Let $T$ be the projection of $D$ on $\overline{EF}$. Then $I,T,X$ are collinear because $T,X$ are Inverses of each other WRT $\odot(I)$. By Reim's $GTFX$ is cyclic. Also $\Delta XFE\stackrel{+}{\sim}\Delta XBC\implies \measuredangle DBX=\measuredangle TFX=\measuredangle DGX\implies B,D,G,X$ are concyclic.

Let $L$ the intersection of $(AI)$ and $(BC)$, and $S$ the intersection of $(AI)$ and $\odot(ABC)$ ($S$ is the South pole). $(EA,EI)=(FA,FI)=90^\circ$ so $AFIEX$ is cyclic and $\odot(AEF)$ is tangent externally to $\odot(IBC)$ at $I$. Invert WRT $\odot(IBC)$, $A \leftrightarrow L, B \leftrightarrow B, C \leftrightarrow C, I \leftrightarrow I, S \leftrightarrow \infty$ so $\odot(ABC) \leftrightarrow (BC)$ and $\odot(AEF) \leftrightarrow \odot(IDL)$, it follows that $X \leftrightarrow D$ so $X,D,S$ are collinear. WRT $(XD)$ and $(BG)$, $(XB)$ and $(AI)$ are antiparallel lines and $(AI)//(DG)$ (both are perpendicular to $(EF)$), hence $(XB)$ and $(DG)$ are anti-parallel lines, it follows that $XBDG$ is cyclic.

some lemmas that kill the problem https://artofproblemsolving.com/community/c946900h1911664_properties_of_the_sharkydevil_point

Here is a solution without inversion. As $AEF$ is isosceles, $AI$ is perpendicular to $EF$. Then $AI$ and $DG$ are parallel. Then $\angle DGB=\angle \frac{A}{2}$.As $\angle BXC=\angle A$, we only need that $XD$ is the bisector of $\angle BXD$ to get the desired concylic. By the spiral similarity center lemma (see Yufei for example), $X$ is the spiral similarity center between $FE$ and $BC$. Then $ \Delta XFB \sim \Delta XEC$, thus $\frac{FB}{EC}=\frac{XB}{XC}$. The tangency with the incircle implies $BF=BD$ and $CE=CD$ and thus $ \frac{XB}{XC}=\frac{DB}{DC}$. Then the angle bisector theorem implies the desired angle bisector and we are done.

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Let $I$ be the incenter of $ABC$, let $H$ be the intersection of $DG$ with $FE$. We start off with a lemma, which is ultra nice. $\color{red}\rule{25cm}{0.5pt}$ Lemma: The points $I,H$ and $X$ are colinear. Proof: Denote with $\varphi$ the inversion around the incircle of $ABC$. Notice that the circumcircle of $ABC$ gets sent to the nine-point circle of $DEF$ under $\varphi$, but also notice that $\overline{FE}$ gets sent to $(FIE)$ under $\varphi$. Since we have that $X$ lies on both $(FIE)$ and $(ABC)$ we must have that the inverse point of $X$ must lie on both the nine-point circle and on line $FE$. Since that can't be the midpoint we have that it must be the foot of the D-altitude of $DEF$, which is $H$. This implies that $I,H$ and $X$ are colinear points. $\color{red}\rule{25cm}{0.5pt}$ Now let's return to the problem at hand here. We have that $\angle DFG=\angle DFE + \angle EFG = 90 + \frac{1}{2}\angle B$ and we have that $\angle FDG = 90 - \angle EFD = \frac{1}{2} \angle C$. Thus we have that $\angle FGD = \frac{1}{2} \angle A$. But notice the following: $$\angle FXH = \angle FXI = \angle FAI = \frac{1}{2} \angle A = \angle FGD = \angle FGH$$this implies that $FXGH$ is a cyclic quadrilateral. This implies that $\angle FXG = 90$. Since we have that $\angle BDG = \angle FDG + \angle FDB = 90 + \frac{1}{2}\left( \angle C - \angle B \right)$. To show that the problem statement is true we must have that $\angle BXG = 180-\angle BDG$. but this implies that we should have that $\angle BXF = \frac{1}{2}\left( \angle B - \angle C \right)$. But notice that we have that $\angle BXF = \angle BXA - \angle FXA = 180 - \angle C - 180 + 90 - \frac{1}{2} \angle A = \frac{1}{2}\left(\angle B - \angle C \right)$. Thus we have that $BXGD$ is a cyclic quadrilateral.

By the properties of the sharky devil point, we know that if $M = XD \cap (ABC)$, $M$ is the midpoint of arc $BC$ in $(ABC)$ and so $\angle BXD = \angle BXM = \angle BAM = \frac{A}{2}$. Also, we know that $\angle AFE = 90 - \frac{A}{2}$ which gives $\angle BDG = \frac{A}{2}$. So, $B, D, G, X$ are concyclic

anyone__42 wrote: Let $ABC$ be an acute triangle with $AC>AB$, Let $DEF$ be the intouch triangle with $D \in (BC)$,$E \in (AC)$,$F \in (AB)$,, let $G$ be the intersecttion of the perpendicular from $D$ to $EF$ with $AB$, and $X=(ABC)\cap (AEF)$. Prove that $B,D,G$ and $X$ are concylic Sharky-Devil point :3. Clearly $X$ is the $A$-SharkyDevil Point on $\triangle ABC$ so its known that $X-D-M_A$ where $M_A$ is the midpoint of the smaller arc $\widehat{BC}$. Now note that $GD \perp EF$ and $AI \perp EF$ and that means $GD \parallel EF$ so the rest is angles :3. $\angle BXM_A=\angle BAM_A=\angle BGD \implies BDGX \; \text{cyclic}$ Thus we are done

anyone__42 wrote: Let $ABC$ be an acute triangle with $AC>AB$, Let $DEF$ be the intouch triangle with $D \in (BC)$,$E \in (AC)$,$F \in (AB)$,, let $G$ be the intersecttion of the perpendicular from $D$ to $EF$ with $AB$, and $X=(ABC)\cap (AEF)$. Prove that $B,D,G$ and $X$ are concylic $\color{blue}\boxed{\textbf{Proof:}}$ $\color{blue}\rule{24cm}{0.3pt}$ Let $P\equiv EF\cap DG$ Let $I$ be the incenter of $\triangle ABC$ Let $Y\equiv EF\cap BC$ $X$ is the A-Sharky-Devil point $$\Rightarrow I, P \text{ and }X \text{ are collinears}$$Since $IA\perp FE$ and $DG\perp FE$: $$\Rightarrow IA//DG$$$$\Rightarrow \angle FAI=\angle FGP...(I)$$Since $AEIFX$ is cyclic: $$\Rightarrow \angle FAI=\angle FXI...(II)$$By $(I)$ and $(II)$: $$\Rightarrow FXGP \text{ is cyclic}$$Since $DG\perp EF$ and $FXGP$ is cyclic: $$\Rightarrow FX\perp XG...(III)$$Since $AXBC$ is cyclic: $$\Rightarrow 180-\angle ACB=\angle BXA=\angle BXF+\angle FXA...(IV)$$Since $AEFX$ is cyclic: $$\Rightarrow \angle FXA=180-\angle AEF...(V)$$By $(IV)$ and $(V)$: $$\Rightarrow \angle AEF=\angle BXF+\angle ACB=\angle BXF+\angle ECB$$$$\Rightarrow \angle BXF=\angle FYB...(VI)$$In $\triangle YPD$: $$\Rightarrow \angle PYD+\angle PDY=90=\angle FYB+\angle PDY$$By $(VI)$: $$\Rightarrow \angle BXF+\angle PDY=90$$$$\Rightarrow \angle BXF=90-\angle PDY=\angle PDI$$$$\Rightarrow \angle BXF+90=\angle PDI+90$$By $(III)$ and $ID\perp BC$: $$\Rightarrow \angle BXF+\angle FXG=\angle PDI+\angle IDG$$$$\Rightarrow \angle BXG=\angle PDC=\angle GDC$$$$\Rightarrow XBDG \text{ is cyclic}_\blacksquare$$$\color{blue}\rule{24cm}{0.3pt}$

Observe that $X$ is the $A-$Sharkydevil Point. Since $GD \perp EF \perp AI$, it follows that $GD \parallel AI$. Simply note that since $XD$ bisects $\angle BXC$ (well known Sharkydevil), we have that, \[2\measuredangle DXB = \measuredangle CXB = \measuredangle CAB = 2\measuredangle IAB = 2\measuredangle DGB\]from which the desired result follows.

There is a much simpler solution by just angle chase and similar triangles.