## Task A-1.2.

Points $A, B, C, D$ and $E$ are connected by segments as shown in the figure. Segments $\overline{A B}$ and $\overline{B C}$ intersect segment $\overline{D E}$ at points $F$ and $G$, respectively. If $\varangle A B C=20^{\circ}$ and if $\varangle D F A=\varangle C G E$, determine $\varangle E A B+\varangle D E A$.

![](https://cdn.mathpix.com/cropped/2024_05_30_ce71b02a7f068198719dg-02.jpg?height=345&width=322&top_left_y=1295&top_left_x=1278)
## Solution.

Let $\varphi=\varangle D F A=\varangle C G E$.

![](https://cdn.mathpix.com/cropped/2024_05_30_ce71b02a7f068198719dg-02.jpg?height=645&width=559&top_left_y=1839&top_left_x=680)

The angles $\varangle B F G$ and $\varangle D F A$ are vertical angles, so $\varangle B F G=\varangle D F A=\varphi$. Similarly, $\varangle B G F$ and $\varangle C G E$ are vertical angles, so $\varangle B G F=\varangle C G E=\varphi$.

The sum of the angles in triangle $B F G$ is $180^{\circ}$, so we have

$$
180^{\circ}=\varangle G B F+\varangle B F G+\varangle B G F=\varangle A B C+2 \varphi=20^{\circ}+2 \varphi,
$$

from which we conclude that $\varphi=80^{\circ}$.

The angles $\varangle D F A$ and $\varangle E F A$ are supplementary, so $\varangle E F A=180^{\circ}-\varphi$.

Finally, the desired sum of the angles is the sum of the two remaining angles in triangle $A F E$:

$$
\varangle E A B+\varangle D E A=\varangle E A F+\varangle F E A=180^{\circ}-\varangle E F A=\varphi=80^{\circ}
$$