It is easy to see that IcY and IbX are concur at the antipode of A in AXY let A' the main claim is A' is circum center of IIbIc and with some easy angle chasing you can observe that ID passes through A' and by using thales we are done

I'll present my trig bash solution. WLOG let $AC>AB$. Let $I_B$ and $I_C$ be the $B$ and $C$-excenters. We define the midpoint of arc $BAC$ as $M$ and $K=I_CY\cap I_BX$. It is trivial that $A,X,Y,K$ are concyclic. If we apply the homothety centered at $A$ with scale factor $2$ we see that $$UT\perp BC\Leftrightarrow KI \perp BC$$holds. We will prove the equivalent condition. Claim. $M\in(AXY).$ Proof. Define $I_A$ as the $A$-excenter. It is well-known that $I$ is the orthocenter of $I_AI_BI_C$. From this fact we obtain $$90-\angle{KI_CI_B}=\angle{YAI_C}=\angle{I_BAC}=90-\frac{\angle{BAC}}{2}=\angle{I_CAB}=\angle{XAI_B}=90-\angle{KI_BI_C}$$$$\Rightarrow KI_B=KI_C.$$This implies that $K$ is on the perpendicular bisector of $I_BI_C$. However, we know that $(ABC)$ is the nine-point circle of $\triangle{I_AI_BI_C}$, so $M$ is the midpoint of $I_BI_C$. This means $\angle{KMA}+\angle{KXA}=180$, which is enough to prove the claim. $\blacksquare$ Define $\angle ABC=B$ and $\angle ACB=C$. From the previous claim it is apparent that $KM\parallel AI\Rightarrow \angle{MKI}=\angle{KIA}$. Moreover, $KI\perp BC\Leftrightarrow \angle{MKI}=\angle{KIA}=180-\angle{BID}-\angle{AIB}=180-(90-\frac{\angle B}{2}+90+\frac{\angle C}{2})=\frac{\angle B-\angle C}{2}.$ We will prove this. Applying LoS while we consider the parallel lines and the isosceles triangle we get that $$\frac{KI}{KI_C}=\frac{\sin{(90-\frac{C}{2})}}{\sin{(90-\frac{B}{2}+\angle KIA)}}=\frac{\cos{\frac{C}{2}}}{\cos{(\frac{B}{2}-\angle KIA)}}\text{ and}$$$$\frac{KI}{KI_B}=\frac{\sin{(90-\frac{B}{2})}}{\sin{(90-\frac{C}{2}-\angle KIA)}}=\frac{\cos{\frac{B}{2}}}{\cos{(\frac{C}{2}+\angle KIA)}}$$holds. These two ratios must be equal. Equating and using identities gives out that $$\frac{1}{2}(\cos{(C+\angle KIA)}+\cos{\angle KIA})=\frac{1}{2}(\cos{(B-\angle{KIA})}+\cos{\angle KIA})$$$$\Rightarrow \cos{(C+\angle KIA)}=\cos{(B-\angle KIA)}\Rightarrow \angle{KIA}=\frac{B-C}{2}.$$holds true. We're done. $\blacksquare$

Let $I_CY \cap I_BX=Z$, then from orthology of $\triangle II_BI_C, \triangle ABC$ we have that $Z,I,D$ colinear and therefore $U$ is midpoint of $Z$ so homothety at $A$ with scale 2 finishes, thus done .

Construct diamater AJ of (AXY), let Z be the midpoint of minor arc BC of (ABC), I_b and I_c be excenters opposite to vertexes B and C. It's easy to see that I_bJ is perpidencular to AB, I_cJ is perpidencular to AC. with a simple angle chase, it shows that figures JI_bI_cA and ZCIBD are similar. Thus, AIJ = DIZ implying J, I, D are collinear. So it should be trivial to point out that UT is perpidencular to BC.

Let $I_CY \cap I_BX=M$. Since $\angle{AYM}=\angle{AXM}=90$, $AXYM$ is cyclic and $M$ is the antipode of $A$ in $(AXYM)$. We will prove that $M, I, D$ are collinear thus $MD \perp BC$ and by homothety $UT \perp BC$. Since the line $I_BAI_C$ is the exterior angle bisector of $\angle A$ we have $\angle I_CAY = \angle I_BAX=90-\angle A/2$, using the 90 degree angles at $X$ and $Y$ it is seen that $MI_B=MI_C$ so there's a circle centered at $M$ which passes through $I_B$ and $I_C$. But since $\angle I_CII_B=90+\angle A/2$ and $\angle I_CMI_B=180-\angle MI_CI_B -\angle MI_BI_C = \angle I_CAY +\angle I_BAX =180-\angle A$ it is seen that $I$ is on the circle with center $M$. Thus $MI_C=MI$ which means $\angle MII_C=\angle MI_CI=\angle MI_CI_B +\angle CI_CI_B = \angle A/2 + \angle CBI_B=\angle A/2 + \angle B/2 = 90 -\angle C/2 = \angle CID$ and we're done since this implies the required collinearity.

Attachments:

Let $\triangle = \overset{\triangle}{J_AJ_BJ_C}$ where $J_A$, $J_B$, $J_C$ are excenters. Let $K$ be the circumcenter of $\triangle$. $I$ is the orthocenter of $\triangle$ and also $B$ and $C$ are feet of altitudes from $J_B$ and $J_C$ respectively. Let $P$ be the antipode of $A$ in $(AXY)$. Lemma. In a triangle $ABC$ with circumcenter $O$ and orthocenter $H$, let $D,E,F$ be the feet of altitudes. If $O$ and $O'$ are symmetric wrt line $BC$, then $O'H\perp EF$. Proof. $OO'\parallel AH$ and $OO'=2\cdot OM=AH$ so $O'H\parallel AO\perp EF$. We will show that $K$ and $P$ are symmetric wrt $J_BJ_C$ and this will mean $P,I,D$ are collinear by the Lemma, and $UT\perp BC$ will follow by Thales. Since $\angle PYA=\angle PXA=90$, $J_C,Y,P$ and $J_B,X,P$ are collinear. We have $\angle PJ_CA=90-\angle J_CAY=\frac{\angle A}{2}$ and similarly $\angle PJ_BA=\frac{\angle A}{2}$. We have $\angle KJ_CA=90-\angle J_CJ_AJ_B=\frac{\angle A}{2}$ and similarly $\angle KJ_BA=\frac{\angle A}{2}$. $\angle PJ_CA=\angle PJ_BA=\angle KJ_CA=\angle KJ_BA$ means $P$ and $K$ are symmetric wrt $J_BJ_C$. [asy][asy] size(15cm); import olympiad; import geometry; pair A = dir(110); pair B = dir(210); pair C = dir(330); pair I = incenter(A, B, C); pair Ja = excenter(B, C, A); pair Jb = excenter(A, C, B); pair Jc = excenter(A, B, C); pair D = foot(I, B, C); pair X = foot(Jb, A, B); pair Y = foot(Jc, A, C); pair P = extension(Jb, X, Jc, Y); pair U = (A + P) / 2; pair T = (A + D) / 2; draw(A -- B -- C -- cycle); draw(Jc -- Y -- A -- X -- Jb); draw(Y -- P -- X); draw(I -- D); draw(P -- A -- D, dashed); draw(Jc -- Jb, blue); dot("$A$", A, dir(20)); dot("$B$", B, dir(180)); dot("$C$", C, dir(0)); dot("$D$", D, dir(270)); dot("$I$", I, dir(70)); dot("$J_B$", Jb, dir(40)); dot("$J_C$", Jc, dir(140)); dot("$X$", X, dir(90)); dot("$Y$", Y, dir(90)); dot("$P$", P, dir(90)); dot("$U$", U, dir(160)); dot("$T$", T, dir(200)); dot("$K$", circumcenter(Ja,Jb,Jc), dir(90)); [/asy][/asy]