Let $TB,TC$ cut $\odot O_2$ again at $B',C'.$ Since $T$ is the exsimilicenter of $\odot O_1 \sim \odot O_2,$ then $B'C' \parallel BCF$ $\Longrightarrow$ $F$ is midpoint of the arc $B'TC'$ of $\odot O_2$ $\Longrightarrow$ $TF \equiv TM$ bisects $\angle B'TC' \equiv \angle BTC$ externally $\Longrightarrow$ $M$ is midpoint of the arc $BTC$ of $\odot O_1$ $\Longrightarrow$ $AM$ is external bisector of $\angle BAC$ and $MB=MC.$ Note that $\odot O_2$ is a Thebault circle of the cevian $AD$ of $\triangle ABC$ externally tangent to its circumcircle $\odot O_1.$ By Sawayama's lemma $EF$ passes through its C-excenter $\Longrightarrow$ $N$ is C-excenter of $\triangle ABC$ $\Longrightarrow$ $N \in AM.$ $\angle BNA=90^{\circ}-\frac{_1}{^2}\angle ACB=90^{\circ}-\frac{_1}{^2}\angle BMN$ $\Longrightarrow$ $\triangle BMN$ is M-isosceles, i.e. $MB=MN.$ Hence $MB=MC=MN$ $\Longrightarrow$ $M$ is circumcenter of $\triangle BCN.$

s372102 wrote: As shown in the figure, $\odot O_1$ and $\odot O_2$ touches each other externally at a point $T$, quadrilateral $ABCD$ is inscribed in $\odot O_1$, and the lines $DA$, $CB$ are tangent to $\odot O_2$ at points $E$ and $F$ respectively. Line $BN$ bisects $\angle ABF$ and meets segment $EF$ at $N$. Line $FT$ meets the arc $\widehat{AT}$ (not passing through the point $B$) at another point $M$ different from $A$. Prove that $M$ is the circumcenter of $\triangle BCN$. As shown in the figure. Let $N$ be the intersection of $AM$ and $EF$. It's fairly obvious $\angle FET = \angle TFK =\angle TBM = \angle TCM = \angle TAM$ So we have $ \angle CBM = \angle CTM = \angle BTF = \angle BCM $ $MB^2=MB\times MF$, and $A,E,N,T$ are concyclic. So $MC=MB$, and $ \angle MNT = \angle TEA = \angle MFN$ Thus $MN^2= MT\times MF $, Consequently $MB=MC=MN$. $ \angle ABN = \angle ABM + \angle MBN$ $= \angle ABM + \frac{180- \angle BMN}{2}$ $=\angle ABM +90 - \frac{\angle BCA}{2}$ $= \angle ABM + 90-\frac { \angle BCM}{2} -\frac{ \angle ACM }{2}$ $=\frac{\angle CMB}{2} +\frac{\angle BCM}{2}+ \frac{\angle ACM}{2}$ $= \frac{\angle CAB}{2} +\frac{\angle BCA}{2}$ $=\frac{\angle ABF}{2}$ Namely $AN$ is the bisector of $ \angle ABF $.

Let AM meet EF at N'. we will prove N' is N. Claim1 : M is center of BCN'. Proof : Let BT meet $\odot O_2$ at B' and CT meet at C'. B'C' || BC so ∠C'B'F = ∠B'FB = ∠FC'B' so F is midpoint of arc C'B'T Note that FT meets $\odot O_1$ at M so M is midpoint of arc CBT so MB = MC. ∠N'ET = ∠TFB = ∠TAM so ATN'E is cyclic. ∠TN'M = ∠TEA = ∠TFN' so MN'^2 = MT.MF = MB^2 so MC = MB = MN'. Claim2 : BN' is angle bisector of ABF. Proof : ∠ABN' = ∠ABM + 90 - ∠BMN'/2 = ∠ABM + 90 - ∠BCA/2 = ∠ABM + 90 - ∠BCM/2 - ∠ABM/2 = ∠ABM/2 + 90 - ∠CBM/2 = ∠ABM/2 + 90 - ∠CBA/2 - ∠ABM/2 = 90 - ∠CBA/2. we're Done.

As shown in the figure, $\odot O_1$ and $\odot O_2$ touches each other externally at a point $T$, quadrilateral $ABCD$ is inscribed in $\odot O_1$, and the lines $DA$, $CB$ are tangent to $\odot O_2$ at points $E$ and $F$ respectively. Line $BN$ bisects $\angle ABF$ and meets segment $EF$ at $N$. Line $FT$ meets the arc $\widehat{AT}$ (not passing through the point $B$) at another point $M$ different from $A$. Prove that $M$ is the circumcenter of $\triangle BCN$.