\(BM=MN=NC\Rightarrow\overrightarrow{BM}=\overrightarrow{MN}=\overrightarrow{NC}=\dfrac{1}{3}\overrightarrow{BC}\)
\(\overrightarrow{BC}=\overrightarrow{BA}+\overrightarrow{AC}=-\overrightarrow{AB}+\overrightarrow{AC}\Rightarrow\overrightarrow{BM}=-\dfrac{1}{3}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}\)
\(\Rightarrow\overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{BM}=\overrightarrow{AB}-\dfrac{1}{3}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}=\dfrac{2}{3}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}\) (đpcm)
b.
\(CP=\dfrac{2}{3}CA\Rightarrow\overrightarrow{CP}=\dfrac{2}{3}\overrightarrow{CA}=-\dfrac{2}{3}\overrightarrow{AC}\)
\(\overrightarrow{NC}=\overrightarrow{BM}=-\dfrac{1}{3}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}\)
\(\Rightarrow\overrightarrow{NP}=\overrightarrow{NC}+\overrightarrow{CP}=-\dfrac{1}{3}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}-\dfrac{2}{3}\overrightarrow{AC}=-\dfrac{1}{3}\overrightarrow{AB}-\dfrac{1}{3}\overrightarrow{AC}\)
Do đó:
\(\overrightarrow{AM}+\overrightarrow{NP}+\overrightarrow{PM}=\overrightarrow{AM}+\overrightarrow{NM}=\overrightarrow{AM}-\dfrac{1}{3}\overrightarrow{BC}\)
\(=\dfrac{2}{3}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}-\dfrac{1}{3}\left(-\overrightarrow{AB}+\overrightarrow{AC}\right)=\overrightarrow{AB}\) (đpcm)
