1. 設\(D\)為\(\triangle ABC\)中\(BC\)邊中點，由分點公式得\(\overset{\rightharpoonup}{OD} = \frac{\overset{\rightharpoonup}{OB} + \overset{\rightharpoonup}{OC}}{2}\)。
2. 由分點公式求\(\overset{\rightharpoonup}{OG}\)：
- \(\overset{\rightharpoonup}{OG} = \frac{\overset{\rightharpoonup}{OA} + 2\overset{\rightharpoonup}{OD}}{3}\)。
- 把\(\overset{\rightharpoonup}{OD} = \frac{\overset{\rightharpoonup}{OB} + \overset{\rightharpoonup}{OC}}{2}\)代入上式，\(\overset{\rightharpoonup}{OG} = \frac{\overset{\rightharpoonup}{OA} + 2\times\frac{\overset{\rightharpoonup}{OB} + \overset{\rightharpoonup}{OC}}{2}}{3} = \frac{\overset{\rightharpoonup}{OA} + \overset{\rightharpoonup}{OB} + \overset{\rightharpoonup}{OC}}{3}\)。
3. 兩邊同乘\(3\)，得\(3\overset{\rightharpoonup}{OG} = \overset{\rightharpoonup}{OA} + \overset{\rightharpoonup}{OB} + \overset{\rightharpoonup}{OC}\)。
綜上，證得對於\(\triangle ABC\)，\(G\)為重心，\(O\)為任意點，有\(3\overset{\rightharpoonup}{OG} = \overset{\rightharpoonup}{OA} + \overset{\rightharpoonup}{OB} + \overset{\rightharpoonup}{OC}\)。
推廣:
$1.若O為原點,則G=\frac{A+B+C}{3}$
$2.求合力時,三股萬有引力的合力=直接對三星球重心的引力\times3$