Czech-Polish-Slovak Mathematical Match competition problems

Denote $\alpha = \angle CAB$, $\beta = \angle ABC$, and $\gamma = \angle BCA$. Let $K$ and $L$ be the second intersections of lines $BE$ and $CE$ with $\omega$, respectively, different from $B$ and $C$. Observe that
$$\angle BAK = \angle BAD + \angle DAK = \angle BAD + \angle DBE = \angle BAD + \gamma = \angle ACD$$
and symmetrically $\angle CAL = \angle ABD$. It follows that arcs $AD$, $BK$, and $CL$ of $\omega$ have equal lengths, so chords $AD$, $BK$, and $CL$ also have equal lengths. In particular, since $E = BK \cap CL$ does not lie on $AD$, these chords are not diameters. It follows that if $O$ is the center of $\omega$, then $O$ does not lie on any of the chords $AD$, $BK$, $CL$, and in particular $O \ne E$. Moreover, $O$ and $E$ lie on the same side of line $AD$. Suppose without loss of generality that $O$ and $E$ lie in triangle $ACD$, for the second case is symmetric.

Observe that
$$\angle BEC = 360^\{\circ\} - \angle DBE - \angle DCE - \angle BDC = 360^\{\circ\} - \beta - \gamma - (180^\{\circ\} - \alpha) = 2\alpha = \angle BOC,$$
which implies that $B, O, E, C$ are concyclic. Further, if we denote $P = AD \cap BC$, then we have
$$\text{aligned DOE = BOE - BOD = 180 - BCE - 2 BAD = 180 - DCE - BAD = 180 - - BAD = APB = EFD, aligned}$$
which implies that $D, F, O, E$ are also concyclic.

Consider triangles $AFO$ and $GFO$. We have $AF = FG$ by assumption, also $OA = OG$ since $G$ lies on $\omega$, hence these two triangles are congruent. In particular $\angle FAO = \angle FGO$. Triangle $AOD$ is isosceles, hence $\angle FAO = \angle FDO$. This implies that $F, O, G, D$ are concyclic as well.

Since points $F, O, D$ are pairwise distinct, this implies that all five points $D, F, O, E, G$ lie on the circumference of triangle $FOD$, so in particular $D, E, F, G$ are concyclic. $\square$