Slovenia — National Math Olympiad

Denote $\angle FAE = \alpha$. Since the point $E$ lies on the bisector of the segment $AC$, its distances to the points $A$ and $C$ are the same. Hence, $E$ is the centre of the circle containing the points $A$, $C$ and $F$ and we have $|AE| = |CE| = |FE|$. So, $\angle EFA = \angle FAE = \alpha$. Since $AB$ and $CD$ are parallel we get $\angle DEA = \angle EAF = \alpha$ and $\angle CEF = \angle EFA = \alpha$.

The right triangles $AED$ and $CEG$ are congruent because they have three common angles and the hypothenuses have the same length.
So, $|ED| = |EG|$ and $|CG| = |AD| = |BC|$. From here we can conclude that the triangle $DEG$ is isosceles and
$$\angle EGD = \frac{\pi - \angle DEG}{2} = \frac{\angle GEC}{2} = \frac{\alpha}{2}.$$

By Pythagoras' theorem we have $|FB|^2 = |FC|^2 - |BC|^2 = |FC|^2 - |GC|^2 = |FG|^2$, so $|FB| = |FG|$. Therefore, $GFB$ is an equilateral triangle and
$$\angle FGB = \frac{\pi - \angle GFB}{2} = \frac{\angle GFA}{2} = \frac{\alpha}{2}.$$
We have shown that $\angle FGB = \angle EGD$, so the points $B$, $G$ and $D$ are congruent.