Let $E'$, $F'$ be the reflection point of $A$ throught $E$, $F$. $M'$ be the midpoint of $\overline{E'F'}$.Then we know $TM_AM'$ is the Newton line of $\mathcal{Q}\{BF',F'C,CE',E'B\}$, that is $T$, $M'$, $M_A$ are collinear. Let $AI$ intersect $\odot(ABC)$ at $M\not= A$. Then \[ \measuredangle TSX=\measuredangle TM_AX \iff \measuredangle AMX=\measuredangle M'M_AX \]So we only need to show that $X, M_a, M, M'$ are concyclic. Notice that $X$ is the miquel point of $\mathcal{Q}\{BF',F'E',E'C,CB\}$. Thus \[ \triangle BXF' \stackrel{+}{\sim}\triangle M_AXM' \stackrel{+}{\sim}\triangle C'XE' \]$\implies X$ is the miquel point of $\mathcal{Q}\{M_AM',M'E',E'C,CM_A\}$. Let $Y:=BC\cap E'F'\implies X, Y, M_A, M'$ are concyclic. But it is obvious that $M, Y, M_A, M'$ are concyclic. \[\implies X, M_a, M, M' \text{ are concyclic}.\]

Official Solution: Let $O$ be the circumcenter of $\triangle ABC$, $J$ be the intersection of $EF$ and $OI$, and $L$ be the second intersection of $AJ$ and $\Omega$. Claim. $O, M_a, X, J, L$ are concyclic. Proof: Since $JO$ is the bisector of $\angle XJA$ and $XO = LO$, we have $J, L, O, X$ are concyclic. On the other hand, Let $M \in OI$ be the midpoint of $AX$, $Y \in \Omega$ such that $XY \parallel BC$. By $A, F, I, M, E$ are concyclic and $AFIE$ are harmonic quadrilateral, we have \[ -1 = (F, E; A, I) \stackrel{M}{=} (F, E; MA \cap EF, J) \stackrel{A}{=} (B, C; X, L) \stackrel{Y}{=} (B, C, \infty_{BC}, BC \cap YL). \]It then follows that $Y, M_a, L$ are collinear, that is, $OM_a$ is the bisector of $\angle LM_aX$. Thus $M_a, X, J, O$ are concyclic, as desired. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(11.5cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -10.409838616257938, xmax = 24.1887818224764, ymin = -13.624034082703128, ymax = 11.99225220366728; /* image dimensions */ /* draw figures */ draw(circle((-0.055794107161586025,1.4431604291360227), 7.582401971278432), linewidth(0.4)); draw(circle((5.576957684285586,-0.5852282522438403), 5.986840016803628), linewidth(0.4) + linetype("4 4")); draw((5.437781235050451,6.669387520785117)--(11.517399413958824,-1.3291405072564817), linewidth(0.4)); draw(circle((8.299594097037167,1.5595712564407571), 9.31210844456331), linewidth(0.4) + linetype("4 4")); draw((5.437781235050451,6.669387520785117)--(-6.659638498772746,-2.2827030822439744), linewidth(0.4)); draw((-6.659638498772746,-2.2827030822439744)--(6.405574813756307,-2.5246514769204365), linewidth(0.4)); draw((6.405574813756307,-2.5246514769204365)--(5.437781235050451,6.669387520785117), linewidth(0.4)); draw((0.9254617064835794,3.330271069645659)--(11.517399413958824,-1.3291405072564817), linewidth(0.4)); draw((-0.6109286318611473,2.193342219270571)--(17.30919271344813,-0.794679463952661), linewidth(0.4)); draw((17.30919271344813,-0.794679463952661)--(0.9254617064835794,3.330271069645659), linewidth(0.4)); draw((-0.055794107161586025,1.4431604291360227)--(11.517399413958824,-1.3291405072564817), linewidth(0.4)); draw((2.273453174918753,8.658936669636306)--(17.30919271344813,-0.794679463952661), linewidth(0.4)); draw((-0.1270318425082193,-2.4036772795822055)--(17.30919271344813,-0.794679463952661), linewidth(0.4) + linetype("4 4")); draw((6.45005905050214,5.337604464549033)--(-2.8480836688231372,-5.6063742120926445), linewidth(0.4) + linetype("4 4")); /* dots and labels */ dot((5.437781235050451,6.669387520785117),linewidth(3pt) + dotstyle); label("$A$", (5.558755432388681,6.850848816792463), NE * labelscalefactor); dot((-6.659638498772746,-2.2827030822439744),linewidth(3pt) + dotstyle); label("$B$", (-6.538664301434515,-2.1012417862366277), NE * labelscalefactor); dot((6.405574813756307,-2.5246514769204365),linewidth(3pt) + dotstyle); label("$C$", (6.526549011094537,-2.3431901809130897), NE * labelscalefactor); dot((2.8342164418025018,0.7508727786741206),linewidth(3pt) + dotstyle); label("$I$", (2.9578101896166937,0.9231131472191462), NE * labelscalefactor); dot((-0.055794107161586025,1.4431604291360227),linewidth(3pt) + dotstyle); label("$O$", (-0.06654474383910534,0.8021389498809152), NE * labelscalefactor); dot((6.025423155672994,1.0867892748710175),linewidth(3pt) + dotstyle); label("$E$", (6.133382869745283,1.2557921898992812), NE * labelscalefactor); dot((0.9254617064835794,3.330271069645659),linewidth(3pt) + dotstyle); label("$F$", (1.0524665815395402,3.524058389991112), NE * labelscalefactor); dot((-0.6109286318611473,2.193342219270571),linewidth(3pt) + dotstyle); label("$M_c$", (-0.39922378651924323,2.7377261072926107), NE * labelscalefactor); dot((5.921678024403379,2.0723680219323404),linewidth(3pt) + dotstyle); label("$M_b$", (6.0426522217416085,2.2538293179396867), NE * labelscalefactor); dot((17.30919271344813,-0.794679463952661),linewidth(3pt) + dotstyle); label("$T$", (17.444470320869968,-0.6193078688432986), NE * labelscalefactor); dot((-0.1270318425082193,-2.4036772795822055),linewidth(3pt) + dotstyle); label("$M_a$", (-0.6109286318611492,-3.250496660949822), NE * labelscalefactor); dot((11.517399413958824,-1.3291405072564817),linewidth(3pt) + dotstyle); label("$J$", (11.637708848634837,-1.1334482075307801), NE * labelscalefactor); dot((2.4735765925266993,-5.704923476191715),linewidth(3pt) + dotstyle); label("$X$", (2.5948875976019976,-5.518762861041653), NE * labelscalefactor); dot((6.45005905050214,5.337604464549033),linewidth(3pt) + dotstyle); label("$L$", (6.556792560429095,5.5201326460719224), NE * labelscalefactor); dot((2.273453174918753,8.658936669636306),linewidth(3pt) + dotstyle); label("$S$", (2.3831827522600917,8.846923072873274), NE * labelscalefactor); dot((3.9556789137885753,0.4822320222967007),linewidth(3pt) + dotstyle); label("$M$", (4.0768215149953395,0.6509212032081265), NE * labelscalefactor); dot((-2.8480836688231372,-5.6063742120926445),linewidth(3pt) + dotstyle); label("$Y$", (-2.7279770852802083,-5.428032213037979), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Back to the main problem. Let $N$ be the midpoint of arc $BC$ which does not include $A$. By Claim we have \[ \measuredangle XJM_a = \measuredangle XOM_a = 2\measuredangle XAN = 2\measuredangle XAI = 2\measuredangle(OI, EF). \]Also, we have $AJ$ and $XJ$ are symmetric w.r.t OI. Therefore, $AJ$ and $M_aJ$ are symmetric w.r.t $EF$. On the other hand, consider the Newton line $\ell$ of $\mathcal{Q} \{CA, AB, EM_c, FM_b\}$. The image of $\ell$ under the homothety with center $A$ and ratio 2 is $TM_a$, so the reflection of $A$ w.r.t $EF$ lies on $TM_a$. Combining the result above we have $M_a, J, T$ are collinear. Finally, \[ \measuredangle TM_aX = \measuredangle JM_aX = \measuredangle JLX = \measuredangle ALX = \measuredangle ASX = \measuredangle TSX, \]so $X, M_a, S, T$ are concyclic.

I'll just add diagram to that beautiful solution of leochang in #7 and rewrite it to my taste. Let $EFT\mapsto E'F'T'$ be homothety $(A;k=2)$. Then $AE'F'X$ - cyclic with center $I$, $X$ is a Miquel point of $BCE'F'$. Let $Y=BC\cap E'F'$ and $M'$ - midpoint of $E'F'$. Then $XYM_AM'$ - cyclic (gliding between $XYBF'$ and $XYCE'$). If $AI$ cut $(ABC)$ at $M$ then $M\in (XYM_AM')$ ($MY$ - diameter). Also $T,M',M_A$ are collinear as Newton line of $AE'T'F'$. Hence we have $$\measuredangle XST=\measuredangle XMA=\measuredangle XM_AM'=\measuredangle XM_AT$$ video of the statement

Attachments:

Let $XM_a$ intersect $\Omega$ again at $Y$. By Reim's theorem, it is equivalent to prove that $TM_a//AY$ And now we claim a general problem: "Let $J$ be a point on the angle bisector of $\angle BAC$, $K$ and $L$ be the foot of perpendicular from $J$ to $AC$ and $AB$ respectively. $M_bL, M_cK$ intersect at $T'$, let $Y'$ be on $\Omega$ s.t. $T'M_a//AY'$. If $Y'M_a$ intersect $\Omega$ again at $Z$, then $AJ=JZ$" Proof: We will prove this with the method of moving points. Let $J$ move on $AI$. Let Z' be the point on $\Omega$ s.t.$AJ=JZ'$, then by an inversion at $A$ it is easy to obtain that $J\rightarrow Z'$ is projective. Now we will prove that $J\rightarrow Z$ is projective. Since $K\rightarrow L$ is projective, by Steiner conic, $T'$ moves on some conic $\Gamma$. When $J$ goes to point of infinity, we get $M_a$ is on $\Gamma$. Hence $J\rightarrow K \rightarrow M_cK\rightarrow T'\rightarrow M_aT'\rightarrow P_{\infty}\rightarrow Y'\rightarrow Z$ is projective. Now we check three points to prove that the two projective functions are the same. When $J=A$, $Z=Z'=A$, when $J$ is on the perpendicular bisector of $AC$, $Z=Z'=C$ when $J$ is on the perpendicular bisector of $AB$, $Z=Z'=B$ Hence the general problem is proved, and this problem is the case when $J=I$.

The following solution is terrible but it explores other properties related with the configuration so I decided to post it. Let $L$ be the foot of perpendicular from $M_a$ to $\overline{EF}$, and let $S$ be the midpoint of segment $EF$. We will start by showing the following claim. Claim 1: $\overline{IL}$ bisects segment $SD$. Proof. Let $\overline{BI}$ and $\overline{CI}$ meet $\overline{EF}$ at $B'$ and $C'$ respectively. Then by Iran lemma, both of them lie on the circle with diameter $BC$. Since the center of this circle is $M_a$, it follows that $L$ is the midpoint of segment $B'C'$. Now notice that $I$ is also the incenter of $\triangle B'C'D$. Let $S'$ be the $D$-extouchpoint in that triangle. Then we are done by noticing that $\overline{IL}$ is the homothetic image of $\overline{DS'}$ centered at $S$ with ratio $\frac{1}{2}$. $\blacksquare$ Claim 2: Let $A'$ be the reflection of $A$ over $\overline{EF}$. Then $\overline{EF}$, $\overline{OI}$, $\overline{A'M_a}$ are concurrent. Proof. Let $H$ be the orthocenter of $\triangle DEF$, and let $\overline{OI}$ meet $\overline{EF}$ at $K$. Then it suffices to show that \[(\overline{M_aK},\overline{M_aA};\overline{M_aS},\overline{M_aL})=-1.\]It is known that $H$ lies on $\overline{OI}$. Using the fact that $\overline{DI}$, $\overline{EF}$ and $\overline{AM_a}$ are concurrent, \[(\overline{M_aK},\overline{M_aA};\overline{M_aS},\overline{M_aL})=(K,\overline{M_aA}\cap\overline{EF};S,L)=(H,D;P_\infty,\overline{IL}\cap\overline{DH}).\]Let $N=\overline{IL}\cap\overline{DH}$. Then by claim 1, $ISND$ is a parallelogram, and since $DH=2IS$, it follows that $N$ is the midpoint of segment $DH$ and we're done. $\blacksquare$ [asy][asy] defaultpen(fontsize(10pt)); size(11cm); pen mydash = linetype(new real[] {5,5}); pair A = dir(135); pair B = dir(220); pair C = dir(320); pair I = incenter(A,B,C); pair D = foot(I,B,C); pair E = foot(I,C,A); pair F = foot(I,A,B); pair B1 = extension(B,I,E,F); pair C1 = extension(C,I,E,F); pair Ma = midpoint(B--C); pair L = foot(Ma,E,F); pair H = orthocenter(D,E,F); pair S = extension(A,I,E,F); pair N = midpoint(H--D); pair K = extension(H,I,E,F); pair A1 = 2*S-A; draw(A--B--C--cycle, black+1); draw(D--E--F--cycle); draw(B--B1); draw(C--C1); draw(A--A1); draw(D--S); draw(D--H); draw(Ma--L); draw(K--E); draw(L--N, mydash); draw(H--K); draw(A1--K, mydash); draw(circumcircle(A,B,C)); draw(circumcircle(B,C,B1), dotted); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$I$", I, dir(270)); dot("$D$", D, dir(270)); dot("$E$", E, 2*dir(170)); dot("$F$", F, dir(180)); dot("$B'$", B1, dir(60)); dot("$C'$", C1, dir(120)); dot("$M_a$", Ma, dir(270)); dot("$L$", L, dir(90)); dot("$H$", H, dir(180)); dot("$S$", S, dir(60)); dot("$N$", N, dir(180)); dot("$K$", K, dir(0)); dot("$A'$", A1, dir(180)); [/asy][/asy] Claim 3: $T$ lies on $\overline{A'M_a}$. Proof. By Desargue's theorem, it suffices to show that $\triangle A'EF$ and $\triangle M_aM_cM_b$ are perspective axially. Let $P=\overline{M_bM_c}\cap \overline{EF}$, $Q=\overline{M_aM_b}\cap\overline{A'F}$ and $R=\overline{M_aM_c}\cap\overline{A'E}$. Since $AEA'F$ is a rhombus, it follows that $EA'\parallel M_aM_b$ and $FA'\parallel M_aM_c$. By Menelaus' theorem, we have \[1=\frac{AM_c}{M_cF}\cdot\frac{FP}{PE}\cdot\frac{EM_b}{M_bA}=\frac{ER}{RA'}\cdot\frac{FP}{PE}\cdot\frac{A'Q}{QF},\]and so by Menelaus' theorem again, $P,Q,R$ are collinear. $\blacksquare$ [asy][asy] defaultpen(fontsize(10pt)); size(10cm); pen mydash = linetype(new real[] {5,5}); pair A = dir(135); pair B = dir(220); pair C = dir(320); pair I = incenter(A,B,C); pair D = foot(I,B,C); pair E = foot(I,C,A); pair F = foot(I,A,B); pair Ma = midpoint(B--C); pair Mb = midpoint(C--A); pair Mc = midpoint(A--B); pair O = circumcenter(A,B,C); pair A1 = 2*foot(A,E,F)-A; pair T = extension(Mb,F,Mc,E); pair P = extension(Mb,Mc,E,F); pair R = extension(A1,E,Ma,Mc); pair Q = extension(A1,F,Ma,Mb); pair K = extension(O,I,E,F); draw(A--B--C--cycle, black+1); draw(Mc--T--F); draw(E--A1--F--cycle); draw(Mb--Ma--Mc--cycle); draw(A1--Q--Ma); draw(K--F); draw(A--A1); draw(P--Q, mydash); draw(K--A1, mydash); draw(circumcircle(D,E,F)); draw(circumcircle(A,B,C)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$I$", I, dir(45)); dot("$D$", D, dir(270)); dot("$E$", E, dir(60)); dot("$F$", F, dir(180)); dot("$M_a$", Ma, dir(315)); dot("$M_b$", Mb, dir(0)); dot("$M_c$", Mc, dir(180)); dot("$A'$", A1, dir(225)); dot("$T$", T, dir(0)); dot("$P$", P, dir(90)); dot("$Q$", Q, dir(270)); dot("$R$", R, dir(225)); dot("$K$", K, dir(0)); [/asy][/asy] Now let $J=\overline{AX}\cap\overline{OI}$. Let $\overline{M_aA'}$ cut $\overline{AB}$ and $\overline{AC}$ at $U$ and $V$ respectively. Since $\overline{AA'}$ is the angle bisector in $\triangle AUV$ and $\overline{EF}$ is its perpendicular bisector, $AK$ must be tangent to $(AUV)$. Now notice that $J$ lies on $(AEF)$. Therefore, \[(U,V;\overline{AJ}\cap\overline{UV},K)\stackrel{A}{=}(F,E;\overline{AJ}\cap\overline{EF},K)\stackrel{J}{=}(F,E;A,I)=-1\]so $\overline{AJ}$ is the $A$-symmedian in $\triangle AUV$. As $J$ is the foot of perpendicular from $K$ to the $A$-symmedian, it must be the case that $J$ is the $A$-Dumpty point on $\triangle AUV$. It is then well-known that $X$ lies on $(AUV)$, so $X$ is the Miquel point of quadrilateral $ABM_aV$. Consequently, $M_a,X,V,C$ are concyclic and \[\measuredangle XM_aT=\measuredangle XCA=\measuredangle XST\]and so $X,M_a,S,T$ are concyclic as desired. [asy][asy] defaultpen(fontsize(10pt)); size(12cm); pen mydash = linetype(new real[] {5,5}); pair A = dir(131); pair B = dir(218); pair C = dir(322); pair I = incenter(A,B,C); pair D = foot(I,B,C); pair E = foot(I,C,A); pair F = foot(I,A,B); pair Ma = midpoint(B--C); pair Mb = midpoint(C--A); pair Mc = midpoint(A--B); pair O = circumcenter(A,B,C); pair K = extension(E,F,O,I); pair U = extension(Ma,K,A,B); pair V = extension(Ma,K,A,C); pair A1 = 2*foot(A,E,F)-A; pair J = foot(A,I,O); pair X = 2*J-A; pair T = extension(Mb,F,Mc,E); pair S = 2*foot(O,A,T)-A; draw(A--B--C--cycle, black+1); draw(B--U); draw(circumcircle(A,B,C)); draw(F--K); draw(A--A1); draw(I--K); draw(A--X, dotted); draw(U--K); draw(A--S); draw(circumcircle(D,E,F)); draw(circumcircle(A,E,F), dotted); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$I$", I, dir(225)); dot("$D$", D, dir(270)); dot("$E$", E, dir(60)); dot("$F$", F, dir(180)); dot("$M_a$", Ma, dir(315)); dot("$O$", O, dir(315)); dot("$K$", K, dir(0)); dot("$U$", U, dir(270)); dot("$V$", V, dir(0)); dot("$A'$", A1, dir(0)); dot("$J$", J, dir(250)); dot("$X$", X, dir(X)); dot("$T$", T, dir(280)); dot("$S$",S,dir(S)); [/asy][/asy]

leochang wrote: Let $E'$, $F'$ be the reflection point of $A$ throught $E$, $F$. $M'$ be the midpoint of $\overline{E'F'}$.Then we know $TM_AM'$ is the Newton line of $\mathcal{Q}\{BF',F'C,CE',E'B\}$, that is $T$, $M'$, $M_A$ are collinear. Let $AI$ intersect $\odot(ABC)$ at $M\not= A$. Then \[ \measuredangle TSX=\measuredangle TM_AX \iff \measuredangle AMX=\measuredangle M'M_AX \]So we only need to show that $X, M_a, M, M'$ are concyclic. Notice that $X$ is the miquel point of $\mathcal{Q}\{BF',F'E',E'C,CB\}$. Thus \[ \triangle BXF' \stackrel{+}{\sim}\triangle M_AXM' \stackrel{+}{\sim}\triangle C'XE' \]$\implies X$ is the miquel point of $\mathcal{Q}\{M_AM',M'E',E'C,CM_A\}$. Let $Y:=BC\cap E'F'\implies X, Y, M_A, M'$ are concyclic. But it is obvious that $M, Y, M_A, M'$ are concyclic. \[\implies X, M_a, M, M' \text{ are concyclic}.\] What is newton line?

hydo2332 wrote: What is newton line? Newton–Gauss line (Wikipedia).

Relabel $M_a, M_b, M_c$ to $L, M, N$; reflect $A$ in $E, F, T$ to $E', F', V$; let $D, R$ be the midpoints of $E'F', AX$ and let $U := BC \cap E'F'$, $W = AI \cap (ABC)$. Clearly $V = BE' \cap CF'$ and $TDL$ is the Newton-Gauss line of $BF'CE'$; since $AR \perp RI$, $R \in (AEF)$ so $X \in (AE'F') \implies X$ is the Miquel point of $BCE'F'$. Then $\triangle XF'D \sim \triangle XBL$, so that $\angle XDU = \angle XLU \implies XDLU$ is cyclic, and $UDLW$ is cyclic as clearly $UD \perp DW, UL \perp LW$ so $XDLW$ is cyclic. Then $\angle TLX = \angle DWX = \angle TSX$.