solved with binsherlo Let $CG\cap EF =L, EF\cap BC=K$ and $N$ be the reflection of $K$ to $L$. \[\angle BGF=\angle BGC-\angle FGC=\angle BCA-\angle KEC=\angle BKF\]Hence $B,F,G,K$ are cyclic.It's easy to show that $\angle CKL=\angle LGK$. We have $LN^2=LK^2=LC.LG=LE.LF$ thus, $(K,N;E,F)=-1$. $BDF\sim CDE$ and $FNE\sim BDC$. \[\frac{DE}{DF}=\frac{CD}{BD}=\frac{EN}{FN}\]Which gives that $\angle FDN=\angle NDE$. So $ND\perp BC\implies LD=LN=LK$. $\angle LDK=\angle LGK$ so $L,D,G,K$ are cyclic. $\angle DGL=\angle DKL=\angle BGH\implies \angle AEF=\angle BGD$.

Let $X$ be the intersection of $CG$ and $EF$, let $Y$ be the intersection of $BC$ and $EF$, and let $D'$ be the fixed point along segment $BC$ of the inversion centered at $X$ preserving $(FECG)$. Now $$\angle GBD'=180^{\circ}-\angle BCG-\angle BGC =\angle BCX-\angle BCA=\angle XCE=\angle GFX$$$$\angle GBF=\angle ABC+\angle GBD'=\angle ACB +\angle XCE=\angle XCD'=\angle GD'X$$These properties uniquely define $D'$ so $D'$ must be the image of $F$ in $\sqrt{xb}$-inversion in triangle $GXB$. As $$\angle ED'C=\angle XFD'+\angle XGD'=\angle EFD'+\angle FGB=\angle EFD'+\angle BGC-\angle CGF=$$$$\angle EFD'+\angle ACB-\angle AEF=\angle EFD'+\angle FYB=\angle FD'B$$Thus we have that $D=D'$ to finish the problem simply notice $$\angle AEF=\angle XGF=\angle BGD$$

Let $P$ and $T$ be the intersections of $EF$ with $CG$ and $BC$, respectively. Note that $$\angle GBT=180^\circ-\angle ECG=\angle GFT,$$so $BGTF$ is cyclic. From this, $\angle TGF=\angle TBF=\angle CGB$, so $\triangle BGC\sim\triangle FGT$, and $$\angle PTC=\angle FGB=\angle TGC,$$implying that $$PT^2=PC\cdot PG=PE\cdot PF.$$After some manipulations, this is equivalent to $\frac{FT}{TE}=\frac{FP}{PT}$. By Menelaus on $\triangle AFE$ with the colinear points $B-C-T$ and the similar triangles $\triangle FBD\sim\triangle ECD$, this is equivalent to $\frac{FP}{PT}=\frac{BD}{DC}$. Hence, $D$ and $P$ are corresponding points in the similar triangles $BGC$ and $FGT$, so the result follows. $\square$

Let $FE$ meet $BC$ at $P$, $FG$ meet $\Omega$ again at $Q$. Then we have $$\angle AEF=\angle CGF=\angle ECQ, $$i.e. $EF\parallel CQ$. This implies $\angle FPB=\angle QCB=\angle FGB$, i.e. $F,B,G,P$ are concyclic. Hence, $$\angle CGP=\angle BCG-\angle BPG=\angle BQG-\angle BFG=\angle FBQ=\angle BCQ=\angle FPB,$$i.e. $FP$ touches $(PCG)$. Let $X$ be a point on line $EF$ such that $DX=XP$. Then by angle chasing we have $$\angle EDX=\angle EDC-\angle XDP=\angle FDB-\angle XPD=\angle DFE,$$i.e. $DX$ touches $(DEF)$. This implies $X$ lies on the radical axis $\ell$ of $(PCG)$ and $(DEF)$. Note that $C,E,F,G$ are also concyclic, implying $EF, CG$ and $\ell$ must concur at $X$. Hence, $DX^2=PX^2=XC\cdot XG$, implying $\angle DGX=\angle XDC=\angle XPD=\angle QGB$. So we obtain that $\angle BGD=\angle QGC=\angle AEF$ as desired.