\(\left\{{}\begin{matrix}\overrightarrow{IJ}=\overrightarrow{IA}+\overrightarrow{AB}+\overrightarrow{BJ}\\\overrightarrow{IJ}=\overrightarrow{ID}+\overrightarrow{DC}+\overrightarrow{CJ}\end{matrix}\right.\)
Cộng vế với vế:
\(2\overrightarrow{IJ}=\left(\overrightarrow{IA}+\overrightarrow{ID}\right)+\left(\overrightarrow{BJ}+\overrightarrow{CJ}\right)+\overrightarrow{AB}+\overrightarrow{DC}=\overrightarrow{AB}+\overrightarrow{DC}\)
\(\Rightarrow\overrightarrow{IJ}=\frac{1}{2}\overrightarrow{AB}+\frac{1}{2}\overrightarrow{DC}\)
b/ Đặt \(\frac{MA}{MB}=\frac{ND}{NC}=k\)
\(\left\{{}\begin{matrix}\overrightarrow{IP}=\overrightarrow{IA}+\overrightarrow{AM}+\overrightarrow{MP}\\\overrightarrow{IP}=\overrightarrow{ID}+\overrightarrow{DN}+\overrightarrow{NP}\end{matrix}\right.\)
\(\Rightarrow2\overrightarrow{IP}=\left(\overrightarrow{IA}+\overrightarrow{ID}\right)+\left(\overrightarrow{MP}+\overrightarrow{NP}\right)+\overrightarrow{AM}+\overrightarrow{DN}=\overrightarrow{AM}+\overrightarrow{DN}\)
\(\Rightarrow2\overrightarrow{IP}=k.\overrightarrow{AB}+k.\overrightarrow{DC}\)
\(\Rightarrow\overrightarrow{IP}=\frac{k}{2}\left(\overrightarrow{AB}+\overrightarrow{DC}\right)=\frac{k}{2}.\overrightarrow{IJ}\Rightarrow P;I;J\) thẳng hàng hay P thuộc IJ
