## Task 1 - 060911

Determine the sum of the sizes of the interior angles at the five points of the five-pointed star shown in the figure, without measuring.

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0903.jpg?height=260&width=394&top_left_y=441&top_left_x=1342)
Let: $\alpha=\angle C A D ; \beta=\angle E B D ; \gamma=\angle A C E ; \delta=\angle B D A ; \epsilon=\angle B E C$. In the triangle $\triangle A C D$, the sum of the interior angles is then given by:

$$
180^{\circ}=\alpha+\gamma+\angle E C D+\angle C D B+\delta
$$

Using the sum of the interior angles in triangle $\triangle C D F$:

$$
=\alpha+\gamma+\delta+\left(180^{\circ}-\angle C F D\right)
$$

Furthermore, $\angle B F E=\angle C F D$, because they are vertical angles. Additionally, the sum of the interior angles in triangle $\triangle B F E$ is:

$$
=\alpha+\gamma+\delta+180^{\circ}-\angle B F E=\alpha+\gamma+\delta+180^{\circ}-\left(180^{\circ}-\beta-\epsilon\right)=\alpha+\gamma+\delta+\beta+\epsilon
$$

This is the sum of all 5 interior angles at the points of the five-pointed star, which thus totals $180^{\circ}$.