VÌ \(G'\) là trọng tâm \(\Delta A'B'C'\) nên ta có
\(3\overrightarrow{GG'}=\overrightarrow{GA'}+\overrightarrow{GB'}+\overrightarrow{GC'}\)
\(=\overrightarrow{GA}+\overrightarrow{AA'}+\overrightarrow{GB}+\overrightarrow{BB'}+\overrightarrow{GC}+\overrightarrow{CC'}\)
\(=\left(\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}\right)+\overrightarrow{AA'}+\overrightarrow{BB'}+\overrightarrow{CC'}\)
\(=\overrightarrow{AA'}+\overrightarrow{BB'}+\overrightarrow{CC'}\) ( VÌ \(G\) là trọng tâm \(\Delta ABC\) \(\Rightarrow\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow{0}\) )
\(=\overrightarrow{AC}+\overrightarrow{CA'}+\overrightarrow{CB}+\overrightarrow{BC'}+\overrightarrow{BA}+\overrightarrow{AB'}\)
\(=\left(\overrightarrow{CA'}+\overrightarrow{BC'}\right)+\overrightarrow{AB'}+\left(\overrightarrow{AC}+\overrightarrow{CB}+\overrightarrow{BA}\right)\)
\(=\left(\overrightarrow{CA'}+\overrightarrow{A'B'}\right)+\overrightarrow{AB'}=\overrightarrow{CB'}+\overrightarrow{AB'}=\overrightarrow{0}\)
vậy \(3\overrightarrow{GG'}=\overrightarrow{0}\) \(\Leftrightarrow G\equiv G'\) (đpcm)