Let $DEF$ be the contact triangle of $ABC$ , $M = AI \cap (ABC)$ and $A'$ be the antipode of $A$ in $(ABC)$. It is well known that $T,D,M$ are collinear, also $T,I,A'$ and $A,O,A'$ are collinear as well. It suffices to show $\angle TDH + \angle DHT = \angle AIX + \angle AXI$. We know $\angle AXI = \angle AOI$ and $\angle AIX = \angle AMO$. Also, $\angle TDH = \angle TMO$ since $OM \parallel ID$. So it suffices to show $\angle TMI + \angle DHT = \angle AOI$. Note also $\angle TMI = \angle TA'O = \angle AOI - \angle OIA'$, thus we only need to show $\angle DHT = \angle OIA'$. This is equivalent to $(ITH)$ being tangent to line $IO$. Now we invert with respect to incircle. Let $B'$ and $C'$ be the midpoints of $AC,AB$ respectively. Since the polar of $H$ with respect to the incircle is $B'C'$, under inversion $H$ maps to $ID \cap B'C'$, also $T$ maps to the foot from $D$ to $EF$. Let $Z$ be the orthocenter of $DEF$, $P$ be the foot from $D$ to $EF$, $Q=ID \cap B'C'$ and $Y$ be the midpoint of the altitude from $A$ to $BC$. Since $Z$ lies on $IO$, we need to show $ZI \parallel PQ$, or $\frac{DZ}{ZP} = \frac{DI}{IQ}$. By the similarity of the contact and the excentral triangle, $\frac{DZ}{ZP}=\frac{II_A}{IA}$, so we need to show $AQ \parallel DI_A$. But since $Y,D,I_A$ are collinear(well known) and $AYDQ$ is a parallelogram(since $AY\parallel DQ$ and $AY=DQ$), we get $AQ \parallel DI_A$, so we are done.

Surprised that angle and length chasing sufficed. Let the reflection of $I$ over $BC$ be $W$, and let the arc midpoint of $BC$ opposite $A$ be $M$. Step 1: $\triangle AIO\sim \triangle IWM$. Proof: Let $AQ$ be the $A$-altitude of $\triangle ABC$. Let $WM\cap AQ = K$, let $A$-intouch in $\triangle ABC$ be $D$ and let $MI\cap BC=P$ and $MD\cap AQ=N$. Then: $$\frac{MD}{MN}=\frac{MI}{MA}=\frac{MP}{MI}$$so since $PD\perp ID$ we have $IN\perp AN$. We also have $N$ midpoint of $AK$ because $D$ is midpoint of $IW$, so: - $IA=IK$ and $OA=OM$ - $\angle KAI=\angle MAO$ because $AO$ and $AQ$ are isogonally conjugate in $\angle BAC$. Therefore, $\triangle AKI\sim\triangle AMO$ so $\triangle IWM\sim \triangle AKM\sim \triangle AIO$ as needed. Step 2: We show that it suffices to prove $\measuredangle AIO=\measuredangle HTM$. Proof: Since $\triangle AIO\sim\triangle IWM$ we have $$\measuredangle AIX=\measuredangle AMO=\measuredangle IMO=\measuredangle MIW=\measuredangle OAI$$so $AX=OI$ meaning $AXOI$ is an isosceles trapezium. Thus, if the reflection of $TH$ over $AI$ is $\ell$, then $\measuredangle(AX,TH)=\measuredangle (\ell,OI)$. The problem is equivalent to $\measuredangle (AX,TH)=\measuredangle AMT=\measuredangle (AI, TM)$ so it suffices for $$\measuredangle(\ell, OI)=\measuredangle (AI,TM)\iff \measuredangle(OI,TM)=\measuredangle (\ell, AI)$$But by reflection, $\measuredangle(\ell,AI)=\measuredangle(AI,TH)$ so it suffices for $\measuredangle (OI,AI)=\measuredangle (TM, TH)$ or $\measuredangle AIO=\measuredangle HTM$. Step 3: $\measuredangle AIO=\measuredangle HTM$. Proof: It is well-known that $T, D, M$ are collinear. By power of a point at $D$, we have $$DT\times DM = DB\times DC=DH\times DW$$so $THMW$ is cyclic. Therefore, $$\measuredangle AIO=\measuredangle IWM=\measuredangle HWM=\measuredangle HTM$$as desired. Diagram: https://cdn.discordapp.com/attachments/990202812849860618/1115887387004895374/Screen_Shot_2023-06-07_at_4.18.00_pm.png

Solved with binsherlo. Define some points. Let $(AIO) \cap (ABC)=Q,AX\cap (ABC)=P,A'$ be the antipode of $A$ and $M,N$ be the midpoints of arcs $BC,ABC$. Let $R$ be the point on $(ABC)$ such that $AR\parallel BC$. Note that $Q$ is the reflection of $M$ with respect to $IO$. Lemma $1$: $X,Q,N$ are collinear. Proof: Let $N'$ be the reflection of $N$ onto $IO$. \[\measuredangle OQX-90=90-\measuredangle XIO=\measuredangle (IO,BC)=\measuredangle N'NM=\measuredangle QMN=90-\measuredangle NQO\]Thus, the claim has proven. Lemma $2$: $H,Q,R$ are collinear. Proof:We will use complex numbers. Let $(ABC)$ be unit circle. $A=a^2,B=b^2,C=c^2$. After some calculations we get $H=b^2+c^2+ab+ac+bc,Q=\frac{-I}{\overline{I}bc},R=\frac{-b^2c^2}{a^2}$. Hence, we get the result. Finally, apply pascal at $RAMYTA'$ in order to get concurrency of $AR,YT,IH$ then pascal at $TPARQY$ which implies the collinearity of $T-P-H \square$