Let $r$ be the radius of $(AB_1C_1)$ By Power of point.We get $2BM^2=AC.CB_1=AB.BC_1=OB^2-r^2=OC^2-r^2 \implies OB=OC$

Let $\measuredangle BAC=\alpha$, since $O$ is circumcenter of $AB_1C_1$ we get $\measuredangle B_1OC_1=2\measuredangle BAC=2\alpha$, also that $\measuredangle B_1MC_1$ $=$ $180^{\circ}$ $-$ $\measuredangle C_1MB$ $-$ $\measuredangle B_1MC$ $=$ $180^{\circ}$ $-$ $2\alpha$ $\Longrightarrow$ $\measuredangle B_1OC_1+\measuredangle B_1MC_1=180^{\circ}$ $\Longrightarrow$ $MB_1OC_1$ is cyclic $\Longrightarrow$ $\measuredangle OMB_1$ $=$ $\measuredangle OC_1B_1$ $=$ $90^{\circ}-\alpha$ $\Longrightarrow$ $\measuredangle OMB_1$ $+$ $\measuredangle B_1MC=90^{\circ}$ $\Longrightarrow$ $OM\perp BC$, hence $OB=OC$.

Basically the same idea as Monster. We just need to prove that $OM\perp BC$. Denothe P,Q the midpoints of $AC_1,AB_1$. By Carnot's Lemma, we just need to prove that $AP^2-PB^2=AQ^2-QB^2$, which is equivalent with $BA*BC_1=CA*CB_1=BM*BC=MC*BC$, which is true.

It is enough to show that $OM\perp BC $. Observe that $\angle BMC_1 = \angle CMB_1 = \angle BAC $. So, $\angle B_1MC_1 = 180 - 2\angle BAC $. But, $\angle B_1OC_1 = 2\angle BAC $. Thus, $\angle B_1OC_1 + \angle B_1MC_1 = 180$. This implies that $MB_1OC_1$ is cyclic. Since, $OC_1 = OB_1$, so, $OM $ bisects $\angle B_1MC_1$. As, $\angle BMC_1 = \angle CMB_1$, so, it follows that $OM\perp BC $. This proves the result. P.S. - The case when $B_1$ or $C_1$ does not lie on segment $AC $ or $AB $ can be dealt in a similar manner with some similar angle chase.

MY SOLUTION Let $Pow_{\omega}(X)$ denote the power of $X$ with respect to $\omega$ Let $\omega=\odot (AC_1B_1)$ By Power of a point, $(BC_1)(BA)=(BM)(BC)=Pow_{\omega}(B)$ $(CB_1)(CA)=(CM)(CB)=Pow_{\omega}(C)=(BM)(BC)$ $\implies$$ Pow_{\omega}(B)=Pow_{\omega}(C)$ $\implies OB=OC$,and we are done $\blacksquare$

$\mathbf {PROOF:}$ Join $MC_1$ and $MB_1$.Denote the midpoint of $B_1C_1$ as point $M_1$.Note that $\angle C_1MB=\angle B_1MC$.(cyclic quads)This implies that $\Delta MBC_1$ is similar to$\Delta MB_1C$.Thus,$\frac{BM}{B_1M}$=$\frac{MC}{C_1M}$.Hence,$B_1M$=$C_1M$.This implies that $M$ lies on the perpendicular bisector of $B_1C_1$.This also implies that $\Delta MC_1B_1$ is isosceles with $\angle MC_1B_1$=$\angle MB_1C_1$=$\angle A$.It immediately follows that $B_1C_1 \parallel BC$.From this, it follows that $O$ lies on the perpendicular bisector of $BC$,thus proving the claim.$\blacksquare$

Let W be circumcircle of AC1B1. We need to prove that " power of C w.r.t W = power of B w.r.t W ". power of C w.r.t W = CB1.CA = CM.CB power of B w.r.t W = BC1.BA = BM.BC BM.BC = CM.CB so we're Done