Answer. The angles $\angle BSE$ and $\angle FSC$ are equal if and only if $\angle BAD = 60^\circ$ or $AB = AD$.

We denote the circumcircle of the trapezoid by $u$, the second intersection point of the line $SB$ with $u$ by $T$ and the centre of $u$ by $M$, see Figure 4. As the trapezoid is inscribed, it is isosceles. As $ABSD$ is a parallelogram by construction, we have $BS = AD = BC$ and $DS = AB$.

Figure 4: Problem 14, Case 1: $B$ between $S$ and $T$

Consider the reflection across the line $MS$. It clearly maps $E$ and $F$ to each other and maps $u$ to itself. We say that the trapezoid meets the *angle condition* if $\angle BSE = \angle FSC$.
The trapezoid meets the angle condition if and only if the reflection maps the rays $SB$ and $SC$ to each other. Equivalently, the intersection points of these rays with $u$ are mapped to each other corresponding to the order of the points on the rays.
We first consider the case that $B$ is between $S$ and $T$, see Figure 4. Then the trapezoid meets the angle condition if and only if the reflection maps $B$ and $C$ to each other. Equivalently, the triangle $BSC$ is isosceles with axis of symmetry $SM$. As $M$ lies on the perpendicular bisector of $BC$ in any case, this is equivalent to $CS = BS$. As $BS = BC$, this is in turn equivalent to the triangle $BSC$ being equilateral. Again by $BS = BC$, this is equivalent to $\angle CSB = 60^\circ$. As $ABSD$ is a parallelogram, the trapezoid meets the angle condition in this case if and only if $\angle BAD = 60^\circ$.
We now consider the case that $T$ lies between $S$ and $B$, see Figure 5. Then the above considerations show that the trapezoid meets the angle condition if and only if the reflection maps $B$ and $D$ to each other. Equivalently, the triangle $BSD$ is isosceles with axis of symmetry $MS$. By the same argument as in the first case, this is equivalent to $SB = SD$. This is equivalent to $AB = AD$.

Figure 5: Problem 14, Case 2: $T$ between $S$ and $B$