Czech-Polish-Slovak Mathematical Match

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 \begin\{aligned\} \angle DOE &= \angle BOE - \angle BOD = 180^\{\circ\} - \angle BCE - 2\angle BAD \\ &= 180^\{\circ\} - \angle DCE - \angle BAD = 180^\{\circ\} - \beta - \angle BAD = \angle APB = \angle EFD, \end\{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. $\square$