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 $\angle ABF$ .