This reminds me with ISL 2011 G4

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Let $BE$ meet $(ABC)$ again at $D$, let $F$ be the intersection point of the perpendicular bisector of $AC$ with $MN$, and let $O$ be the circumcenter of triangle $ABC$. We will preform $\sqrt{\tfrac{ac}{2}}$-inversion about $B$. Notice that $A$ and $M$ map to each other and $B$ and $N$ map to each other. Also $H$ and $O$ map to each other and $D$ and $E$ map to each other. The point $T$ is mapped to the point $T'$ on line $MN$ such that $AT'C$ is tangent to $MN$ thus clearly $T'=F$. As $F$, $O$, and $D$ are collinear, we are done.

Attachments:

Let $B$ symmetric about $A$ at $B_1$ and about $C$ at $B_2$. $BE \cap (ABC)=F$ Inversion Around B and $r^2=AB*BC$ $A=>C , N=>B_2, M=>B_1, (ABC)=>AC, (MNT)=>(B_1B_2T’)$ and AC tanget $(B_1B_2T’)$ at T’ but by mideline $AC||B_1B_2$ so T lies on perpendicular bisector of $B_1B_2$. See that $H=>H’$ and $H’$ is antipode of B and $E=>E’$ where $E’$ is symmetric to $B$ about $F$. We have to show that $E’H’$ lies on perpendicular bisector of $B_1B_2$ and by inversion $EHTB$ will be cycle. Homothety centre A and coefficient $\frac{1}{2}$ $B_1B_2=>AB , H’=>O$ (O is circumcenter) $E’=>F$ but O and F lies on perpendicular bisector of AB so we are finish.

$T$ is the $B$-Why Point of $(ABC)$. Hence from 2011 G4, $T,H,D$ are collinear where $D$ is the point on $(ABC)$ such $BD\parallel AC$. Now, $\measuredangle HTB=\measuredangle DCB=\angle C-\angle A$. However, $\measuredangle BEH=2\measuredangle BEN=\angle B+2\angle A$ Hence, $\measuredangle HTB+\measuredangle BEH=\angle C-\angle A+2\angle +\angle B=180^\circ$. Thus $BEHT$ is cyclic.