Firstly, let's suppose that $B, G, F$ were collinear. Then, by Reim on $DFEG, DFAB$ we would have that $DFEG$ is cyclic. Hence, let's set out to show that $B, G, F$ are collinear. Let $G' = BF \cap (\triangle BDC).$ We claim that $G = G'$. Observe that $$\angle EFG' = \angle EFB = \angle FBC = 180 - \angle G'DC = \angle G'DE,$$so that $EFDG'$ is cyclic. Now, observe that $\angle EG'F = \angle EDF = \angle ABF$, which implies $EG' || AB$. Therefore, $G = G'$, and we're done. $\square$

https://artofproblemsolving.com/community/c6h34316p213011

Same as above but still, Let $F' = BG \cap (\triangle ABD).$ Claim 1. $D,E,F',G$ is cyclic By converse Reim of $(\triangle ABD)$ wrt parallel lines $AB$ and $EG$. Claim 2.$EF'||BC$ Angle chasing gives, $$ \angle EF'B = \angle EF'G = \angle EDG = \angle CBG = \angle CBF' $$So $F=F'$ and we are done

Here's the proof which I found quite quickly. It's easy to have something you want to work off when you are aware of Reim's theorem. Denote $G' = BF \cap (BDC)$. We would like to show that $EG' || AB$, which will prove that $G' = G$ as required. First, note that indeed we have that $ \measuredangle{EFG} = \measuredangle{EFB} = \measuredangle{FBC} = \measuredangle{GBC} = \measuredangle{GDE}$ hence $D,G',F,E$ concyclic. Now, by the converse of Reim's theorem we have that $EG' || AB$ as required. Well in hindsight I could have easily applied Reim's for the first part as well, but whatever.

Let BF meet circumcircle of BDC at G' we will prove G' is G. ∠DG'F = ∠DCB = ∠DEF so DFEG' is cyclic so ∠DEG' = ∠DFG' = ∠DAB so ∠EG' || AB so G' is G. Now that we know G is G' and DFEG' is cyclic we have DFEG is cyclic as wanted.