## Task A-1.5.

Let $A B C D E$ be a regular pentagon and let $F$ be a point inside it such that triangle $A B F$ is equilateral. Determine the angles of triangle $D E F$.
## Solution.

The sum of the measures of all angles in a regular pentagon is $(5-2) \cdot 180^{\circ}=540^{\circ}$. Therefore, each interior angle in a regular pentagon is $108^{\circ}$.

Since triangle $A B F$ is equilateral, we have $\varangle E A F=\varangle E A B-\varangle F A B=$ $108^{\circ}-60^{\circ}=48^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_30_a374c05c2f1967f14be6g-04.jpg?height=702&width=733&top_left_y=1054&top_left_x=593)

Furthermore, since $|A E|=|A B|=|A F|$, triangle $A E F$ is isosceles, so $\varangle A E F=\varangle A F E=\frac{180^{\circ}-48^{\circ}}{2}=66^{\circ}$. It follows that

$$
\varangle D E F=\varangle D E A-\varangle A E F=108^{\circ}-66^{\circ}=42^{\circ}
$$

Triangles $D E F$ and $D C F$ are symmetric with respect to the line $D F$, so they are congruent.

It follows that $2 \varangle F D E=\varangle F D E+\varangle F D C=\varangle C D E=108^{\circ}$, or $\varangle F D E=54^{\circ}$.

It remains to calculate $\varangle D F E=180^{\circ}-\varangle D E F-\varangle F D E=180^{\circ}-42^{\circ}-54^{\circ}=84^{\circ}$.

The angles of triangle $D E F$ are $42^{\circ}, 54^{\circ}$, and $84^{\circ}$.

Note: Instead of the congruence of triangles $D E F$ and $D C F$, it can be argued that line $D F$ is the angle bisector of $\varangle A F B$, and then calculate $\varangle D F E=180^{\circ}-66^{\circ}-30^{\circ}$, and finally determine $\varangle F D E$.