Any solution

$\measuredangle TBC = 180^{\circ} - \measuredangle BCT - \measuredangle BTC = 180^{\circ} - \measuredangle YCI_A - \measuredangle CYI_A = \measuredangle CI_AY$. Similarly $\measuredangle SCB = \measuredangle BI_AX$. Applying trig ceva in $BCYI_A$ we get: $\frac{ \sin \angle BCT}{ \sin \angle TCY} \cdot \frac{ \sin \angle CYT}{ \sin \angle TYI_A} \cdot \frac{ \sin \angle YI_AT}{ \sin \angle TI_AB} \cdot \frac{ \sin \angle I_ABT}{ \sin \angle TBC} = 1 \Longrightarrow \frac{ \sin \angle CYT}{ \sin \angle TYI_A} = \frac{ \sin \angle TI_AB}{ \sin \angle I_ABT}$ $(\star)$ On the other hand $\measuredangle CYT + \measuredangle TYI_A = \measuredangle AYI_A = \measuredangle BTI_A = \measuredangle TI_AB + \measuredangle I_ABT$ $(\star \star)$. Thus by $(\star)$, $(\star \star)$ and Two Equal Angles Lemma, we get $\measuredangle CYT = \measuredangle TI_AB$ and $\measuredangle TYI_A = \measuredangle I_ABT$. Similarly $\measuredangle SXI_A = \measuredangle SCI_A$ and $\measuredangle SXB = \measuredangle SI_AC$. From here we get $TY \parallel SX$. On the other hand applying the general angle bisector theorem in $\triangle STI_A$ and Trig Ceva in $BCTS$ we get: $\frac{TZ}{ZS} = \frac{TI_A}{SI_A} \cdot \frac{KT}{KS} = \frac{TI_A}{SI_A} \cdot \frac{ \sin \angle KSB}{ \sin \angle SBK} \cdot \frac{ \sin \angle KBC}{ \sin \angle BCK} \cdot \frac{ \sin \angle KCT}{ \sin \angle CTK} = \frac{TI_A}{SI_A} \cdot \frac{ \sin \angle TI_AY}{ \sin \angle I_AYT} \cdot \frac{ \sin \angle SXI_A}{ \sin \angle SI_AX} = \frac{YT}{XS}$. Thus $X$, $Y$ and $Z$ must be collinear.

Let $CS\cap I_aX=X',BT\cap I_aY=Y'$. By easy angle chasing (see picture) we get $BSX'X,CYY'T,BX'I_aY'C$ are cyclic and $\triangle BXS\sim TYC\sim BI_aC\Longrightarrow \angle BXS=\angle CYT=\angle BI_aC=90-\frac{A}{2}\Longrightarrow XS\parallel YT$ hence if $Z'=ST\cap YX$ then: $$\frac{SZ'}{Z'T}=\frac{XS}{YT}\ (1)$$note that: $$\angle SCI_a=\angle I_aBX'=\angle SXI_a\ , \ \angle TBI_a=\angle TBI_a=\angle \angle Y'CI_a=\angle TYI_a\ \clubsuit$$Observe that by ceva_sin theorem: $$\frac{SZ}{ZT}=\frac{SI_a}{TI_a}.\frac{\sin BI_aK}{\sin CI_aK}=\frac{SI_a}{TI_a}.\frac{\sin I_aBK}{\sin I_aCK}.\frac{\sin BCX'}{\sin CBY'}\stackrel{\clubsuit}{=}\frac{SI_a}{TI_a}.\frac{\sin I_aYT}{\sin I_aXS}.\frac{\sin BI_aX}{\sin CI_aY}=\frac{SI_a}{TI_a}.\frac{TI_a}{YT}.\frac{XS}{I_aS}=\frac{XS}{YT}$$$$\Longrightarrow \frac{SZ}{ZT}=\frac{XS}{YT}\ (2)$$combining $(1),(2)$ we conclude that $Z\equiv Z'$. Q.E.D

Nice! By fact 5, $I_AX=I_AY \Longrightarrow BX+CY=BC$. Let $L$ be the point on the line $BC$ such that $BX=BL$ and $CY=CL$. Note that $B, L, T, I_A$ and $C, L, S, I_A$ are concyclic. Since $\angle TLS=\angle A=180-2\angle BI_AC$ and $XS, LS$ are reflections in $I_AB$; $YT, TL$ are reflections in $CI_A$, we get $XS \parallel YT$. Since $L$ is the miquel point of quadrilateral $SKTI_A$ we conclude that $LZ$ bisects angle $TLS$. By the angle bisectors theorem, we obtain $$\frac{TZ}{ZS}=\frac{TL}{LS}=\frac{YT}{XS}$$so $XS \parallel YT \Longrightarrow Z \in XY$ as desired.

Let the excircle be tangent to $AB, AC$ at $P, Q$, and assume wlog that $X$ lies on segment $AP$. Then $AXI_aY$ cyclic quickly implies that triangles $I_aPX$ and $I_aQY$ are congruent and $Q$ lies on segment $AY$. Let $R$ be the reflection of $X$ about $BI_a$, which evidently lies on $BC$. Then $$CR = BC - BR = BP + CQ - BX = CQ + PX = CQ + QY = CY$$ So $R$ is also the reflection of $Y$ about $CI_a$. Also notice that we have $T = I_aC \cap (I_aBR)$ and $S = I_aB \cap (I_aCR)$. Thus the problem is reduced to the following: Let $ABC$ be a triangle and $P$ a point on side $BC$. Let $(ABP)$ and $(ACP)$ cut segments $AC$ and $AB$ at $D$ and $E$ respectively and let $Q = BD \cap CE$. Let $R = AQ \cap DE$, then $R$ lies on the line through the reflections of $P$ about $AB$ and $AC$. To prove this, note by angle chasing that $ADQE$ is cyclic and let $O$ be its center. Notice that $P$ is the Miquel point of $ADQE$ and hence $ODPE$ is cyclic and $O, R, M$ are collinear by well-known properties of the Miquel Point. Let $X = (RPE) \cap AB$, then angle chasing using the two previously mentioned properties shows that $RX$ is perpendicular to $AC$. Analogously if $Y = (RPD) \cap AC$ then $RY$ is perpendicular to $AB$. Moreover, it is evident that $AXPY$ is cyclic, and hence we just want to show that $R$ lies on the Steiner line of $P$ wrt $\triangle AXY$, which is true as $R$ is the orthocenter of $AXY$.

Here is my solution for this problem Solution We have: ($TB$; $TI_a$) $\equiv$ ($XI_a$; $XB$) $\equiv$ ($YI_a$; $YA$) $\equiv$ ($SC$; $SI_a$) (mod $\pi$) So: $K$, $T$, $I_a$, $S$ lie on a circle Then: ($KB$; $KS$) $\equiv$ ($I_aC$; $I_aB$) $\equiv$ $\dfrac{\pi}{2}$ $-$ $\dfrac{(AB; AC)}{2}$ (mod $\pi$) But: ($CB$; $CT$) $\equiv$ ($CI_a$; $CY$) (mod $\pi$) and ($TC$; $TB$) $\equiv$ ($TI_a$; $TB$) $\equiv$ ($TI_a$; $TK$) $\equiv$ ($SI_a$; $SK$) $\equiv$ ($YC$; $YI_a$) (mod $\pi$) so: $\triangle$ $BCT$ $\stackrel{+}{\sim}$ $\triangle$ $I_aCY$ or $\triangle$ $BCI_a$ $\stackrel{+}{\sim}$ $\triangle$ $TCY$ Then: $\dfrac{BI_a}{TY}$ = $\dfrac{BC}{TC}$ Similarly: $\dfrac{CI_a}{SX}$ = $\dfrac{BC}{BS}$ Hence: $\dfrac{TY}{SX}$ . $\dfrac{CI_a}{BI_a}$ = $\dfrac{TC}{BS}$ or $\dfrac{TY}{SX}$ = $\dfrac{BI_a}{CI_a}$ . $\dfrac{TC}{BS}$ = $\dfrac{TI_a}{TS}$ . $\dfrac{KI_a}{KS}$ . $\dfrac{KT}{KI_a}$ . $\dfrac{ST}{SI_a}$ = $\dfrac{KT}{KS}$ . $\dfrac{I_aT}{I_aS}$ = $\dfrac{ZT}{ZS}$ But: ($TZ$; $TY$) $\equiv$ ($TZ$; $TI_a$) + ($TI_a$; $TY$) $\equiv$ ($KS$; $KZ$) + ($TC$; $TY$) $\equiv$ ($KS$; $KB$) + ($KB$; $KZ$) + ($BC$; $BI_a$) $\equiv$ $-$ $\dfrac{\pi}{2}$ + $\dfrac{(AB; AC)}{2}$ + ($SZ$; $SB$) + $\dfrac{(BC; BA)}{2}$ $\equiv$ $\dfrac{\pi}{2}$ $-$ $\dfrac{(CA; CB)}{2}$ + ($SZ$; $SB$) $\equiv$ ($SB$; $SX$) + ($SZ$; $SB$) $\equiv$ ($SZ$; $SX$) (mod $\pi$) so: $TY$ $\parallel$ $SX$ or $X$, $Y$, $Z$ are collinear

Nice Problem! Also I think the main idea behind the proof is same in almost all posts, but since I worked hard on this, I'll anyways post Iran TST 2017 P3 wrote: In triangle $ABC$ let $I_a$ be the $A$-excenter. Let $\omega$ be an arbitrary circle that passes through $A,I_a$ and intersects the extensions of sides $AB,AC$ (extended from $B,C$) at $X,Y$ respectively. Let $S,T$ be points on segments $I_aB,I_aC$ respectively such that $\angle AXI_a=\angle BTI_a$ and $\angle AYI_a=\angle CSI_a$.Lines $BT,CS$ intersect at $K$. Lines $KI_a,TS$ intersect at $Z$. Prove that $X,Y,Z$ are collinear. Solution: Let $E$ be a point such that $BXI_AE$ is a parallelogram $$\implies \angle BXI_A=\angle BTI_A=\angle BEI_A \implies T \in \odot (BEI_A)$$Let $\odot (BEI_A) \cap BC=G$ $\implies$ $XBI_A=\angle BI_AE=\angle I_ABG$ $\implies$ $GI_A$ $=$ $BE$ $=$ $XI_A$ $=$ $YI_A$ $\implies$ $G,S$ $\in$ $\odot (CFI_A)$, where $F$ is such a point, $CYI_AF$ is parallelogram, $G$ is the miquel point of $BKCTI_AS$ and since, $SKTI_A$ is cyclic $\implies$ $ZG \perp BC$ and, $ZG$ bisects $\angle SGT$ and $\angle KGI_A$ $$\angle STY=\angle STG+\angle GTY=\angle GTI_a-\angle STI_A+2\angle GTC=180^{\circ}+\angle GTC-\angle STI_A$$$$=\angle 270^{\circ}-\frac{\angle B}{2}- \left(180^{\circ}-\angle BI_AC-\angle TSI_A \right)=90^{\circ}-\frac{1}{2}B+180^{\circ}-\angle BIC+\angle TSI_A$$$$=\frac{C}{2}+90^{\circ}+\angle TSI_A =\angle XST \implies XS || TY \implies \Delta XSZ \sim \Delta YTZ \implies X - Z -Y$$

Nice problem . Iranian TST 2017, first exam, day1, problem 3 wrote: In triangle $ABC$ let $I_a$ be the $A$-excenter. Let $\omega$ be an arbitrary circle that passes through $A,I_a$ and intersects the extensions of sides $AB,AC$ (extended from $B,C$) at $X,Y$ respectively. Let $S,T$ be points on segments $I_aB,I_aC$ respectively such that $\angle AXI_a=\angle BTI_a$ and $\angle AYI_a=\angle CSI_a$.Lines $BT,CS$ intersect at $K$. Lines $KI_a,TS$ intersect at $Z$. Prove that $X,Y,Z$ are collinear. Solution: Firstly we'll give a construction for $S$. Let $Y'$ be the reflection of $Y$ over $I_AC$. We have that $S=\odot CI_AY' \cap BI_A$ . Similairly for $T$. Now we'll animate $X$ on $AB$. Let $XY \cap KI_A=W$ and $XY \cap ST=W'$ Our goal is to prove $W=W'$. Easy to see that $X \mapsto W$ is projective . Now $X\mapsto Y$ is projective. And also since $Y'$ is the reflection of $Y$ along a fixed line $CI_A$ together with the fact that $S\in \odot(Y'CI_A)$ we have $X\mapsto Y \mapsto Y' \mapsto S\mapsto W'$ is projective. So it suffices to check for three values of $X$. $\bullet X=B$ Notice that in this case $S,K=A$ so done $\square$. $\bullet X$ is a point such that $CX \perp I_AB$ Notice that in this case $Y,Z,K,T=C$ so done again $\square$. $\bullet X$ is a point such that $X,B,Y',I_A$ are concyclic. By easy angle chase we have $ST \equiv XY$ and we are done $\blacksquare$.

Firstly since $\angle BSC=180^{\circ}-\angle CSI_A=180^{\circ}-\angle AYI_A=\angle BXI_A$ and $\angle XBI_A=\angle SBI_C$, $$\triangle BXI_A\sim \triangle BSC$$similarly$$\triangle CYI_A\sim\triangle CTB$$Therefore$$\triangle BXS\sim\triangle BI_AC\sim\triangle TYC$$Hence$$\angle BSK=\angle BXI_A=180^{\circ}=\angle AYI_A=\angle 180^{\circ}-\angle CTB=\angle KTI_A$$which implies $S,K,T,I_A$ are concyclic. Now$$\angle XST=\angle BSX+\angle BST=\angle BCT+\angle BST$$$$\angle STY=180^{\circ}-\angle CTY+180^{\circ}-\angle STC=\angle BCT+\angle BST$$hence $XS\|TY$ Now notice that from the similarities $$\frac{XS}{BS}=\frac{CI_A}{BC}$$$$\frac{TY}{TC}=\frac{BI_A}{BC}$$Hence$$\frac{XS}{TY}=\frac{BS}{TC}\cdot\frac{CI_A}{BI_A}$$Now $$\frac{BS}{BK}=\frac{\sin\angle BKS}{\sin\angle BSK}=\frac{\sin\angle CKT}{\sin\angle BTC}=\frac{CT}{CK}$$Hence $$\frac{BS}{TC}\cdot\frac{CI_A}{BI_A}=\frac{BK}{BI_A}\cdot\frac{CI_A}{CK}$$Now $$\frac{BK}{BI_A}=\frac{\sin\angle SI_AZ}{\sin\angle ZSI_A}=\frac{SZ}{ZI_A}$$similarly $$\frac{CI_A}{CK}=\frac{ZI_A}{ZT}$$Hence $$\frac{XS}{TY}=\frac{SZ}{ZT} $$together with $XS\|TY$, $X,Y,Z$ are collinear.

My solution. Quite interesting one. Let $Z_1$ be the point on $BC$, such that $ZZ_1\perp BC$. Let $I=BT\cap I_aY$ and $H=CS\cap I_aX$. Claim. $SKTI_a$ is cyclic. $$\angle KTI_a+\angle KSI_a=\angle BTI_a+\angle CSI_a=\angle AXI_a+\angle AYI_a=180^{\circ} \qquad \square$$Also, in the same manner, it is easy to see that $BSHX$ and $CYIT$ are cyclic. By those three concyclicity, we as well have that $BCII_aH$ is cyclic. Claim. $Z_1$ is a Miquel's point of $SKTI_a$. For this, see EGMO Theorem 10.12 and its corollaries. (This is inversion+angle chase+Brocard's theorem basically.) Claim. $XS\parallel YT$. We have, \begin{align*} \angle XSZ&= \angle XSI_a+\angle I_aSZ\\ &= 180^{\circ}-\angle XSB+\angle I_aSZ\\ &= 180^{\circ}-\angle XHB+\angle I_aSZ\\ &= \angle BHI_a+\angle I_aSZ\\ &= \angle BHI_a+\angle ZKT\\ &= 180^{\circ}-(90^{\circ}-\frac{\angle C}{2})+\angle I_aST\\ &= 90^{\circ}+\frac{\angle C}{2}+\angle I_aST\\ &= 90^{\circ}+\frac{\angle C}{2}+\angle I_aST. \end{align*}On the other hand, \begin{align*} \angle YTZ&= \angle CTY+\angle ZTC\\ &= \angle CIY+\angle ZTC\\ &= 180^{\circ}-\angle CII_a+\angle ZTC\\ &= \angle I_aBC+\angle ZTC\\ &= 90^{\circ}-\frac{\angle B}{2}+180^{\circ}-\angle STI_a. \end{align*}Subtracting those we get that $$\angle XSZ-\angle YTZ=90^{\circ}+\frac{\angle C}{2}+\angle I_aST-(90^{\circ}-\frac{\angle B}{2}+180^{\circ}-\angle STI_a)=\frac{\angle B}{2}+\frac{\angle C}{2}+\angle I_aST+\angle STI_a-180^{\circ}=\frac{\angle B}{2}+\frac{\angle C}{2}-\angle BI_aC=0.\qquad \square$$ Claim. $SZ_1YL$ is cyclic with $Z_1$ being the angle bisector of $\angle SZ_1T$. Inversion around $SKTI_a$ takes $S-Z-T$ collinearity to $S-Z_1-T-L$ concyclity. Since $LS=LT$, we have $\angle SZ_1L=\angle STL=\angle LST=\angle LZ_1T$. $\qquad \square$ Claim. $\frac{SZ_1}{Z_1T}=\frac{SX}{TY}$. By Law of Sines on $\triangle BSZ_1$ and $\triangle BSZ_1$, we obtain that $$\frac{SZ_1}{SX}=\frac{\frac{SB\cdot \sin{\angle SBC}}{\sin{\angle SZ_1B}}}{\frac{SB\cdot \sin{\angle SBX}}{\sin{\angle SXB}}}=\frac{\sin{\angle SXB}}{\sin{\angle SZ_1B}}$$and Law of Sines on $\triangle CTZ_1$ and $\triangle CTY$ gives $$\frac{Z_1T}{TY}=\frac{\frac{CT\cdot \sin{\angle BCT}}{\sin{\angle CZ_1T}}}{\frac{CT\cdot \sin{\angle YCT}}{\sin{\angle CYT}}}=\frac{\sin{\angle CYT}}{\sin{\angle CZ_1T}}.$$Since $\angle SXB=\angle CXB=\angle CIB=\angle CIT=\angle CYT$ and $\angle SZ_1B=180^{\circ}-\angle SZ_1C=\angle SI_aC=\angle BI_aT=180^{\circ}-\angle BZ_1T=\angle CZ_1T$, the claim follows. Therefore, using the last claim and the angle bisector theorem on $\triangle SZ_1T$, we have $$\frac{SZ}{ZT}=\frac{SZ_1}{Z_1T}=\frac{SX}{TY},$$hence $\triangle XSZ\sim \triangle YTZ$, hence $\angle XZS=\angle YZT$, which means that $X,Y,Z$ are collinear.

Here is a little bashy solution which I found

This is such a beautiful problem! I probably overcomplicated the solution, but here goes. By Ptolemy's theorem, we have $AX+AY=2AI_A\cos\frac{\alpha}{2}=2s$, so we can choose a point $P$ on segment $BC$, such that $BP=BX$ and $CP=CY$. Since $\overline{BI_A}$ is the external angle bisector of $\angle B$, it follows that $P$ is the reflection of $X$ over $\overline{BI_A}$. Hence from the given condition, it follows that $P$ lies on both $(BTI_A)$ and $(CSI_A)$, and so it is the Miquel point of quadrilateral $SI_ATK$. Since $P$ lies on $\overline{BC}$, this quadrilateral is cyclic. Let its center be $O$. Now let $E$ and $F$ be points on $\overline{I_AS}$ and $\overline{I_AT}$, such that $Z$ is the orthocenter of $\triangle I_AEF$. By the Miquel point properties, $OSPT$ is cyclic, so \[\measuredangle ZES=90^\circ-\measuredangle SI_AT=\measuredangle OTS = \measuredangle OPS=\measuredangle ZPS\]and so $ZPES$ is cyclic too. Similarly, $ZPFT$ is also cyclic. Consequently, \[\measuredangle PEI_A=\measuredangle PZS=\measuredangle PZT=\measuredangle PFI_A\]and thus $I_AEPF$ is also cyclic. Now to finish, since $\overline{XY}$ is the Steiner line of $P$ with respect to $\triangle I_AEF$, it passes through its orthocenter, which is $Z$. [asy][asy] defaultpen(fontsize(10pt)); size(10cm); pen mydash = linetype(new real[] {5,5}); pair A = dir(120); pair B = dir(220); pair C = dir(320); pair I = incenter(A,B,C); pair I1 = 2*circumcenter(I,B,C)-I; real s = 0.3; pair P = s*B+(1-s)*C; pair X = 2*foot(P,B,I1)-P; pair Y = 2*foot(P,C,I1)-P; pair S = 2*foot(circumcenter(C,P,I1),I1,B)-I1; pair T = 2*foot(circumcenter(B,P,I1),I1,C)-I1; pair K = extension(B,T,C,S); pair Z = extension(I1,K,S,T); draw(A--B--C--cycle, black+1); draw(B--X); draw(C--Y); draw(B--I1); draw(C--I1); draw(B--T); draw(C--S); draw(S--T); draw(K--I1); draw(X--Y, mydash); draw(P--X, dotted); draw(P--Y, dotted); draw(circumcircle(A,X,Y)); draw(circumcircle(B,P,T)); draw(circumcircle(C,P,S)); draw(rightanglemark(P, foot(P,B,I1), B, 2)); draw(rightanglemark(P, foot(P,C,I1), I1, 2)); dot("$A$", A, dir(A)); dot("$B$", B, dir(180)); dot("$C$", C, dir(30)); dot("$X$", X, dir(270)); dot("$Y$", Y, dir(0)); dot("$P$", P, dir(90)); dot("$S$", S, dir(225)); dot("$T$", T, dir(10)); dot("$K$", K, dir(90)); dot("$I_A$", I1, dir(270)); dot("$Z$", Z, dir(315)); [/asy][/asy] [asy][asy] defaultpen(fontsize(10pt)); size(10cm); pen mydash = linetype(new real[] {5,5}); pair K = dir(35); pair S = dir(95); pair I1 = dir(195); pair T = dir(345); pair B = extension(I1,S,T,K); pair C = extension(S,K,I1,T); pair Z = extension(S,T,I1,K); pair O = (0,0); pair P = extension(O,Z,B,C); pair E = extension(Z,foot(Z,I1,T),I1,B); pair F = extension(Z,foot(Z,I1,S),I1,C); pair X = 2*foot(P,B,I1)-P; pair Y = 2*foot(P,C,I1)-P; draw(K--S--I1--T--cycle); draw(S--B); draw(K--B); draw(T--C); draw(K--C); draw(S--T); draw(I1--K); draw(B--C); draw(O--P, dotted); draw(foot(Z,I1,T)--E, mydash); draw(foot(Z,I1,S)--F, mydash); draw(circumcircle(K,S,T)); draw(circumcircle(Z,F,T), dotted); draw(circumcircle(Z,E,S), dotted); draw(rightanglemark(Z,P,C,2)); draw(rightanglemark(Z,foot(Z,I1,T),I1,2)); draw(rightanglemark(Z,foot(Z,I1,S),S,2)); draw(I1--E--F--cycle, fuchsia+1); draw(circumcircle(I1,E,F), fuchsia+1); dot("$I_A$", I1, dir(I1)); dot("$S$", S, dir(S)); dot("$T$", T, dir(315)); dot("$K$", K, dir(45)); dot("$E$", E, dir(90)); dot("$F$", F, dir(315)); dot("$B$", B, dir(90)); dot("$C$", C, dir(0)); dot("$O$", O, dir(270)); dot("$Z$", Z, dir(180)); dot("$P$", P, dir(45)); [/asy][/asy]

Solved with erkosfobiladol. Let $Q$ be the miquel point of $SKTI_A$. Note that $\angle I_AQB=\angle I_ATB=\angle BXI_A$ and $\angle QBI_A=\angle XBI_A$ hence $I_AQ=I_AX$. Similarily $I_AQ=I_AY$. $X$ and $Y$ are the reflections of $Q$ with respect to $I_AB$ and $I_AC$ respectively. We will prove a new statement which proves this one. Lemma: If $ABCD$ is a cyclic quadrilateral whose miquel point is $M$ and and $M_1,M_2,M_3,M_4$ are the reflections of $M$ with respect to $AB,BC,CD,DA$, then $M_1,M_2,M_3,M_4$ are collinear and this line passes through the intersection of diagonals. Proof: $AB\cap CD=P,AD\cap BC=Q,AC\cap BD=R$. Let $H_1,H_2,H_3,H_4$ be the orthocenters of $QAB,QDC,PBC,PDA$. $\overline{M_2M_3M_4},\overline{M_1M_3M_4},\overline{M_1M_2M_4},\overline{M_1M_2M_3}$ are steiner lines of $M$ on $(QAB),(QDC),(PBC),(PAD)$ hence $M_1,M_2,M_3,M_4$ are collinear. Let $E,F$ be the altitudes from $D,C$ to $QC,QD$ and $H_1$ be the orthocenter of $QCD$. Since $Pow_{(AC)}(H_1)=H_1D.H_1E=H_1C.H_1F=Pow_{(BD)}(H_1)$. Similarily, $H_1,H_2,H_3,H_4$ lie on the radical axis of the circles with diameter $AC$ and $BD$ where $R$ also belongs to. Also since steiner line of $M$ passes through the orthocenters, we get that $M_1,M_2,M_3,M_4,H_1,H_2,H_3,H_4,R$ are collinear as desired.$\blacksquare$

Let $M$ be the Miquel's point of cyclic complete quadrilateral $I_AKST$. Let the lines through $Z$ parallel to $BT$ and $CS$ meet $BI_A$ and $CI_A$ at $D$ and $E$, respectively. By construction $M$ is the center of the spiral symmetry that sends $C$ to $S$ and that sends $T$ to $B$. As $$\frac{CE}{ET}=\frac{SZ}{TZ}=\frac{DS}{BD}$$the spiral symmetry also sends $E$ to $D$. Then $$\Delta DZE\sim\Delta BKC\sim \Delta SMC\sim \Delta DME$$so $Z$ and $M$ are reflections about $ED$. Notice that $EDI_AM$ is cyclic as $\angle MDE=\angle MSC=\angle MI_AE$. As $X$ and $Y$ can be observed to just be the reflections of $M$ about $BI_A$ and $CI_A$, the points $X$, $Y$, and $Z$ all lie on the steiner line of $M$ with respect to triangle $\Delta DEI_A$.