If $J$ is tangent point of $GH$ with incircle of $FGH$ and $D$ is intersection of $F$-mixtilinear inclrcle with $(FGH)$, then $\angle FGH=\angle GDJ$. Let $FGH$ be a triangle with circumcircle $A$ and incircle $B$, the latter with touchpoint $J$ in side $GH$. Let $C$ be a circle tangent to sides $FG$ and $FH$ and to $A$, and let $D$ be the point where $C$ and $A$ touch, as shown here.

Prove that $\angle FGH = \angle GDJ$.
 A: 
Notations:
Write $a:=GH$, $b:=HF$, $c:=FG$, and $s:=\frac{a+b+c}{2}$.  Let $\Omega$ and $\omega$ be the circumcircle and the incircle of $FGH$, respectively.  The circle internally tangent to $FG$, $FH$, and $\Omega$ is denoted by $\Gamma$.  Suppose that $\Gamma$ intersects $HF$ and $FG$ at $P$ and $Q$, respectively.  Denote by $\omega_a$ the excircle opposite to $F$ of $FGH$, which touches $GH$ at $T$.  Extend $FT$ to meet $\Omega$ again at $S$. Finally, $\theta:=\angle GFD$.

Proof:
Let $i$ be the inversion at $F$ with radius $FP=FQ$.  Then, $i(\Gamma)=\Gamma$, whereas $i(\Omega)$ is the tangent to $\Gamma$ at the point $E$, where $E$ is the second intersection between $\Gamma$ and $FD$.  Suppose that $i(\Omega)$ meets $HF$ at $G'$ and $FG$ at $H'$.  As $FG'H'$ and $FGH$ are similar triangles and $\Gamma$ is the excircle opposite to $F$ of $FG'H'$, it follows that $$\angle HFS=\angle HFT=\angle H'FE=\angle GFD=\theta\,.$$
Consequently, the minor arcs $HS$ and $GD$ of the circle $\Omega$ subtend the same angle $\theta$ at the circumference, so they are equal.  Ergo, $HS=GD$.  Since $TH=s-b=JG$ and $$\angle THS=\angle GHS=\angle GFS=\angle HFD=\angle HGD=\angle JGD\,,$$ we conclude that $GDJ$ and $HST$ are congruent triangles.  Thence, $$\angle GDJ=\angle HST=\angle HSF=\angle FGH\,.$$

P.S.:
It can be shown, using Casey's Theorem, that the center of $\omega$ is the midpoint of $PQ$.  Also, one can see that the internal angular bisector of $\angle FGH$ meets the line $DP$ at a point on $\Omega$, at which the tangent line $\ell_b$ is parallel to $HF$.  Likewise, the internal angular bisector of $\angle GHF$ meets the line $DQ$ at a point on $\Omega$, at which the tangent line $\ell_c$ is parallel to $FG$.  Finally, if $Z$ is the point of intersection between $\ell_b$ and $\ell_c$, then $Z$, $F$, $D$ are collinear.
A: This is NOT an answer but is an as-accurate-as-possible re-sketch of the original figure after guessing. Please let me know if there is any mis-interpretation.
[Note: The previous diagrams have been incorrectly drawn and were therefore deleted. The one below is the most updated version. Sorry for giving some misleading info.]

This time Geogebra shows that the two mentioned angles are actually equal.
PS: Maybe the OP can disclose the source of the question.
A: Using the Evan Chen's mixtilinear incircle article in here, the results become trivial.
I will change some notations.
In $\triangle ABC$, let the incircle hit $BC$ at $D$ and the $A$-mixtilinear incircle hit the circumcircle of $\triangle ABC$ at $E$. Prove that $\angle DEB = \angle ABC$. 
Since (9) holds, we have $\angle DTM_A = \angle AFB=180-\angle B - \frac{1}{2}\angle A$, where $F = AI \cap BC$.
Since $\angle BTM_A = 180-\frac{1}{2} \angle A$, we have $$\angle DTB = \angle BTM_A - \angle DTM_A = \angle B = \angle ABC$$ as desired.
A: The pictures I made with Cinderella Geometry show that the problem is stated correctly. The answer is that indeed $x=y$. (Of course, a proof is needed.) 

A different triangle, angles again are equal. 

A: Here's part of a solution based on a comment from @Blue.
Here's the picture showing a few additional points and lines. 

Angle equivalence
As per the suggestion from @Blue, we first reflect the point $F$ about the perpendicular bisector of $\overline{GH}$.  Then we draw a segment $\overline{F'D}$.  Since $\overline{F'G} = \overline{FH}$, the law of sines says that $\sin{\angle{x}} = \sin{\angle{y}}$.  Since both are interior angles of a triangle, $x < \pi, y < \pi$, so either $y = x$ or $y = \pi -x$.
Next, we note that by the law of sines, $\sin(\angle{GFH})=\sin(\angle{HDG})$. Since neither angle can be greater than $\pi$, either $\angle{GFH}=\angle{HDG}$ or $\angle{GFH}=\pi-\angle{HDG}$.  
If one could prove that $F'$ and $J$ and $D$ are collinear, that would be the end of the proof, but it is not yet obvious to me how one would do that.  
