Here is a very weird solution. Let the perpendicular bisector of $\overline{AD}$ intersect $\overline{BC}$ at $E$. Then to show the concurrency, it suffices to show that $(E,X;B,C) = -1$. Let $K$ and $N$ be the midpoints of minor and major arcs $BC$. We first claim that $S$, $K$, $X$ are collinear. Note that $T$ is the $A$-mixtilinear touch point, so $T$, $I$, $N$ are collinear. Moreover, by $\sqrt{bc}$ inversion at $A$, $\triangle AID \sim \triangle ATI_A$, so \[ \measuredangle DIK = \measuredangle I_ATA = \measuredangle STA = \measuredangle SKI, \]and hence $\overline{SK} \parallel \overline{DI}$. Meanwhile, $\overline{KM} \parallel \overline{ID}$ due to midpoints, so the collinearity is proved. Now we just need to show that $S$, $N$, $E$ are collinear, since $(N, K; B, C) = -1$ as it is a kite. Consider the $\sqrt{bc}$ inversion at $A$, and let $F$ be the image of $S$ under the inversion. Then since $S$, $T$, $I_A$ are collinear, we see that $F$, $D$, $I$, $A$ are concyclic; let this circle be $\Gamma$. Our goal now is to show that $E$, $A$, $N$, $F$ are concyclic since this would imply that \[ \measuredangle ANE = \measuredangle AFC = \measuredangle ABS = \measuredangle ANS. \] Let $I'$ be the reflection of $I$ over $N$. Then $I_AI' \parallel KN$, so $I_A$, $D$, $I'$ are collinear. Moreover, \[ \measuredangle II'D = \measuredangle INK = \measuredangle TAK = \measuredangle IAD, \]so $I'$ lies on $\Gamma$. Also, $(A,F; I,I') \stackrel{D}{=} (A, \overline{BC} \cap \overline{AI_A}; I, I_A) = -1$, so $I'I$ is the symmedian of $\triangle AIF$. Since $N$ is the midpoint of the symmedian, it follows that $A$, $N$, $F$ and the center of $\Gamma$ are concyclic. Finally, note that this implies \[ \measuredangle ANF = 2 \measuredangle ADF = 2 \measuredangle ADE = \measuredangle AEF, \]as desired.

Midbase, midbase, midbase!. Let $YZ \cap BC=W$, $N,L$ midpoints of minor and mayor arc $BC$ in $(ABC)$ respectivily, let $NT \cap BC=K$ and let $IL \cap DI_A=P$ and let $AL \cap (AID)=Q$ Claim 1: $S,X,N,M$ are colinear. Proof: Let $S_1$ the inverse of $S$ w.r.t. $(BIC)$ inversion, notice that by $\sqrt{bc}$ inversion $T$ is the A-mixtilinear intouch point and $T,I,L$ are colinear, as well was $T,K,N$ by the inversion over $(BIC)$, now notice that $KS_1NI_A$ is cyclic but becuase of what i said that $T,I,L$ are colinear using PoP and tangencies we get that $\angle AIK=90$ so $KIDI_A$ is cyclic so by reim's over $(KIDI_A), (KS_1NI_A)$ we get that $SN \parallel ID$ but by midbase $NM \parallel ID$ so $S,N,M$ are colinear and this is enough. Claim 2: $L,S,W$ are colinear. Proof: Denote $'$ the inverse of a point w.r.t. $\sqrt{bc}$ inversion, note that $(AID) \cap BC=S'$ and now by angle chase $\angle ALI=\angle ADB=\angle S'IN$ so $(ALI)$ is tangent to $IS'$ hence $\angle LIS'=90$ so $P,A,I,D,S'$ is cyclic and now note that by midbase since $N$ is midpoint of $II_A$ by I-E lemma we get that $L$ is midpoint of $IP$ but now by reims over $(ALI), (AID)$ we have that $LI \parallel QS'$ so in general as $L$ is midpoint of $IP$ we get that $AL,LS'$ are symetric w.r.t. $IP$ so $\angle ALT=\angle S'LT$ but remember that by the similarities that the inversion gives we havs $\angle W'LT=\angle W'AT=\angle WAD=\angle ADB=\angle ALT$ so $L,S',W'$ are colinear and now reflecting over $AI$ we get that $L,S,W$ are colinear as desired. Finishing: Just note that $-1=(L, N; B, C) \overset{S}{=} (W, X; B, C)$ so $AX,BY,CZ$ are concurrent as desired .

Really nice one! (I'm too lasy to translate English)

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Let $R,N$ be the midarc of $BC$ of $\odot(ABC)$ containing and not containing $A$, respectively, and $I$ be the incenter of $\triangle ABC$. Claim 1: $S,M,N$ are collinear. Proof: Not hard to see that $\sqrt{bc}$ inversion sends $B \to C, I \to I_A, D \to T$, so $\triangle AID \overset+\sim \triangle ATI_A$. Therefore, $\measuredangle ANS = \measuredangle ATS = \measuredangle ATI_A = \measuredangle AID$. Now we have $ID \parallel SN$. Note that $N,M$ are midpoints of $DI_A,II_A$, so $ID\parallel NM$ too. Thus, $NM \parallel SN$ and $M,N,S$ are collinear. Let $P$ be the midpoint of $BC$, $E$ be the reflection of $D$ across $P$ (known that $IE \perp BC$), $F$ be a point such that $\square EIFP$ is a rectangle, and $SR$ intersect $BC,ID$ at $Q,U$, respectively.. It's known that $AD$ passes through the reflection of $E$ across $E$. Hence, $A,F,D$ are colliear. Because $MN \parallel DI$, $ID \perp SR$. Since $A,R,U,I,F$ and $I,E,Q,U$ are concyclic on the circles with diameter $RI,IQ$ respectively, by radical axis, $A,F,P,Q$ are also concyclic. $\angle QAD = \angle FED = \angle FDE$. Thus, $Q$ lies on the perpendicular bisector of $AD$, so $Q,Y,Z$ are collinear. Because $(B,C;X,Q) \overset S = (B,C;N,R) = -1$, $AX,BY,CZ$ are concurent as desired. [asy][asy] size(300); import geometry; import olympiad; import graph; pair A,B,C,R,N,M,IA,Q,I,E,D,T,P,S,U,F; A = dir(115); B = dir(213); C = dir(-33); N = dir(-90); P = B/2 + C/2; I = incenter(A,B,C); IA = 2*N - I; D = foot(IA,B,C); E = foot(I,B,C); M = D/2 + IA/2; R = dir(90); pair[] SS = intersectionpoints(line(M,N),circumcircle(A,B,C)); S = SS[0]; Q = extension(R,S,B,C); U = extension(D,I,R,Q); F = I + P - E; pair[] TT = intersectionpoints(line(S,IA),circumcircle(A,B,C)); T = TT[1]; fill(A--B--C--cycle, 0.1*cyan + 0.9 * white); draw(unitcircle); draw(R--T,red); draw(M--S,green); draw(D--U,green); draw(R--Q,magenta); draw(C--A--B); draw(C--Q); draw(A--D,orange); draw(S--IA,purple); draw(I--IA,royalblue,StickIntervalMarker(2,2,royalblue)); draw(D--IA,royalblue,StickIntervalMarker(2,2,royalblue)); draw(circumcircle(A,F,I),dashed+pink); draw(circumcircle(Q,U,I),dashed+blue); dot("$A$",A,dir(A)); dot("$B$",B,dir(-110)); dot("$C$",C,dir(C)); dot("$P$",P,dir(-80)); dot("$M$",M,dir(C)); dot("$N$",N,dir(-120)); dot("$I_A$",IA,dir(-30)); dot("$I$",I,dir(-150)); dot("$U$",U,1.5*dir(125)); dot("$S$",S,dir(160)); dot("$Q$",Q,dir(-170)); dot("$R$",R,dir(90)); dot("$D$",D,dir(-45)); dot("$E$",E,dir(-70)); dot("$F$",F,dir(C)); dot("$T$",T,dir(-110)); [/asy][/asy]

We first prove that $MS$ passes through the midpoint $N_a$ of arc $\overarc{BC}$ (not containing $A$). Let $N_a^{\prime}$ be the midpoint of arc $\overarc{BAC}$. It is known that $N_a^{\prime}$, $I$, $T$ are collinear. Let $U$ be the second intersection of $N_aD$ and $\omega$. Consider the following projection: \[ -1 = (D, I_a; M, \infty_{BC_\perp}) \stackrel{N_a}{=} (U, A; \, \cdot\, , N_a^{\prime}). \] Then proving that $S$, $M$, $N_a$ is collinear is equivalent to proving that $(U, A; S, N_a^{\prime}) = -1$. Let $K = AN_a \cap BC$. Notice that by the well-known property of the inversion, we have $\triangle ADN_a \stackrel{+}{\sim} \triangle AKT$, so $\measuredangle ATK = \measuredangle AN_aD = \measuredangle AN_aU = \measuredangle ATU$, implying that $T$, $K$, $U$ are collinear. Now we have \[ (U, A; S, N_a^{\prime}) \stackrel{T}{=} (K, A; I_a, I) = -1. \]The second equality is trivial by the property of incenter and excenter. Therefore we deduce that $M$, $S$, $N_a$ are collinear. Let $V = YZ \cap BC$. Then the statement is equivalent to showing that $(B, C; X, V) = -1$, or equivalently, $N_a^{\prime}$, $S$, $V$ are collinear. Let $E$, $F$ be the tangent point of $A$-excircle at line $AB$, $AC$, respectively, and $P = EF \cap BC$, $Q = AN_a^{\prime} \cap BC$. Notice that $AV = PV$, $V$ lies on the radical axis $\ell$ of $A$-excircle and the point $A$ (circle with center $A$ and radius $0$). Moreover, $\ell$ passes through the midpoints of $AE$ and $AF$, and is also perpendicular to $AI_a$. Notice that both $EF$ and $AN_a^{\prime}$ are perpendicular to $AI_a$, we can deduce that $\ell$ is the "median line" of $EF$ and $AN_a^{\prime}$. In particular, the midpoint of $PQ$ should lie on $\ell$, which is exactly $V$. Notice that $AD$, $BE$, $CF$ are collinear by Ceva's theorem, we have $(B, C; D, P) = -1$. By, \[ -1 = (B, C; N_a, N_a^{\prime}) \stackrel{U}{=} (B, C, D, N_a^{\prime}U \cap BC), \]we get $N_a^{\prime}$, $U$, $P$ are collinear. Now we have the following projection \[ -1 = (P, Q; V, \infty_{BC}) \stackrel{N_a^{\prime}}{=} (U, A; \, \cdot \,, N_a^{\prime}) \]But we have already proved that $(U, A; S, N_a^{\prime}) = -1$, so $N_a^{\prime}$, $S$, $V$ are collinear, as desired.

If $YZ \cap BC = Q$ then by Ceva it suffices to show that $Q, B, X, C$ are harmonic. Let $M_A, N_A$ be the midpoints of minor and major $\widehat{BC}$; then by Brocard on $B, C, M_AQ \cap (ABC), N_AQ \cap (ABC)$, $Q, B, C$, and the orthocenter of $\triangle{QM_AN_A}$ are harmonic, so it suffices to show that $M_AX \perp N_AQ$. Also, we claim that $M_AX \parallel ID$; by force-overlay inversion at $A$ swapping $B$ and $C$, $S^* = (IAD) \cap BC$, so $\angle{IAS} = \angle{IDQ}$. The angle between $SM_A, BC$ is equal to $\angle{IAS}$ by angle chasing, and $M_AM \parallel ID$, so $S, X, M_A, M$ are collinear, and it suffices to show that $M_AM \perp N_AQ$. Thus, it suffices to show that $\triangle{II_AD} \sim \triangle{N_AWQ}$ if $W = AN_A \cap BC$; note that we already know $\angle{II_AD} = \angle{N_AWQ}$. Furthermore, $Q$ lies on the radical axis of point $A$ and the $A$-excircle, so if $T_B, T_C$ are the tangency points of the $A$-excircle on $AB, AC$, and $T_BT_C \cap BC = V$, then $Q$ is the midpoint of $WV$, and it suffices to show that $\triangle{N_AWV} \sim \triangle{M_AI_AD}$. Because we already know $N_AW \perp M_AI_A$ and $WV \perp I_AD$, it suffices to show that $M_AD \perp N_AV$. Note that this is equivalent to $B, D, C, V$ being harmonic. If $I_AB, I_AC \cap T_BT_C = B', C'$ respectively, then $I_ADV = 90^{\circ}$, so projecting from $I_A$ to $T_BT_C$ gives that it suffices to show $DI_A$ bisects $\angle{B'DC'}$, which is obvious since $\angle{AT_BT_C} = \angle{BDB'} = \angle{CDC'} = \angle{AT_CT_B}$. Thus, $\triangle{N_AWV} \sim \triangle{M_AI_AD} \implies M_AM \perp N_AQ$, and we are done. $\square$

probably the longest solution to this problem It is well-known that $T$ is the $A$-mixtilinear touchpoint. Let $Sd$ be the $A$-Sharkydevil point and $Sd'$ be its reflection across the perpendicular bisector of $BC$, $I$ be the incenter, and $D'$ be the $A$-intouch point. Also let $F$ be the midpoint of arc $BAC$ and $E$ be its antipode. Let $K = AE \cap BC$. Then $SdD'$ and $Sd'D$ pass through $E$ and $TI$ passes through $F$. First, it is well-known that $T$ is the Miquel point of $ABD'C$, so \[ \frac{CD}{DB} = \frac{BD'}{D'C} = \frac{BD'/D'T}{D'C/D'T} = \frac{AC/CT}{BA/BT}=\frac{KC \cdot BT}{KB \cdot CT} \]implies $(B,C;K,D)=(B,C;T,E)$, so $Sd'T$ passes through $K$. Now $(A,Sd';F,S)=(A,K;I,I_A)=-1$, so projecting from $E$ onto $DI_A$ we find that $SE$ passes through $M$. Let $X'$ be the point on $BC$ such that $(B,C;X,X')=-1$; as $SX$ passes through $E$, we get that $SX'$ passes through $F$. Extend $SF$ to meet the tangents to the circumcircle from $A$ and $Sd'$ at point $Q$ (as $(A,Sd';F,S)=-1$). We claim that $SFQ$ is the radical axis of the point-circle $A$ and the circle $\omega$ through $D,Sd'$ and tangent to $BC$ and the circumcircle (this is known to exist as $E$ is the midpoint of arc $BC$). Firstly, $QA^2 = QSd'^2$, so $Q$ lies on the radical axis; for point $F$, we prove that it instead lies on the radical axis of $(A)$ and the reflection $\omega'$ of $\omega$ across the perpendicular bisector of $BC$ (which passes through $Sd$ and $D'$). Letting the $B$ and $C$-intouch points be $D_B,D_C$ and $P$ be the projection of $E$ on $D_BD_C$, it is well-known that $Sd$ and $P$ are swapped under inversion w.r.t. the incircle. Also let $J$ be the projection of $A$ on $ID$; then the image $J'$ os $J$ under inversion w.r.t. the incircle is well-known to be $D_BD_C \cap ID$. So \[ \measuredangle IJSd = \measuredangle S'PI = 90^\circ - \measuredangle IPD = 90^\circ - \measuredangle SdDI \]implies that $\angle DSdJ = 90^\circ$, so $J$ lies on $\omega'$ and $SJ$ passes through $F$. Finally, \[ \measuredangle JAF = \measuredangle (BC,AF) = \measuredangle (EF,AE) = \measuredangle FSdA \]since $BC \perp EF$ and $AF \perp AE$, which implies that $FA^2=FSd \cdot FJ$, so $F$ lies on the desired radical axis. Since $X'$ lies on $BC$, from the radical axis we obtain $X'A^2 = X'D^2$, so $X'$ lies on the perpendicular bisector of $AD$, so $X',Y,Z$ collinear. By harmonic bundle properties, we get that $AX,BY,CZ$ concurrent.

Let $P,N$ be the midpoints of the arcs $BAC,BC$ respectively. $AN\cap BC=E, \ ND\cap (ABC)=L, \ YZ\cap BC=K$. The problem asks for $(K,X;B,C)\overset{?}{=}-1$. Note that $T$ is the tangency point of $A-$mixtilinear with $(ABC)$. Claim: $T,E,L$ are collinear. Proof: $NT\cap BC=T'$. Take the inversion centered at $N$ with radius $NB$. \[\measuredangle ATC=\measuredangle B=\measuredangle ABD \ \ \text{and} \ \ \measuredangle CAT=\measuredangle DAB\]Hence we get that $ABD\sim ATC$. Thus, $\measuredangle T'NA=\measuredangle TNA=\measuredangle TCA=\measuredangle BDA=\measuredangle T'DA$ which gives that $T',A,D,N$ are concylic. This yields $T,E,L$ are collinear.$\square$ Claim: $S,M,N$ are collinear. Proof: \[(S,P;L,A)\overset{T}{=}(I_A,I;N,A)=-1=(NM,NP;ND,NI_A)\overset{(ABC)}{=}(NM\cap (ABC),P;L,A)\]Thus, $S=NM\cap (ABC)$ which is the desired result.$\square$ Claim:$K,S,P$ are collinear. Proof: Take $\sqrt{bc}$ inversion and reflect the diagram over $AI$. $P^*$ is the intersection of $AP$ and $BC$. $D$ and $T$ swaps. $K^*$ is on $(ABC)$ such that $TA=TK^*$. $S^*$ is the intersection of $(IAD)$ with $BC$ for second time. Let $R$ be the midpoint of $BC$. \[\measuredangle RS^*I= \measuredangle DS^*I=\measuredangle DAI=\measuredangle NAT=\measuredangle NPT=\measuredangle RPI\]So $R,S^*,I,P$ are concyclic which is equavilent to $\measuredangle PIS^*=90$. \[90-\measuredangle S^*PI=\measuredangle IS^*P=\measuredangle NIT=90-\measuredangle TNI=90-\measuredangle TK^*A=90-\measuredangle K^*AT=90-\measuredangle K^*PI\]Thus, $P,K^*,S^*$ are collinear. We get $PA.PP^*=PB^2=PK^*.PS^*$ hence $A,P^*,K^*,S^*$ are concyclic.$\square$ \[(K,X;B,C)\overset{S}{=}(P,N;B,C)=-1\]As desired.$\blacksquare$

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