A (convex) trapezoid $ABCD$ is called good if it has a circumcircle, the sides $AB$ and $CD$ are the parallel sides, and $CD$ is shorter than $AB$. For a good trapezoid $ABCD$, the following notations are defined.

- The line through $B$ parallel to $AD$ intersects the extension of side $CD$ at point $S$.
- The two tangents from $S$ to the circumcircle of the trapezoid touch the circle at $E$ and $F$, respectively, with $E$ lying on the same side of line $CD$ as $A$.

Provide the simplest equivalent condition (expressed in terms of the side lengths and/or angles of the trapezoid) for when, in a good trapezoid $ABCD$, the two angles $\angle BSE$ and $\angle FSC$ are equal.

(Walther Janous)

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

![](https://cdn.mathpix.com/cropped/2024_04_17_5ac0678f2eaae27c667fg-03.jpg?height=548&width=623&top_left_y=103&top_left_x=725)

Figure 1: Problem 2, Solution 1, Case 1: $B$ between $S$ and $T$
We denote the circumcircle of the trapezoid by $u$, the second intersection point of the line $S B$ with $u$ by $T$, and the center of $u$ by $M$, see Figure 1. Since the trapezoid has a circumcircle, it is isosceles. Since by construction $A B S D$ is a parallelogram, it follows that $B S = A D = B C$ and $D S = A B$.

We consider the reflection across the line $M S$. This clearly maps $E$ and $F$ onto each other and $u$ onto itself. We say that the trapezoid satisfies the angle condition if the angles $\angle B S E$ and $\angle F S C$ are equal.

The trapezoid satisfies the angle condition if and only if the reflection maps the rays $S B$ and $S C$ onto each other. This is equivalent to the intersection points of these rays with $u$ being mapped onto each other, and in the corresponding order.

We first consider the case where $B$ lies between $S$ and $T$, see Figure 1. Then, according to the above reasoning, the trapezoid satisfies the angle condition if and only if $B$ and $C$ are mapped onto each other. This is equivalent to the triangle $B S C$ being isosceles with the symmetry axis $S M$. Since $M$ as the circumcenter lies on the perpendicular bisector of $B C$, this is equivalent to $C S = B S$. Since $B S = B C$, this is equivalent to the triangle $B S C$ being equilateral. Again, because $B S = B C$, this is equivalent to $\angle C S B = 60^\circ$. Since $A B S D$ is a parallelogram, the trapezoid satisfies the angle condition in this case if and only if $\angle B A D = 60^\circ$.

![](https://cdn.mathpix.com/cropped/2024_04_17_5ac0678f2eaae27c667fg-03.jpg?height=605&width=625&top_left_y=1725&top_left_x=721)

Figure 2: Problem 2, Solution 1, Case 2: $T$ between $S$ and $B$

We now consider the case where $T$ lies between $S$ and $B$, see Figure 2. Then, according to the above reasoning, the trapezoid satisfies the angle condition if and only if $B$ and $D$ are mapped onto each other. This is equivalent to the triangle $B S D$ being isosceles with the symmetry axis $M S$. By the same argument as in the first case, this is equivalent to $S B = S D$. This is exactly the case when $A B = A D$.

Remark. The equation $C S = B S$ or $C S = T S$, which is equivalent to the angle condition, can also be justified by the fact that the two tangents are rotated by the same angle inward.

(Theresia Eisenkölbl)