Very suitable IMO1 angle chasing problem! Denote $\angle ABC = \beta$, $\angle ACB = \gamma$. Observe that $\angle ABD = \frac{\beta + \gamma}{2}$ due to $AB = AD$, so $\angle IBD = \angle ICD = \frac{\gamma}{2}$ and $BIDC$ is cyclic. Next, $FD = FG$ implies $\angle GFB = \gamma = \angle BCD = \angle EID = \angle EIG$, so $BIGF$ is also cyclic. We now bravely claim that the second intersection point $T$ of the circumcircles of $CEI$ and $BIGF$ lies on both $AG$ and $EF$, which would complete the proof. On one hand, $\angle IFT = \angle IBF = \angle ICE = \angle ITE$, so $E$, $F$, $T$ are collinear. On the other hand, $\angle FTG = \angle FIG = \frac{\beta - \gamma}{2}$, so the collinearity of $A$, $G$, $T$ is now equivalent to $\angle ETA = \frac{\beta-\gamma}{2}$, i.e. (as $\angle CTE = \angle CIE = \frac{\beta + \gamma}{2}$ to $\angle CTA = \beta$ - that is, $ABTC$ to be cyclic. However, $\angle BTI + \angle CTI = \angle BFI + 180^{\circ} - \angle CEI = \frac{\beta}{2} + \angle IBC + \angle ECB = \beta + \gamma = 180^{\circ} - \angle BAC$ and so we are done.

Let $X$ be the Miquel point of complete quadrilateral $IEDF$. Claim: $BIDC$ is concyclic $$\angle BID=180^{\circ}-2\angle BIA=\angle DCB$$Claim: $E$, $F$, and $X$ are collinear It is sufficient to show that both $XF$ and $XE$ bisect $\angle IXD$. $$\angle IXF=\angle IBF=\angle FCD=\angle FXD$$$$\angle IXE=\angle ICE=\angle DBE=\angle DXE$$Claim: $X$ lies on the circumcircle of $ABC$ $$\angle BXC=\angle FIE+\angle FDE=\angle FBC+\angle ACB+\angle ABD=180^{\circ}-\angle BAC$$Claim: $BGIFX$ is concyclic $$\angle IGF=\angle IDF=\angle IBF$$Claim: $A$, $G$, and $X$ are collinear $$\angle GXB=\angle GFB=\angle ACB=\angle AXB$$

My solution is a bit different (Lovely problem by the way ) Let $\angle BAC=\alpha , \angle ABC=\beta , \angle ACB=\gamma , \odot (ABC)=\Gamma , \odot(CEI)=\Omega , EF \cap \Omega =\{X\}$ So $\alpha +\beta+\gamma=180 , \angle ABE=\angle CBE=\frac{\beta}{2} , \angle ACI=\angle BCI=\frac{\gamma}{2}$ $\textbf{Claim:}$ Points $B,D,I$ and $C$-are concyclic $\textbf{Proof:}$ From the isosceles triangle $\triangle ABD$ we have: $\angle ABD+ \angle ADB+\angle BAD=180 \implies 2\angle ABD +\alpha =180 \implies \angle ABD=90-\frac{\alpha}{2}=\frac{\beta+\gamma}{2}$ $\implies \frac{\beta+\gamma}{2}=\angle ABD=\angle ABE+\angle EBD=\frac{\beta}{2}+\angle EBD \implies \angle EBD=\frac{\gamma}{2} \implies \angle ICD=\frac{\gamma}{2}=\angle EBD \equiv IBD \implies$ $\angle ICD=\angle IBD \implies$ Points $B,D,I$ and $C$-are concyclic $\square$. Let $\odot(BDIC)=\sigma$ $\textbf{Claim:}$ Points $F,D,X$ and $C$-are concyclic $\textbf{Proof:}$ $\angle FDC \equiv \angle BDC \stackrel{\sigma}{=} \angle BIC=180-\angle EFC \stackrel{\Omega}{=} 180-\angle EXC \equiv 180- \angle FXC \implies \angle FDC=180-\angle FXC \implies$ $ \angle FDC+\angle FXC =180 \implies$ Points $F,D,X $ and $C$-are concyclic $\square$. Let $\odot(FDXC)=\omega_1$ $\textbf{Claim:}$ Points $B,I,F$ and $X$-are concyclic $\textbf{Proof:}$ By Power Of The Point Theorem (POP) we have: $EI \cdot EB=Pow(E,\sigma)=ED \cdot EC=Pow(E,\omega_1)=EF \cdot EX \implies EI \cdot EB= EF \cdot EX$ which by the converse of POP means that Points $B,I,F$ and $X$-are concyclic $\square$. Let $\odot(BIFX)=\omega_2$ $\textbf{Claim:}$ $G \in \omega_2$ $\textbf{Proof:}$ Since $FD=FC$ we get $\angle FGD=\angle FDG$ so: $180-\angle IGF=\angle FGD=\angle FDG \equiv \angle BGI \stackrel{\sigma}{=} \angle BCI =\frac{\gamma}{2}=\angle EBC \equiv IBF \implies 180-\angle IGF =\angle IBF \implies$ $ \angle IGF+\angle IBF=180 \implies$ Points $B,I,F$ and $G$-are concyclic $\iff G \in \odot(BIF) \iff G \in \omega_2 \square$ $\textbf{Claim:}$ $X \in \Gamma \iff X \in \odot (ABC)$ $\textbf{Proof:}$ $\angle BXC=\angle BXF+\angle FXC \stackrel{\omega_1}{=} \angle BXF + (180-\angle FDC) \stackrel{\omega_2}{=} (180-BIF)+(180-\angle FDC)=$ $=360-\angle BIF-\angle FDC \equiv 360-\angle BIC-\angle BDC \stackrel{\sigma}{=} 360-\angle BDC-\angle BDC=360-2\angle BDC=360-2(180-\angle BDA)=$ $=360-360+2\angle BDA=2\angle BDA=2\angle ABD=2 \cdot \frac{\beta+\gamma}{2}=\beta+\gamma \implies \angle BXC=\beta+\gamma$ So $\angle BAC+\angle BXC=\alpha +(\beta+\gamma)=\alpha+\beta+\gamma=180 \implies \angle BAC+\angle BXC =180 \implies$ Points $A,B,C$ and $X$ -are concyclic $\iff X \in (ABC) \iff X \in \Gamma \square$ $\textbf{Claim:}$ Points $A-G-X$ are collinear $\textbf{Proof:}$ $\angle GXB \stackrel{\omega_2}{=} \angle GFB=180-\angle GFD \stackrel{\triangle GFD}{=} \angle FGD+\angle FDG \stackrel{FD=FG}{=} \angle FDG+\angle FDG=2\angle FDG \equiv 2\angle BDI \stackrel{\sigma}{=} 2\angle BCI=2 \cdot \frac{\gamma}{2}=\gamma=$ $=\angle ACB \stackrel{\Gamma}{=} \angle AXB \implies \angle GXB=\angle AXB \implies$ Points $A-G-X$ are collinear $\blacksquare$

sami1618 wrote: Let $X$ be the Miquel point of complete quadrilateral $IEDF$. Claim: $BIDC$ is concyclic $$\angle BID=180^{\circ}-2\angle BIA=\angle DCB$$Claim: $E$, $F$, and $X$ are collinear It is sufficient to show that both $XF$ and $XE$ bisect $\angle IXD$. $$\angle IXF=\angle IBF=\angle FCD=\angle FXD$$$$\angle IXE=\angle ICE=\angle DBE=\angle DXE$$Claim: $X$ lies on the circumcircle of $ABC$ $$\angle BXC=\angle FIE+\angle FDE=\angle FBC+\angle ACB+\angle ABD=180^{\circ}-\angle BAC$$Claim: $BGIFX$ is concyclic $$\angle IGF=\angle IDF=\angle IBF$$Claim: $A$, $G$, and $X$ are collinear $$\angle GXB=\angle GFB=\angle ACB=\angle AXB$$ Thing is real, if you could not notice that $X$ lies on the $ABC$.