Lol, I think this was an exercise in class haha.

We are being asked to show that $X_B,X_C,B,C$ are concylic.Since $AX_C,AX_B$ both pass through $A$,it implies $Q,X_B,X_C$ are collinear.Incenter $I$ lies on $BX_B$,it is obvious that $\sphericalangle X_BBX_C=90^\circ$,so $X_BX_CBC$ is cyclic with diameter $X_BX_C$.

The problem is equivalent to proving 2 things, first that X_C, X_B, C, and B are cyclic, and then showing that Q is the center of the circle which passes through them. We will define X_A similarly to how X_B and X_C are defined in the problem. Then we see that X_A, X_B, and X_C are the vertices of a triangle that has ABC as its orthic triangle, meaning that A, B, and C are the feet of the altitudes from X_A, X_B, and X_C respectively. Since the circumcircle of triangle ABC is the circumcircle of the orthic triangle of triangle X_A X_B X_C, it is the 9-point circle, so it also passes through the midpoint of side X_C X_B of triangle X_A X_B X_C (this is a well known property of the 9-point circle). This implies the desired result, i.e. not only are X_C, X_B, C, and B cyclic, but the center of their circumcircle is the point Q, so we are done.

[asy][asy] size(7cm); defaultpen(fontsize(12pt)); pair A,B,C,Xa,Xb,Xc,I,Q; Xa=dir(120);Xb=dir(210);Xc=dir(330);Q=(Xb+Xc)/2;I=extension(Xa,A,Xb,B); A=foot(Xa,Xb,Xc);B=foot(Xb,Xc,Xa);C=foot(Xc,Xa,Xb);I=incenter(A,B,C); filldraw(Xa--Xb--Xc--cycle,white+white+white+white+yellow+white+white+yellow+5*white,red); draw(circumcircle(A,B,C),blue); dot("$X_A$",Xa,dir(Xa));dot("$X_B$",Xb,dir(Xb));dot("$X_C$",Xc,dir(Xc)); dot("$A$",A,dir(A));dot("$B$",B,dir(B+7));dot("$C$",C,dir(C));dot("$Q$",Q,dir(Q));dot("$I$",I,dir(-50)); draw(Xa--A,heavygreen);draw(Xb--B,heavygreen);draw(Xc--C,heavygreen); [/asy][/asy] It is well-known that $(ABC)$ is the nine-point circle of $\triangle X_AX_BX_C$, so $Q$ is the midpoint of $\overline{X_BX_C}$, done!

Note that $(ABC)$ is the nine-point circle of $X_AX_BX_C$, so $Q$ is the midpoint of $X_BX_C.$ Furthermore, $$\angle X_CBX_B=\angle X_CCX_B=90,$$so $X_BX_CBC$ is cyclic with the midpoint of $X_BX_C$, which we have established is $Q$, as its circumcenter, so we are done.

Note that the circumcircle of $\triangle ABC,$ is the nine point circle of $X_A X_B X_C,$ so it follows that $Q$ is the midpoint of $X_B X_C $ so $X_C BC X_B,$ is cyclic with circumcenter $Q \blacksquare$