Gọi $N$ là trung điểm $AC$
$\to \overrightarrow{BN}=\dfrac12(\overrightarrow{BA} +\overrightarrow{BC})$
$\to \overrightarrow{BG} = \dfrac23\overrightarrow{BN}=\dfrac13(\overrightarrow{BA} +\overrightarrow{BC})\qquad (1)$
Ta lại có:
$\overrightarrow{MA}-\overrightarrow{MB}+\overrightarrow{MC}=\overrightarrow{0}$
$\to \overrightarrow{MB} +\overrightarrow{BA}-\overrightarrow{MB}+\overrightarrow{MB}+\overrightarrow{BC}=\overrightarrow{0}$
$\to \overrightarrow{MB} =-(\overrightarrow{BA} +\overrightarrow{BC})\qquad (2)$
$(1)(2)\Rightarrow \overrightarrow{MB} = -3\overrightarrow{BG}$
$\to M, B,G$ thẳng hàng
