Cho hình tứ diện ABCD. Gọi G là trọng tâm tam giác ABC. Chứng minh rằng: \(\overrightarrow{DA}+\overrightarrow{DB}+\overrightarrow{DC}=3\overrightarrow{DG}.\)
Ta có: \(\overrightarrow{DA}+\overrightarrow{DB}+\overrightarrow{DC}\)
\(=(\overrightarrow{DG}+\overrightarrow{GA})+ (\overrightarrow{DG}+\overrightarrow{GB})+ (\overrightarrow{DG}+\overrightarrow{GC})\)
\(=3\overrightarrow{DG}+ (\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC})= 3\overrightarrow{DG}+\overrightarrow{0}=3\overrightarrow{DG}\) (đpcm).
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