Let $AD, BE, CF$ be the altitudes and $H$ be the orthocenter of triangle $ABC$. Denote by $\omega_A$ the circle centered at $M_A$ with radius $M_AH_B=M_AH_C$ and define circles $\omega_B$ and $\omega_C$ similarly. Applying the radical axis theorem on the circles $\omega_A, \omega_B,$ and $\omega_C,$ we conclude that the lines through $H_A, H_B, H_C$ perpendicular to the corresponding sides are concurrent at a point $X$. Note that $X \ne H $ means that $X$ lies inside one of the six planar regions which the altitudes divide the plane into. Suppose $X$ lies in the unbounded planar region made by rays $HB$ and $HF$. Then, $H_C$ lies on the segment $BF$ and $H_B$ lies on the extension of $CE$ beyond $E$. We conclude that $$M_AH_B>M_AE=M_AF>M_AH_B,$$a contradiction.