Giải thích các bước giải:
Gọi M là trung điểm BC, ta có:
\(\begin{array}{l}
\overrightarrow {AG} = \dfrac{2}{3}\overrightarrow {AM} = \dfrac{1}{3}\left( {\overrightarrow {AB} + \overrightarrow {BM} + \overrightarrow {AC} + \overrightarrow {CM} } \right) = \dfrac{1}{3}\left( {\overrightarrow {AB} + \overrightarrow {AC} } \right)\\
\Rightarrow A{G^2} = \dfrac{1}{9}\left( {A{B^2} + 2\overrightarrow {AB} .\overrightarrow {AC} + A{C^2}} \right)\\
= \dfrac{1}{9}\left( {{b^2} + {c^2} + 2bc.\cos A} \right)\\
= \dfrac{1}{9}\left( {{b^2} + {c^2} + 2bc.\dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}} \right)\\
= \dfrac{1}{9}\left( {2{b^2} + 2{c^2} - {a^2}} \right)\\
{T^2} \Rightarrow G{B^2} = \dfrac{1}{9}\left( {2{a^2} + 2{c^2} - {b^2}} \right);\,\,\,\,G{C^2} = \dfrac{1}{9}\left( {2{a^2} + 2{b^2} - {c^2}} \right)\\
\Rightarrow G{A^2} + G{B^2} + G{C^2} = \dfrac{1}{3}\left( {{a^2} + {b^2} + {c^2}} \right)
\end{array}\)
