Assuming $D,E,F,M$ are the midpoints of $BC,CA,AB,AH$. Let $H_A,H_B,H_C$ be the altitudes from $A,B,C$ to $BC,CA,AB$ and $S$ be on $(ABC)$ such that $AS$ is symedian. Let $H_BH_C\cap BC=P$. Invert at $H$ with radius $\sqrt{-HA.HH_A}$. $D^*$ is $A-$queue point and $M^*=(ABC)\cap AH$. We have $-1=(M^*P,M^*A;B,C)$ hence $P,M^*,S$ are collinear. We have $PM^*.PS=MB.MC=PH_C.PH_B$ thus, $M^*,S,H_B,H_C$ are concyclic. \[(D^*,M^*;B,C)\overset{A}{=}(P,H_A;B,C)\overset{S}{=} (M^*,SH_A\cap (ABC);B,C)\]So we see that $D^*,H_A,S$ are collinear. We have proved $(H_BH_CM^*),(ABC),D^*H_A$ concur which is equavilent to the concurrency of $(DEF),(MBC),(HAD)$ as desired.$\blacksquare$

@above I fixed it.

bin_sherlo wrote: What is $M$? $4$ points are midpoints of $3$ sides, am i missing something? I guess $D$ belongs to $BC$ because of specifying $AHD$

Solved with @Enigma714 Let $AA',BB',CC'$ be the altitutes in $\triangle ABC$. Now, invert with center $H$ and radius $\sqrt{-HA.HA'}$, where $R'$ is the inverse of $R$ for every point $R$, we now want to prove that $A'D'$, $\odot(B'M'C')$ and $\odot(ABC)$ concur. (Because $\odot(A'B'C')=\odot(DEF)$) Let $X$ be such that $\{X,D'\}=DA'\cap \odot(ABC)$. By radical axis in $\odot(AB'HC'D')$, $\odot (AD'BC)$ and $\odot(BCB'C')$ we have $AD'$, $B'C'$ and $BC$ concur, say in $Y$. On the other hand, as $\angle M'XD'=\angle M'AD'=\angle A'DD'=\angle A'DM'$, we have $A'DM'X$ is cyclic. Now, by radical axis in $\odot(A'DXM')$, $\odot(AD'A'D)$ and $\odot(AD'M'X)$, we have $AD'$, $M'X$ and $DA'$ concur in $Y$. So by power point in the cyclics, $YM'\cdot YX=YA'\cdot YD=YA\cdot YD'=YC'\cdot YB'$, so $M'XB'C'$ is cyclic, so $X$ is the point we wanted.