easy to see that M is on circumcircle of ABC and MO_1O_2N lie on circle so MN goes from midpoint of arc. BC

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(13cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(11); defaultpen(dps); /* default pen style */ real xmin = -2.819669650405851, xmax = 4.771315809166914, ymin = -2.3201967697763464, ymax = 1.6405169252728058; /* image dimensions */ pen qqwwzz = rgb(0.,0.4,0.6); pen zzttqq = rgb(0.6,0.2,0.); pen xfqqff = rgb(0.4980392156862745,0.,1.); pair A = (-0.5617864724056979,0.8272822731220354), B = (-0.8620870041921642,-0.5067602956062949), C = (0.869910791720445,-0.49320909809968894), D = (-0.27805176545419663,-0.5021907881336911), O_1 = (-0.5750291752977188,0.1294423215439413), O_2 = (0.2895790163424109,0.3139660693596121), M = (-0.19049122214793984,0.981688898931116); draw(A--B--C--cycle, linewidth(2.) + zzttqq); draw(A--O_1--O_2--cycle, linewidth(2.) + xfqqff); /* draw figures */ draw(circle((0.,0.), 1.), linewidth(1.2)); draw(A--B, linewidth(1.2) + zzttqq); draw(B--C, linewidth(1.2) + zzttqq); draw(C--A, linewidth(1.2) + zzttqq); draw(A--D, linewidth(1.2)); draw(M--B, linewidth(1.2)); draw(M--C, linewidth(1.2)); draw(A--O_1, linewidth(1.2) + xfqqff); draw(O_1--O_2, linewidth(2.) + xfqqff); draw(O_2--A, linewidth(1.2) + xfqqff); draw(D--O_1, linewidth(1.2)); draw(D--O_2, linewidth(1.2)); draw(M--(0.007823787528280793,-0.9999693937059835), linewidth(1.2) + red); draw(circle((-0.19599249721428083,0.4712950815452703), 0.5104234642461604), linewidth(1.2)); /* dots and labels */ dot(A,qqwwzz); label("$A$", (-0.5662392630719768,0.8878436390757479), NE * labelscalefactor,qqwwzz); dot(B,qqwwzz); label("$B$", (-0.9609337911997022,-0.5348924506869835), NE * labelscalefactor,qqwwzz); dot(C,qqwwzz); label("$C$", (0.9069810105210449,-0.5119450944004879), NE * labelscalefactor,qqwwzz); dot(D,qqwwzz); label("$D$", (-0.2679236313475332,-0.585376634517274), NE * labelscalefactor,qqwwzz); dot(O_1,qqwwzz); label("$O_1$", (-0.6993339295336517,0.11222299659219435), NE * labelscalefactor,qqwwzz); dot(O_2,qqwwzz); label("$O_2$", (0.3516549883878497,0.3462860307144502), NE * labelscalefactor,qqwwzz); dot(M,qqwwzz); label("$M$", (-0.17154473494425143,1.025527776794722), NE * labelscalefactor,qqwwzz); dot((-0.08945766174102598,-0.027886690641715173),qqwwzz); label("$N$", (-0.0705763672836705,0.01584410018891255), NE * labelscalefactor,qqwwzz); dot((0.007823787528280793,-0.9999693937059835),qqwwzz); label("$M'$", (0.025802529119611287,-0.9525343351012047), NE * labelscalefactor,qqwwzz); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] easy to see that $\triangle AO_1B \sim \triangle AO_2C$ which is telling that $A$ is the miquel point of $DO_1O_2C$, So $AMO_2O_1$ and $AMBC$ are cyclic then it is obvious that $\angle O_1NO_2=180-\angle O_1AO_2=120^{\circ}$ so $N$ is on the circle $AMO_1O_2$ and since $NO_1=NO_2$ we conclude that $MN$ is the bisector of arc $BC$ so it always pass through the midpoint of the arc $BC$.

sororak wrote: In triangle $ABC$, $\angle A=60^{\circ}$. The point $D$ changes on the segment $BC$. Let $O_1,O_2$ be the circumcenters of the triangles $\Delta ABD,\Delta ACD$, respectively. Let $M$ be the meet point of $BO_1,CO_2$ and let $N$ be the circumcenter of $\Delta DO_1O_2$. Prove that, by changing $D$ on $BC$, the line $MN$ passes through a constant point. Observe $ABCM$ is Concyclic and as $\angle O_1DO_2=180^\circ -(180^\circ -\angle BAD -\angle DAC )\implies \angle O_1DO_2=60^\circ$ hence $\angle O_1NO_2=120^\circ $ so $MO_1NO_2$ Is also Concyclic where $O_1N=O_2N \implies NM$ bisects $\angle O_1MO_2=60^\circ$ Hence $MN$ Passes from Midpoint of arc $BC$.