9.2. Parallelogram $A B C D$ is such that $\angle B<90^{\circ}$ and $A B<B C$. Points $E$ and $F$ are chosen on the circle $\omega$ circumscribed around triangle $A B C$ such that the tangents to $\omega$ at these points pass through $D$. It turns out that $\angle E D A=\angle F D C$. Find the angle $A B C$.

(A. Yakubov)
Answer: $60^{\circ}$.

Solution. Let $\ell$ be the bisector of angle $E D F$. Since $D E$ and $D F$ are tangents to $\omega$, the line $\ell$ passes through the center $O$ of the circle $\omega$.

Perform a reflection about $\ell$. Since $\angle E D A = \angle F D C$, the ray $D C$ will be transformed into the ray $D A$. Since $\ell$ passes through $O$, the circle $\omega$ will map to itself; hence, point $C$ will map to a point $C^{\prime}$ lying on $D A$ and on $\omega$. At the same time, since $A D \neq D C$, points $C^{\prime}$ and $A$ are distinct (see Fig. 1).

![](https://cdn.mathpix.com/cropped/2024_05_06_a5e6654ee00a4b45c107g-2.jpg?height=400&width=587&top_left_y=181&top_left_x=461)

Fig. 1

From the same reflection, we have $\angle D C C^{\prime} = \angle D C^{\prime} C$. Since points $A, B, C$, and $C^{\prime}$ lie on $\omega$, we have $\angle D C^{\prime} C = \angle A B C = \angle A D C$. Therefore, all three angles of triangle $D C C^{\prime}$ are equal, from which it follows that $\angle A B C = \angle C D C^{\prime} = 60^{\circ}$.