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Question

Cho $\Delta ABC$.Gọi I,J là hai điểm thỏa $\underset{IA}{\rightarrow}=2\underset{IB}{\rightarrow}$ ; $3\underset{JA}{\rightarrow}+2\underset{JC}{\rightarrow}=\underset{0}{\rightarrow}$. Chứng minh IJ qua trọng tâm G của $\Delta ABC$

score 1 · Answer 1

Ta có:

$\overrightarrow{IA}=2 \overrightarrow{IB}\Rightarrow \overrightarrow{IG}+\overrightarrow{GA}=2(\overrightarrow{IG}+\overrightarrow{GB}) \Rightarrow \overrightarrow{IG}=\overrightarrow{GA}-2 \overrightarrow{GB}$

$3\overrightarrow{JA}+2 \overrightarrow{JC}=\overrightarrow{0}\Rightarrow 3(\overrightarrow{JG}+\overrightarrow{GA})+2(\overrightarrow{JG}+\overrightarrow{GC})=\overrightarrow{0}\Rightarrow 5\overrightarrow{GJ}=3\overrightarrow{GA}+2 \overrightarrow{GC}$

Mà $\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow{0} \Rightarrow \overrightarrow{GA}-2\overrightarrow{GB}=3\overrightarrow{GA}+2 \overrightarrow{GC}$

Từ đó suy ra: $\overrightarrow{IG}=5 \overrightarrow{GJ} \Rightarrow IJ$ đi qua $G$.