2. In $\square A B C D$, $A B<A C<B C$. From point $D$ draw tangents to the circumcircle $\Gamma$ of $\triangle A B C$, the points of tangency are $E$ and $F$. If segment $A D$ intersects $C E$, and $\angle A B F=\angle D C E$, find $\angle A B C$.
2. Since point $D$ is outside the circle $\Gamma$, $\angle A B C$ is an acute angle.
Let $A^{\prime}$ be the second intersection point of line $D C$ with circle $\Gamma$.
From $B C>A C$, we know
$$
\angle D C A=\angle C A B>\angle C B A=\angle D A^{\prime} A \text {. }
$$

This indicates that point $A^{\prime}$ lies on the extension of segment $D C$ in the direction of point $C$.
$$
\begin{array}{l}
\text { Also, } \overparen{E C A}^{\circ}=2\left(180^{\circ}-\angle E C A^{\prime}\right) \\
=2 \angle E C D=2 \angle A B F=\overparen{A C F}^{\circ},
\end{array}
$$

Thus, $\overparen{A E}=\overparen{A^{\prime} F}$.
Therefore, points $A$ and $A^{\prime}$, and points $E$ and $F$ are symmetric with respect to the angle bisector of $\angle A D C$.
$$
\begin{array}{l}
\text { Hence, } \angle D A A^{\prime}=\angle D A^{\prime} A \\
=\angle A B C=\angle A D A^{\prime} .
\end{array}
$$

Therefore, $\angle A B C=\angle A D A^{\prime}=60^{\circ}$.