If $J$ is tangent point of $GH$ with incircle of $FGH$ and $D$ is intersection of $F$-mixtilinear inclrcle with $(FGH)$, then $\angle FGH=\angle GDJ$.

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.

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.

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.

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.