Question 103
In triangle $ABC$ there are points $M_1, M_2, M_3$ – midpoints of sides $BC$, $AC$, $AB$ correspondingly. Point $K$ is symmetrical to $M_2$ with respect to line $BC$, $AH$ is the altitude of triangle $ABC$. Prove that line $KM_3$ splits the segment $HM_1$ in half.