## Task 2 - 150732

In the depicted figure, the following applies:

$$
\begin{aligned}
& \angle A B C=\angle B A C=\alpha \\
& \angle A D E=\angle B D C=\beta \\
& \angle A C D=\angle B C F=\gamma \\
& \angle B C D=\delta, \angle A E D=\epsilon \\
& \angle C E D=\eta, \angle E D C=\psi
\end{aligned}
$$

It is given that $\delta=70^{\circ}$.

Determine $\alpha, \beta, \gamma, \epsilon, \eta$, and $\psi$!

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0413.jpg?height=434&width=851&top_left_y=1730&top_left_x=934)
The three angles with vertex $C$ form a straight angle. Therefore, it holds that

$$
2 \gamma+70^{\circ}=180^{\circ} \Rightarrow \gamma=55^{\circ}
$$

The angle $\angle B C F$ is an exterior angle of the isosceles triangle $A B C$. Thus, according to the exterior angle theorem,

$$
\gamma=2 \alpha \quad ; \quad \alpha=\frac{\gamma}{2}=27.5^{\circ}
$$

Now, applying the angle sum property of a triangle to $\triangle B D C$:

$$
\alpha+\beta+\gamma=180^{\circ} \Rightarrow \beta=180^{\circ}-\alpha-\gamma=180^{\circ}-27.5^{\circ}-70^{\circ}=82.5^{\circ}
$$

The three angles with vertex $D$ also form a straight angle. Therefore, it holds that

$$
2 \beta+\psi=180^{\circ} \Rightarrow \psi=180^{\circ}-2 \beta=180^{\circ}-2 \cdot 82.5^{\circ}=15^{\circ}
$$

Applying the angle sum property of a triangle to $\triangle E C D$:

$$
\eta+\gamma+\psi=180^{\circ} \Rightarrow \eta=180^{\circ}-\gamma-\psi=180^{\circ}-55^{\circ}-15^{\circ}=110^{\circ}
$$

and thus, according to the supplementary angle theorem,

$$
\epsilon=180^{\circ}-110^{\circ}=70^{\circ}
$$

Thus, the angle measures $\alpha, \beta, \gamma, \epsilon, \eta$ and $\psi$ have been determined.