Clearly $AA'$ is the radical axis of $\gamma_B,\gamma_C,$ thus it follows that $BCNM$ and $BCM'N'$ are cyclic $\Longrightarrow$ $\angle BM'N'=\angle BCC'=\angle BB'C$ $\Longrightarrow$ $MN,M'N'$ are antiparallel to $BC$ WRT $AB,AC,$ i.e. $MN \parallel M'N' \parallel B'C'.$ But if $\gamma_B$ cuts $BC$ again at $Y,$ we have $(YM' \parallel CA) \perp BB'$ $\Longrightarrow$ $\angle AMM'=\angle BYM'=\angle BCA=\angle AC'B'$ $\Longrightarrow$ $MM' \parallel B'C'.$ Consequently $M,M',N,N'$ are collinear on a parallel to $B'C'.$

[asy][asy] import graph;import geometry;size(8.5cm); pen label=fontsize(10);defaultpen(label); pair A=(7,6),B=(0,0),C=(10,0),H,X,Xd,M,N,Md,Nd,Ad,Bd,Cd; H=orthocenter(A,B,C); Ad=extension(A,H,B,C); Bd=extension(B,H,C,A); Cd=extension(C,H,A,B); X=A+(1.4)*(Ad-A); Xd=2*Ad-X; path gammab=circumcircle(B,X,Xd);path gammac=circumcircle(C,X,Xd); pair [] x=intersectionpoints(A--B,gammab); if (x[0]==B) { M=x[1]; } else { M=x[0]; } pair [] y=intersectionpoints(A--C,gammac); if (y[0]==C) { N=y[1]; } else { N=y[0]; } pair [] z=intersectionpoints(Bd--B,gammab); if (z[0]==B) { Md=z[1]; } else { Md=z[0]; } pair [] w=intersectionpoints(Cd--C,gammac); if (w[0]==B) { Nd=w[1]; } else { Nd=w[0]; } draw(A--B--C--cycle,blue);draw(X--H--C^^B--Bd^^H--Cd^^H--A,heavymagenta);draw(gammab^^gammac,orange);draw(M--N,purple+dashed);draw(circumcircle(B,M,C),darkgreen+dotted);draw(circumcircle(B,Md,Nd),darkgreen+dotted);draw(circumcircle(A,M,H),darkgreen+dotted); dot(A);dot(H);dot(B);dot(C);dot(M);dot(N);dot(Md);dot(Nd);dot(Xd);dot(X);dot(H);dot(Ad);dot(Bd);dot(Cd);label("$N$",N,NE);label("$M$",M,NW);label("$M'$",Md,SSW);label("$N'$",Nd,SSW);label("$A$",A,NE);label("$B$",B,W);label("$C$",C,ESE);label("$H$",H,NNE);label("$X$",X,S);label("$X'$",Xd,SSW);label("$A'$",Ad,SW);label("$B'$",Bd,NNE);label("$C'$",Cd,NNW);dot(circumcenter(B,X,Xd));dot(circumcenter(C,X,Xd));label("$\gamma _B$",2*(circumcenter(B,X,Xd))-Xd,NE);label("$\gamma _C$",2*(circumcenter(C,X,Xd))-Xd,NE); [/asy][/asy] Suppose a reflection about $BC$ takes $X$ to $X'$; clearly $X'\in AA'$. Now since the centers of $\gamma_B$ and $\gamma_C$ are on $BC$, this reflection keeps them fixed. Thus they pass through $X'$ as well. So by power of point, $$AM\cdot AB=AX\cdot AX'=AN\cdot AC\implies BMNC\text{ is cyclic.}$$On the other hand, $$HM'\cdot HB=HX\cdot HX'=HN'\cdot HC\implies BM'N'C\text{ is cyclic.}$$Now we have \begin{align*} BM'\cdot BH=&(BH-HM')\cdot HB\\ =& BH^2-HM'\cdot HB=BH^2-HX'\cdot HX\\ =& BH^2-(HA-X'A')(HA+XA')=BH^2-HA^2+XA'^2\qquad\text{[Since }X'A'=XA']\\ =& BA'^2+XA'^2\\ =& BA^2-AA'^2+XA'^2=BA^2-(AA'+XA')(AA'-X'A')\\ =& BA^2-AX\cdot AX'\\ =& BA^2-AM\cdot AB=BA\cdot(BA-AM)\\ =&BA\cdot BM.\end{align*}Therefore $BM'\cdot BH=BM\cdot BA\implies AMM'H$ is cyclic. Therefore $$\angle MM'B=\angle BAH=\angle HCB=\angle HM'N'.$$This implies $M,M',N'$ are collinear. Similarly, we can prove $N,M',N'$ are collinear, so that $M,M',N,N'$ are collinear, as desired. $\blacksquare$

This configuration is like 2013 G1

Luis González wrote: $\angle BM'N'=\angle BCC'=\angle BB'C$ I'm sorry but maybe it is a typo.