Do G là trọng tâm ABC \(\Rightarrow\overrightarrow{BG}=\dfrac{1}{3}\overrightarrow{BA}+\dfrac{1}{3}\overrightarrow{BC}\)

I đối xứng B qua G \(\Rightarrow\) \(\overrightarrow{BI}=2\overrightarrow{BG}=\dfrac{2}{3}\overrightarrow{BA}+\dfrac{2}{3}\overrightarrow{BC}=\dfrac{2}{3}\overrightarrow{BA}+\dfrac{2}{3}\left(\overrightarrow{BA}+\overrightarrow{AC}\right)\)

\(\Rightarrow\overrightarrow{BI}=\dfrac{4}{3}\overrightarrow{BA}+\dfrac{2}{3}\overrightarrow{AC}=-\dfrac{4}{3}\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{AC}\)

\(\Rightarrow\overrightarrow{CI}=\overrightarrow{CB}+\overrightarrow{BI}=\overrightarrow{CA}+\overrightarrow{AB}-\dfrac{4}{3}\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{AC}\)

\(\Rightarrow\overrightarrow{CI}=-\dfrac{1}{3}\overrightarrow{AB}-\dfrac{1}{3}\overrightarrow{AC}\)

H đối xứng B qua G \(\Rightarrow\overrightarrow{BH}=2\overrightarrow{BG}=2\left(\dfrac{1}{3}\overrightarrow{BA}+\dfrac{1}{3}\overrightarrow{BC}\right)=-\dfrac{2}{3}\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{BC}\)

\(\overrightarrow{AH}=\overrightarrow{AB}+\overrightarrow{BH}=\overrightarrow{AB}-\dfrac{2}{3}\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{BC}=\dfrac{1}{3}\overrightarrow{AB}+\dfrac{2}{3}\left(\overrightarrow{BA}+\overrightarrow{AC}\right)\)

\(=\dfrac{1}{3}\overrightarrow{AB}-\dfrac{2}{3}\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{AC}=\dfrac{2}{3}\overrightarrow{AC}-\dfrac{1}{3}\overrightarrow{AB}\)

\(\overrightarrow{CH}=\overrightarrow{CA}+\overrightarrow{AH}=-\overrightarrow{AC}+\dfrac{2}{3}\overrightarrow{AC}-\dfrac{1}{3}\overrightarrow{AB}=-\dfrac{1}{3}\overrightarrow{AB}-\dfrac{1}{3}\overrightarrow{AC}\)

\(\overrightarrow{MH}=\overrightarrow{MA}+\overrightarrow{AH}=-\dfrac{1}{2}\overrightarrow{AB}-\dfrac{1}{2}\overrightarrow{AC}+\dfrac{2}{3}\overrightarrow{AC}-\dfrac{1}{3}\overrightarrow{AB}\)

\(=-\dfrac{5}{6}\overrightarrow{AB}+\dfrac{1}{6}\overrightarrow{AC}\)

\(a,\) \(\overrightarrow{IA}=2\overrightarrow{IB}-4\overrightarrow{IC}\)

\(\overrightarrow{IA}=2\overrightarrow{IB}-2\overrightarrow{IC}-2\overrightarrow{IC}=2\overrightarrow{CB}-2\overrightarrow{IC}\)

\(=2\left(\overrightarrow{AB}-\overrightarrow{AC}\right)-2\left(\overrightarrow{AC}-\overrightarrow{AI}\right)\)

\(\overrightarrow{IA}=2\overrightarrow{AB}-2\overrightarrow{AC}-2\overrightarrow{AC}+2\overrightarrow{AI}\)

\(\overrightarrow{IA}=\dfrac{2}{3}\overrightarrow{AB}-\dfrac{4}{3}\overrightarrow{AC}\)

\(b,\overrightarrow{IJ}=\overrightarrow{AJ}-\overrightarrow{AI}=\dfrac{2}{3}\overrightarrow{AB}+\overrightarrow{IA}=\dfrac{2}{3}\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{AB}-\dfrac{4}{3}\overrightarrow{AC}=\dfrac{4}{3}\left(\overrightarrow{AB}-\overrightarrow{AC}\right)\left(1\right)\)

\(\overrightarrow{JG}=\overrightarrow{AG}-\overrightarrow{AJ}=\dfrac{2}{3}\overrightarrow{AM}-\dfrac{2}{3}\overrightarrow{AB}\)\((\) \(\) \(M\)  \(trung\) \(điểm\) \(BC)\)

\(\overrightarrow{JG}=\dfrac{\overrightarrow{AB}+\overrightarrow{AC}}{3}-\dfrac{2}{3}\overrightarrow{AB}=-\dfrac{1}{3}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}=-\dfrac{1}{3}\left(\overrightarrow{AB}-\overrightarrow{AC}\right)\left(2\right)\)

\(\left(1\right)\left(2\right)\Rightarrow\overrightarrow{IJ}=-4\overrightarrow{JG}\Rightarrow I,J,G\) \(thẳng\) \(hàng\)