My solution: Let $H$ be the orthocenter of $ABC$. Then, with symmetry with respect to $BC$, the circle $(ABC)$ turns into $(BHC),$ so $BHM_1C$ is inscribed. Now the statement of the problem is obviously obtained by a homothetic with the center $A$ and the coefficient $1/2$. In this case, the circle $(BHC)$ turns into a circle of 9 points, that is, the middle of $AM_1$ lies on a circle of 9 points, which is what they wanted to prove.