woah, impressive

Another fact that can be used is : fact wrote: If $DO \cap MN = L$, $DO \cap (DEF) = T'$, let $P$ be the the foot of altitude from $A$ to $BC$. then $PN_9T'L$ is cyclic Note that this is true without any conditions. After this the solution should be simple angle chase.

Solved with cursed_tangent1434 Let $R,S$ be the midpoints of $AC,AB$ respectively and $Q$ be the midpoint of $RS$.Since $D$ is orthocenter of $ASR$ and $O$ is antipode of $A$ in $(ASR)$ it follows that midpoint of $DO$ is $Q$. $\Delta MQN \sim \Delta BOC$ since homothety at $D$ with ratio $2$ maps $\Delta MQN$ to $\Delta BOC$.Since $D$ is circumcenter of $AEF$ we have $\Delta FDE \sim \Delta BOC$.So $\measuredangle MTN=\measuredangle FTE=\measuredangle FDE=\measuredangle BOC=\measuredangle MQN \implies MTNQ$ is cyclic. We have $SR \parallel BC \parallel MN$ ; $SM \parallel AD \parallel RN$ and $AD \perp BC$ so it follows that $SMNR$ is a rectangle. This implies that $Q$ lies on the perpendicular bisector of $MN$ and hence $Q$ is midpoint of arc $MN$ in $(MTN)$. $D$ is midpoint of arc $EF$ in $(FTE)$ , hence $T-Q-D$ are collinear as they lie on the angle bisector of $\angle FTE$.Hence $D-Q-O-T$ lie on same line. So we are done

Let $O$ be the circumcenter of $(ABC)$ and $J$ be the midpoint of $DO$. Now it is sufficient to prove that $TD$ and $TJ$ coincide. We first prove that $M,N,T,J$ are concyclic. \[\angle MJN=\angle BOC=2\angle BAC=\angle EDF=180-\angle FTE=\angle 180-\angle NJM\]This proves our claim! Now note that $JM=\frac{OB}{2}=\frac{OC}{2}=JN$, so $TJ$ is the angle bisector of $\angle NTM=\angle ETF$. But, $D$ is the midpoint of arc $EF$ in the nine-point circle, so $TD$ is the angle bisector of $\angle ETF$ as well. Thus, $TD$ and $TJ$ must coincide!

A little bit of complex bash actually works nicely on this question - with a lot of help from Rijul saini Firstly consider all points under the homothety with center at $H$ and ratio $2$, we note that as $DEF$ is the $9 point circle$, this mapping takes $T$ to $T'$ which lies on the circumcircle. Let $P'$ denote $P$ after the mapping for all points $P$ Let the Circumcircle be the unit circle $D' = a$ $F'$ being the reflection of $H$ on line $AB$ we get $F' = \frac{-ab}{c}$ and similarly $E'=\frac{-ac}{b}$ meanwhile as $M$ is midpoint of $M'H$ we get that $M' = \frac{b-c}{2}$ and similarly $N' = \frac{c-b}{2}$ and finally we get $O' = -a-b-c$ now since $F'T'M'$ are collinear with $T'$ lying on unit circle we get $-F'T' = \frac{\frac{c^2-bc-2ab}{2c}}{\frac{-2c^2+ab-ac}{2abc}}$ or $T' = \frac{-c(c^2-bc-2ab)}{2c^2 - ab + ac}$ - eqn(i) Similarly on equating $T'$ with the fact that $E'N'T'$ are collinear we get $T' = \frac{-b(b^2 - bc - 2ac)}{2b^2 - ac + ab}$ - eqn(ii) and finally we get $AO' \cap (ABC) = \frac{-bc(2a+b+c)}{2bc+ab+ac}$ - eqn(iii) But now note that on multiplying numerator and denominator of eqn(i) with $b$ and eqn(ii) with $c$ and subtracting numerators and denominators using continued proportion we get eqn(iii) which gives us that if 2 statements are true then the third is implied which concludes our proof.