2. In $\square A B C D$, $\angle B<90^{\circ}, A B<B C$. From point $D$ draw tangents to the circumcircle $\Gamma$ of $\triangle A B C$, the points of tangency are $E$ and $F$. It is known that $\angle E D A=\angle F D C$. Find $\angle A B C$
2. Let $A D$ intersect the circle $\Gamma$ at point $C^{\prime}$.

Since $\angle E D C^{\prime}=\angle F D C$, points $E$ and $F$ as well as $C^{\prime}$ and $C$ are symmetric with respect to the angle bisector of $\angle E D F$, so,
$$
C^{\prime} D=C D=B A=C C^{\prime} \text {. }
$$

Thus, $\triangle C^{\prime} C D$ is an equilateral triangle.
Therefore, $\angle A B C=\angle A D C=60^{\circ}$.