## Task A-1.4. (4 points)

A regular pentagon $A B C D E$ is given. The lines $B C$ and $D E$ intersect at point $F$. Determine the angles of triangle $B E F$.
## First solution.

The angles at the vertices of a regular pentagon are $108^{\circ}$.

Therefore, $\varangle F D C=\varangle D C F=180^{\circ}-108^{\circ}=72^{\circ}$, so

$$
\varangle E F B=\varangle D F C=180^{\circ}-2 \cdot 72^{\circ}=36^{\circ}
$$

![](https://cdn.mathpix.com/cropped/2024_05_30_13589bc3b5dd31895527g-02.jpg?height=468&width=797&top_left_y=1748&top_left_x=561)

Triangle $D C F$ is isosceles and $|D F|=|F C|$. Since $|E D|=|B C|$, it follows that $|E F|=|E D|+|D F|=|B C|+|C F|=|B F|$.

Thus, triangle $B E F$ is isosceles, so

$$
\varangle F E B=\varangle F B E=\frac{1}{2}\left(180^{\circ}-\varangle E F B\right)=72^{\circ}
$$

1 point

The angles of triangle $B E F$ are $\varangle B E F=\varangle E B F=72^{\circ} \text{ and } \varangle B F E=36^{\circ}$