Proof- Apply Pascal to $A_1MC_1C_2B_2A_2$ to get that $XZ\parallel A_1A_2$, similarly $XY\parallel A_1A_2$,hence $X,Y,Z$ are collinear. Hence proved.

We of course set $(A_1B_1C_1) $ as the unit circle, but moreover, by a suitable rotation we let $A_1A_2$ ,$B_1B_2$, $C_1C_2$ lie perpendicular to the real axis. Thus we obtain $a_2=\overline{a_1}$ and so on. Since $X$ is the intersction of $A_1M$ and $B_2C_2$, we get $ x=\frac{ma_1(\overline{b_1+c_1})-\overline{b_1c_1}(m+a_1)}{ma_1 -\overline{b_1c_1}}=\frac{ma_1(b_1+c_1)-(m+a_1)}{ma_1b_1c_1-1}$ Similar for $Y$ and $Z$ . Thus $X,Y,Z $ are collinear if and only if $$ \begin{vmatrix} x & \overline{x} & 1 \\ y & \overline{y} & 1 \\ z & \overline{z} & 1 \end{vmatrix}=0$$ Reducing row$ 1$ by row$2 $ and row $2$ by row$ 3$ and ignoring $ma_1b_1c_1-1$ we must prove $(a_1-b_1)(c_1-b_1)-(b_1-a_1)(b_1-c_1)=0$,which clearly is , hence the coclusion .

[asy][asy] import graph; size(13.598446204086947cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(7); defaultpen(dps); /* default pen style */ pen dotstyle = black; real xmin = -2.866936780339897, xmax = 3.9322863217035766, ymin = -1.33150104637219, ymax = 2.2160984319007917; pair A = (-0.6,0.8), B = (0.6,0.8), F = (-0.9453531412929731,0.3260482146056132), M = (-0.2554424727087989,0.9668242565929004), C = (0.9453531412929733,0.32604821460561295), D = (0.6785206970400812,-0.7345812846024886), X = (1.2954713630384196,1.7177294819763782), Z = (-0.06871941244821182,1.7177294819763782), Y = (0.5530420712528862,1.7177294819763782); draw(circle((0.,0.), 1.), linewidth(0.8)); draw(A--B, linewidth(0.8)); draw(F--C, linewidth(0.8)); draw((-0.6785206970400813,-0.7345812846024884)--D, linewidth(0.8)); draw(Z--(-0.6785206970400813,-0.7345812846024884), linewidth(0.8)); draw(C--Z, linewidth(0.8)); draw(Z--X, linewidth(0.8)); draw(X--D, linewidth(0.8)); draw(A--X, linewidth(0.8)); draw(Y--D, linewidth(0.8)); draw(Y--F, linewidth(0.8)); draw(circle((0.24216132940233717,1.241756388724116), 0.5685043721500076), linewidth(0.8)); draw(circle((0.6133759752951039,1.1494486881627706), 0.887804696201889), linewidth(0.8)); draw(B--M, linewidth(0.8)); draw(C--M, linewidth(0.8)); dot(A,dotstyle); label("$A$", (-0.5854562352164824,0.8430993290697653), NE * labelscalefactor); dot(B,dotstyle); label("$B$", (0.6148902858214224,0.8430993290697653), NE * labelscalefactor); dot(F,dotstyle); label("$F$", (-0.930761398802729,0.3662493412601871), NE * labelscalefactor); dot((-0.6785206970400813,-0.7345812846024884),dotstyle); label("$E$", (-0.6635609745990858,-0.6943308040404262), NE * labelscalefactor); dot(M,dotstyle); label("$M$", (-0.3675851200965887,0.9664226017791389), NE * labelscalefactor); dot(C,linewidth(4.pt) + dotstyle); label("$C$", (0.9601954494076689,0.3580277897462289), NE * labelscalefactor); dot(D,linewidth(4.pt) + dotstyle); label("$D$", (0.6929950252040257,-0.7025523555543844), NE * labelscalefactor); dot(X,linewidth(4.pt) + dotstyle); label("$X$", (1.3137221645078736,1.751580771362151), NE * labelscalefactor); dot(Z,linewidth(4.pt) + dotstyle); label("$Z$", (-0.05105538680919605,1.751580771362151), NE * labelscalefactor); dot(Y,linewidth(4.pt) + dotstyle); label("$Y$", (0.569671752494652,1.751580771362151), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] \[\angle YBZ=\angle CBD=\angle FME= \angle ZMY \quad \text{so TYMB is cyclic}\]so we get that $ZT \mid \mid AB$ by the same reasoning $ZX \mid \mid AB$ so $X,Y,Z $ are collinear.