In triangle $ABC$, $AB = AC > BC$. A circle is drawn around vertex $B$ with radius $BC$, intersecting side $AC$ at point $D$ (different from $C$) and side $BA$ at point $E$. Another circle with the same radius is drawn around point $D$, intersecting side $AB$ at $E$ again, side $AC$ at $F_{1}$, and the extension of side $AC$ at $F_{2}$. What are the measures of angles $B F_{1} E$ and $B F_{2} E$?
Based on the first circle, $B E = B D$. Based on the second circle, the segment $D E$ is also equal, so $\triangle B D E$ is equilateral, $\angle B D E = 60^{\circ}$.

The angles we are looking for in the second circle are the inscribed angles subtending the arc $B E$, so they are half the size of the central angle corresponding to the arc $B E$, their size is $30^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_bea8d08ad951741ee8e3g-1.jpg?height=349&width=889&top_left_y=346&top_left_x=607)

Márta Kafka (Budapest, Apáczai Csere J. Secondary School, 8th grade)

Note. The calculation can also be completed without using the theorem of inscribed and central angles, for example, as follows: $\triangle B C D$ and $\triangle A B C$ are both isosceles, so

\[
\begin{gathered}
180^{\circ} = \angle C B D + 2 \angle B C D = \angle C B D + 2 \angle A B C = \angle C B D + \\
+ 2 (\angle C B D + 60^{\circ});
\end{gathered}
\]

from this, $\angle C B D = 20^{\circ}$, $\angle A B C = \angle B C D = \angle B D C = 80^{\circ}$, $\angle B D A = \angle B D F_{1} = 100^{\circ}$. From the isosceles triangle $\triangle B D F_{1}$, $\angle B F_{1} D = 40^{\circ}$; $\angle E D A = 40^{\circ}$, from the isosceles triangle $\triangle E D F_{1}$, $\angle E F_{1} D = 70^{\circ}$, and thus $\angle B F_{1} E = \angle E F_{1} D - \angle B F_{1} D = 30^{\circ}$, finally, similarly from the isosceles triangles $\triangle D B F_{2}$ and $\triangle D E F_{2}$, $\angle B F_{2} E = \angle B F_{2} D - \angle E F_{2} D = 50^{\circ} - 20^{\circ} = 30^{\circ}$.

László Takács (Sopron, Széchenyi I. Secondary School, 1st grade)