Ta có:
$\overrightarrow{MN} = \overrightarrow{MC} + \overrightarrow{CN}$
$=\overrightarrow{CB} + \dfrac{1}{3}\overrightarrow{CA}$
$= -\overrightarrow{BC} - \dfrac{1}{3}\overrightarrow{AC}$
$\overrightarrow{MP} = \overrightarrow{MB} + \overrightarrow{BP}$
$=2\overrightarrow{CB} + \dfrac{1}{2}\overrightarrow{BA}$
$=-2\overrightarrow{BC} + \dfrac{1}{2}(\overrightarrow{BC} - \overrightarrow{AC})$
$= -\dfrac{3}{2}\overrightarrow{BC} -\dfrac{1}{2}\overrightarrow{AC}$
$\Rightarrow \overrightarrow{MP} = \dfrac{3}{2}\overrightarrow{MN}$
$\Rightarrow M,N,P$ thẳng hàng
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Áp dụng định lý $Menelaus$ ta được:
$\dfrac{PA}{PB}\cdot\dfrac{MB}{MC}\cdot\dfrac{NC}{NA} = 1\cdot2\cdot\dfrac12 = 1$
$\Rightarrow\overline{M,N,P}$
