Question

Cho tam giác ABC trọng tâm G CMR: vecto MG = 1/3( vecto MA + vecto MB + vecto MC) với M bất kì

Answer 1

Xét ΔABC có G là trọng tâm

nên \(\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow{0}\)

\(\dfrac{1}{3}\left(\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}\right)\)

\(=\dfrac{1}{3}\left(\overrightarrow{MG}+\overrightarrow{GA}+\overrightarrow{MG}+\overrightarrow{GB}+\overrightarrow{MG}+\overrightarrow{GC}\right)\)

\(=\dfrac{1}{3}\left(3\cdot\overrightarrow{MG}+\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}\right)\)

\(=\dfrac{1}{3}\cdot3\cdot\overrightarrow{MG}=\overrightarrow{MG}\)