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.