Points $A, B, C$, and $D$ lie on a line in this order. For a point $E$ outside the line,

$$
A E B \angle=B E C \angle=C E D \angle=45^{\circ} .
$$

Let the midpoint of segment $A C$ be $F$, and the midpoint of segment $B D$ be $G$. What is the measure of angle $F E G$?
Solution. According to the notation in the figure, $\angle AEC = \angle AEB + \angle BEC = 45^{\circ} + 45^{\circ} = 90^{\circ}$, so $\triangle AEC$ is a right triangle.

![](https://cdn.mathpix.com/cropped/2024_05_02_ec2e00444edc8e1ab123g-1.jpg?height=233&width=569&top_left_y=260&top_left_x=775)

Similarly, $\triangle BED$ is also a right triangle. For the angles in the right triangles,

$$
\angle ACE = 90^{\circ} - \alpha \text{, and } \angle DBE = 90^{\circ} - \beta \text{.}
$$

In $\triangle AED$, the angle $\angle AED$ is $135^{\circ}$ according to the conditions, so $\alpha + \beta = 45^{\circ}$.

Now, let's draw the segment $EF$. Point $F$ is the midpoint of the hypotenuse of the right triangle $\triangle AEC$, and by Thales' theorem, it is also the center of the circumscribed circle. Therefore, $\triangle AFE$ is an isosceles triangle, and $\angle AEF = \alpha$.

Similarly, after drawing $EG$, it can be seen that $\triangle EDG$ is also isosceles, and $\angle GED = \beta$.

The desired $\angle FEG$ is, based on the figure and the above, 

$$
\begin{aligned}
\angle FEG & = \angle AEB + \angle BEC + \angle CED - \angle AEF - \angle GED = \\
& = 45^{\circ} + 45^{\circ} + 45^{\circ} - \alpha - \beta = 135^{\circ} - 45^{\circ} = 90^{\circ}.
\end{aligned}
$$