Angles $\angle ACB$ and $\angle ADB$ are equal, therefore points $C$ and $D$ are either both in the interior of the circumcircle of triangle $ABE$, or both on this circle, or both outside this circle. Thus we distinguish three cases:

1. $F \in (EC)$ and $G \in (ED)$,
2. $F = C$, $G = D$ (in this case the statement of the problem is obvious),
3. $C \in (EF)$, $D \in (EG)$.

We only treat the first case, the last one being similar. We have
$$\angle GDC = \angle GDB + \angle BDC = \angle GEA + \angle BAC = \frac{AG}{2} + \angle ABE = \frac{AG+AE}{2} = \angle GFE,$$
which shows that the quadrilateral $FGDC$ is cyclic.

We notice that line $AB$ is the radical axis of the circumcircles of triangle $ABE$ and quadrilateral $ABCD$, line $FG$ is the radical axis of the circumcircles of triangle $ABE$ and quadrilateral $FGDC$, and line $CD$ is the radical axis of the circumcircles of the quadrilaterals $ABCD$ and $FGDC$. It is well known that the radical axis of three circles are either parallel (if the three centers are collinear) or concurrent (in the radical center of the three circles).