Giả sử \(\frac{IA}{AB}=k\Rightarrow\frac{IB}{AB}=1-k\)
\(\Rightarrow\left\{{}\begin{matrix}\overrightarrow{IA}=-k\overrightarrow{AB}\\\overrightarrow{IB}=\left(1-k\right)\overrightarrow{AB}\end{matrix}\right.\)
\(\Rightarrow IB.\overrightarrow{IA}+IA.\overrightarrow{IB}=\left(1-k\right).AB.\left(-k\right)\overrightarrow{AB}+k.AB.\left(1-k\right)\overrightarrow{AB}\)
\(=\left(k^2-k\right)AB.\overrightarrow{AB}+\left(k-k^2\right)AB.\overrightarrow{AB}\)
\(=\left(k^2-k+k-k^2\right).AB.\overrightarrow{AB}=\overrightarrow{0}\)
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