2. If a trapezoid has a circumcircle, it is called "good". Given that trapezoid $ABCD$ is good, and $AB \parallel CD$, $CD < AB$, construct the following figure:
(1) Draw a line through point $B$ parallel to $AD$ intersecting line $CD$ at point $S$;
(2) Draw two tangents from point $S$ to the circumcircle of trapezoid $ABCD$, with the points of tangency being $E$ and $F$, where points $E$ and $A$ are on the same side of line $CD$.

If $\angle BSE = \angle FSC$, describe the good trapezoid $ABCD$ (by describing the side lengths and the interior angles of the trapezoid), and try to make the conclusion as simple as possible.
2. If and only if $\angle B A D=60^{\circ}$ or $A B=A D$, then $\angle B S E=\angle F S C$.

Let the circumcircle of the trapezoid be $\odot M$, and the second intersection point of line $S B$ with $\odot M$ be $T$, as shown in Figure 1.
Since the trapezoid is a cyclic quadrilateral, $A D=B C$.
Also, quadrilateral $A B S D$ is a parallelogram, so $B S=A D=B C, D S=A B$.
Consider the symmetric structure with respect to $M S$.
Clearly, points $E$ and $F$ are symmetric about $M S$, and $\odot M$ is symmetric about $M S$. If the condition $\angle B S E=\angle F S C$ is satisfied, then the trapezoid is said to satisfy the "angle condition".
Clearly, the trapezoid satisfies the angle condition if and only if lines $S B$ and $S C$ are symmetric about $M S$ (as shown in Figure 1, this is the case where point $B$ is between points $S$ and $T$), i.e., $B$ and $C$ are symmetric about $M S$, which means $\triangle B S C$ is an isosceles triangle with $M S$ as the axis of symmetry. Therefore, $M$ lies on the angle bisector of $\angle B S C$.
Furthermore, by $B S=B C$, we get $\angle C S B=60^{\circ}$.
Since quadrilateral $A B S D$ is a parallelogram, the trapezoid satisfies the angle condition if and only if $\angle B A D=60^{\circ}$.
Now consider the case where point $T$ is between points $S$ and $B$, as shown in Figure 2.
From the previous analysis, the trapezoid satisfies the angle condition if and only if points $B$ and $D$ are symmetric about line $M S$, which is equivalent to $\triangle B S D$ being an isosceles triangle with $M S$ as the axis of symmetry.
Similar to the first case, we get $S B=S D$, i.e., $A B=A D$.