10.2. Given a parallelogram $A B C D$, where $A B<A C<B C$. 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 $D$; moreover, segments $A D$ and $C E$ intersect. It turns out that $\angle A B F=\angle D C E$. Find the angle $A B C$.

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

Solution. Since $D$ lies outside $\omega$, angle $A B C$ is acute. Let $A^{\prime}$ be the second intersection point of $D C$ and $\omega$. Since $B C>A C$, we have $\angle D C A=\angle C A B>\angle C B A=\angle D A^{\prime} A$; thus, $A^{\prime}$ lies on the extension of segment $D C$ beyond point $C$. Note that $E_{E C A^{\prime}}=$ $=2\left(180^{\circ}-\angle E C A^{\prime}\right)=2 \angle E C D=2 \angle A B F=\overline{A C F}$ (see Fig. 2).

![](https://cdn.mathpix.com/cropped/2024_05_06_6466072584c468d2174ag-1.jpg?height=498&width=599&top_left_y=1227&top_left_x=451)

Fig. 2

Let $\ell$ be the bisector of angle $E D F$. Since $D E$ and $D F$ are tangents to $\omega$, line $\ell$ passes through the center $O$ of circle $\omega$. Perform a reflection about $\ell$; under this reflection, $\omega$ maps to itself. Since $\overline{E C A^{\prime}}=\overline{A C F}$, points $A$ and $A^{\prime}$ map to each other under this reflection. Therefore, $\angle D A A^{\prime}=\angle D A^{\prime} A$. On the other hand, since point $A^{\prime}$ lies on $\omega$, we have $\angle A A^{\prime} C=\angle A B C=\angle A D A^{\prime}$. Thus, all three angles of triangle $D A A^{\prime}$ are equal, from which it follows that $\angle A B C=\angle A D A^{\prime}=60^{\circ}$.