## Task B-1.4.

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_3d5a33db4db92485fe9eg-03.jpg?height=363&width=326&top_left_y=612&top_left_x=1276)
## Solution.

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

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 $\varphi=80^{\circ}$.

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

Finally, the sum of the desired angles is precisely the sum of the two remaining angles in triangle AFE:

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

Thus, the desired sum is $80^{\circ}$.