Let us draw through point $M$ lines that are parallel to sides $ABC$. Let them intersect $AB$, $BC$, $AC$ at $C_1$, $C_2$; $A_1$, $A_2$; $B_1$, $B_2$ (Fig. 30).

Hence, $C_1A_2 \parallel AC$, $A_1B_2 \parallel BA$, $B_1C_2 \parallel BC$, so $C_1A_2$, $A_1B_2$ and $B_1C_2$ intersect at $M$. Consider $\triangle A_1MA_2$. Obviously, this triangle is equilateral. Line $M_1M$ contains its altitude, because it is perpendicular to $BC$. So $M_1M$ is doubled median,

hence $\overrightarrow{MA_1} + \overrightarrow{MA_2} = \overrightarrow{MM_1}$, similarly $\overrightarrow{MB_1} + \overrightarrow{MB_2} = \overrightarrow{MM_2}$ and $\overrightarrow{MC_1} + \overrightarrow{MC_2} = \overrightarrow{MM_3}$. On the other hand $MC_1AB_2$ is a parallelogram, so $\overrightarrow{MA} = \overrightarrow{MC_1} + \overrightarrow{MB_2}$. Similarly $\overrightarrow{MB} = \overrightarrow{MC_2} + \overrightarrow{MA_1}$ and $\overrightarrow{MC} = \overrightarrow{MA_2} + \overrightarrow{MB_1}$. Hence:

$$\overrightarrow{MM_1} + \overrightarrow{MM_2} + \overrightarrow{MM_3} = \overrightarrow{MA_1} + \overrightarrow{MA_2} + \overrightarrow{MB_1} + \overrightarrow{MB_2} + \overrightarrow{MC_1} + \overrightarrow{MC_2} = \overrightarrow{MA} + \overrightarrow{MB} + \overrightarrow{MC}.$$