Since $BC$ is perpendicular bisector of $OT$ we get $HO\cap AT=N_9$. And we know that circumcenter and $D- \text{excenter}$ of $DEF$ are $N_9$ and $O$,respectively. Let $D- \text{excenter}$ of $DEF$ intersects with $DE$ and $DF$ at $S$ and $K$. $MN\perp AT \iff MN\perp AN_9$. So it's enough to prove that $MK^2-MF\cdot MD=NS^2-ND\cdot NE$. If sides of $DEF$ are $x,y,z$, then all of $\{MK,MF,MD,NS,ND,NE\}$ can be written as a value of $x,y,z$. So calulating them gives that it's true. So $AT\perp MN$.

leibnitz wrote: $MN$ seems to be the radical axis of $N_9$ and $\odot BHC$. Can someone prove this? If this is proved then we can finish since as above $HO\cap AT=N_9$ Yes it can be proved like this: $Pow(M,(HBC))=MH\cdot MB=MF\cdot MD=Pow(M,(DEF))$. Do same thing for $N$ and get the reuslt. Very short and nice!

Oh! I didn't see that. Thank you for completing the solution I believe there should also be a solution with harmonic bundles since there seem to be many harmonic bundles and points like $(DX,DA,DF,DE),(X,D,B,C)$ etc (X is the intersection of BC and EF).