Xét $MA^{2}+MB^{2}+MC^{2}=\overrightarrow{MA}^{2}+\overrightarrow{MB}^{2}+\overrightarrow{MC}^{2}$
=$(\overrightarrow{MG}+\overrightarrow{GA})^{2}+(\overrightarrow{MG}+\overrightarrow{GB})^{2}+(\overrightarrow{MG}+\overrightarrow{GC})^{2}$
= $3\overrightarrow{MG}^{2}+\overrightarrow{GA}^{2}+\overrightarrow{GB}^{2}+\overrightarrow{GC}^{2}+2.\overrightarrow{MG}.(\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC})$
= $3{MG}^{2}+{GA}^{2}+{GB}^{2}+{GC}^{2}$
(vì G là trọng tâm tam giác ABC nên $\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow{0}$)
Suy ra: $MA^{2}+MB^{2}+MC^{2}=3MG^{2}+GA^{2}+GC^{2}+GC^{2}$