints of $O C$ and $O D$ respectively, we have
$$
\begin{array}{l}
F G=\frac{1}{2} C D=\sqrt{3} a=2 a \sin \frac{1}{2} \angle F E G . \\
\text { Hence } \frac{1}{2} \angle F E G=60^{\circ}, \angle F E G=120^{\circ} \text {. }
\end{array}
$$