Let $GF$ cut $BC$ at $P.$ Circle $(K)$ with diameter $\overline{BP}$ is tangent to $AB$ at $B,$ hence reflection $AF$ of $AB$ on $AE$ is tangent to $(K)$ through $F.$ Let $Q$ be the 2nd intersection of $BG$ with $(K).$ By Pascal theorem for $BBQFFP,$ the intersections $A \equiv BB \cap FF,$ $G \equiv BQ \cap FP$ and $QF \cap PB$ are collinear $\Longrightarrow$ $C \in QF$ $\Longrightarrow$ $\angle DFC \equiv \angle BFQ=\angle ABG.$

My solution: Let $ X \in AB $ such that $ CX \parallel GB $ and $ Y=AE \cap BG, M=AE \cap CX $ . From $ \frac{XM}{MC}=\frac{BY}{YG}=\frac{BE}{EF}=1 \Longrightarrow MF=MB=MC=MX $ $ \Longrightarrow B, C, F, X $ are concyclic $ \Longrightarrow \angle CFD=\angle CXB=\angle GBA $ . Q.E.D