[tex]\it \overrightarrow{MB}=-2\overrightarrow{MC} \Rightarrow \overrightarrow{MB}=2\overrightarrow{CM} \Rightarrow MB=2CM\Big|_{+CM} \Rightarrow BC=3CM\Big|_{:2} \Rightarrow \\ \\ \\ \Rightarrow \dfrac{BC}{2}=\dfrac{3}{2}CM \Rightarrow DC=\dfrac{3}{2}CM \Rightarrow DM+CM=\dfrac{3}{2}CM\Big|_{-CM} \Rightarrow \\ \\ \\ \Rightarrow DM=\dfrac{1}{2}CM \Rightarrow \dfrac{DM}{CM}=\dfrac{1}{2}\ \ \ \ \ (1)[/tex]
Punctul G = centrul de greutate pentru Δ ABC, deci vom avea:
[tex]\it \dfrac{DG}{GA}=\dfrac{DG}{2DG}=\dfrac{1}{2}\ \ \ \ \ (2)\\ \\ \\ (1),\ (2)\ \stackrel{R.T.Thales}{\Longrightarrow}\ \ GM||AC[/tex]
