Slovenia — National Math Olympiad

Let $\angle DCA = \gamma$. Since the quadrilateral $AFCD$ is cyclic, we have
$$\angle DFA = \angle DCA = \gamma.$$
Since $ABCG$ is also a cyclic quadrilateral, we have $\angle GBA = \angle GCA = \gamma$. From here we conclude that the lines $DF$ and $GB$ are parallel.

In the cyclic quadrilateral $AFCD$ we have $\angle FAD = \pi - \angle DCF$ and in the cyclic quadrilateral $GEFC$ we have $\angle GEF = \pi - \angle GCF$. Since $\angle GCF = \angle DCF$ we get $\angle FAD = \angle GEF$, so the lines $AD$ and $GE$ are parallel.

Let $A'$ be a point on the line $AD$ and let $B'$ be a point on the line $DF$, such that $GA' \parallel EA$ and $GB' \parallel FB$. Obviously, the points $G$, $B'$ and $A'$ are collinear. The quadrilateral $AEGA'$ is a parallelogram, so $|A'G| = |AE|$. The quadrilateral $FBGB'$ is also a parallelogram, so $|B'G| = |FB|$.

If $|AE| = |FB|$, we have $|A'G| = |B'G|$. Since the point $G$ clearly does not lie between $A'$ and $B'$, we conclude that $A' = B'$. This is only possible when $A' = B' = D$. In this case, the line $GD$ is parallel to the line $AE$, which means that the segments $AB$ and $CD$ are parallel.

Conversely, if $AB$ and $CD$ are parallel, then the quadrilaterals $AEGD$ and $FBGD$ are parallelograms, since they have two pairs of parallel sides each. So, $|AE| = |DG| = |FB|$.