一、对ΔABC重心O来讲有
$\stackrel{\rightharpoonup }{\mathrm{OA}}+\stackrel{\rightharpoonup }{\mathrm{OB}}+\stackrel{\rightharpoonup }{\mathrm{OC}}=\stackrel{\rightharpoonup }{0}$
证明：延长CO与线段$\overline{AB}$交于点D，
根据A、D、B三点共线公式
$\stackrel{\rightharpoonup }{\mathrm{OD}}=m\stackrel{\rightharpoonup }{\mathrm{OA}}+n\stackrel{\rightharpoonup }{\mathrm{OB}}$(其中m+n=1)，因为D是线段$\overline{AB}$的中点，所以有
$\stackrel{\rightharpoonup }{\mathrm{OA}}+\stackrel{\rightharpoonup }{\mathrm{OB}}=2\stackrel{\rightharpoonup }{\mathrm{OD}}$
又因$\stackrel{\rightharpoonup }{\mathrm{OC}}=2\stackrel{\rightharpoonup }{\mathrm{DO}}$，
所以$\stackrel{\rightharpoonup }{\mathrm{OA}}+\stackrel{\rightharpoonup }{\mathrm{OB}}+\stackrel{\rightharpoonup }{\mathrm{OC}}=\stackrel{\rightharpoonup }{0}$，得证。
反过来，如果
$\stackrel{\rightharpoonup }{\mathrm{OA}}+\stackrel{\rightharpoonup }{\mathrm{OB}}+\stackrel{\rightharpoonup }{\mathrm{OC}}=\stackrel{\rightharpoonup }{0}$
则$\stackrel{\rightharpoonup }{\mathrm{OA}}+\stackrel{\rightharpoonup }{\mathrm{OB}}+\stackrel{\rightharpoonup }{\mathrm{OC}}=(\stackrel{\rightharpoonup }{\mathrm{OD}}+\stackrel{\rightharpoonup }{\mathrm{DA}})+(\stackrel{\rightharpoonup }{\mathrm{OD}}+\stackrel{\rightharpoonup }{\mathrm{DB}})+\stackrel{\rightharpoonup }{\mathrm{OC}}$
$=(\stackrel{\rightharpoonup }{\mathrm{OC}}+2\stackrel{\rightharpoonup }{\mathrm{OD}})+(\stackrel{\rightharpoonup }{\mathrm{DA}}+\stackrel{\rightharpoonup }{\mathrm{DB}})$
$=m\stackrel{\rightharpoonup }{\mathrm{OD}}+n\stackrel{\rightharpoonup }{\mathrm{DA}}=\stackrel{\rightharpoonup }{0}$，
因$\stackrel{\rightharpoonup }{\mathrm{OD}}$与$\stackrel{\rightharpoonup }{\mathrm{DA}}$线性无关，所以上式要取得$\stackrel{\rightharpoonup }{0}$只有

$\stackrel{\rightharpoonup }{\mathrm{OC}}+2\stackrel{\rightharpoonup }{\mathrm{OD}}=\stackrel{\rightharpoonup }{0}$并且$\stackrel{\rightharpoonup }{\mathrm{DA}}+\stackrel{\rightharpoonup }{\mathrm{DB}}=\stackrel{\rightharpoonup }{0}$，
可得$\overline{DA}$=$\overline{BD}$，以及$\overline{CO}$=2$\overline{OD}$，即D是线段$\overline{AB}$的中点，O为ΔABC的重心。

二、对ΔABC内心O来讲有
$a\stackrel{\rightharpoonup }{\mathrm{OA}}+b\stackrel{\rightharpoonup }{\mathrm{OB}}+c\stackrel{\rightharpoonup }{\mathrm{OC}}=\stackrel{\rightharpoonup }{0}$
证明：延长CO与线段$\overline{AB}$交于点D，

因为$\overline{CD}$是∠ACB的角平分线，
根据角平分线性质，线段
$\overline{OA}/\overline{OB}=\overline{CA}/\overline{CB}=\overline{DA}/\overline{DB}=b/a$，
并且
$\overline{CO}/\overline{OD}=\overline{CA}/\overline{AD}=\overline{CB}/\overline{BD}=(\overline{CA}+\overline{CB})/(\overline{AD}+\overline{BD})=(a+b)/c$，
而$\overline{CO}$与$\overline{OD}$共线，长度比为$(a+b)/c$，故
$(a+b)\stackrel{\rightharpoonup }{\mathrm{OD}}+c\stackrel{\rightharpoonup }{\mathrm{OC}}=\stackrel{\rightharpoonup }{0}$，
再根据A、D、B三点共线性质有
$a\stackrel{\rightharpoonup }{\mathrm{OA}}+b\stackrel{\rightharpoonup }{\mathrm{OB}}=(a+b)\stackrel{\rightharpoonup }{\mathrm{OD}}$，所以

$a\stackrel{\rightharpoonup }{\mathrm{OA}}+b\stackrel{\rightharpoonup }{\mathrm{OB}}+c\stackrel{\rightharpoonup }{\mathrm{OC}}=(a+b)\stackrel{\rightharpoonup }{\mathrm{OD}}+c\stackrel{\rightharpoonup }{\mathrm{OC}}=\stackrel{\rightharpoonup }{0}$，得证。
反之，若已知$a\stackrel{\rightharpoonup }{\mathrm{OA}}+b\stackrel{\rightharpoonup }{\mathrm{OB}}+c\stackrel{\rightharpoonup }{\mathrm{OC}}=\stackrel{\rightharpoonup }{0}$，则
$a\stackrel{\rightharpoonup }{\mathrm{OA}}+b\stackrel{\rightharpoonup }{\mathrm{OB}}+c\stackrel{\rightharpoonup }{\mathrm{OC}}=$
$a(\stackrel{\rightharpoonup }{\mathrm{OD}}+\stackrel{\rightharpoonup }{\mathrm{DA}})+b(\stackrel{\rightharpoonup }{\mathrm{OD}}+\stackrel{\rightharpoonup }{\mathrm{DB}})+c(\stackrel{\rightharpoonup }{\mathrm{OD}}+\stackrel{\rightharpoonup }{\mathrm{DC}})=$
$(a+b+c)\stackrel{\rightharpoonup }{\mathrm{OD}}+c\stackrel{\rightharpoonup }{\mathrm{DC}}+(a\stackrel{\rightharpoonup }{\mathrm{DA}}+b\stackrel{\rightharpoonup }{\mathrm{DB}})=$
$(a+b)\stackrel{\rightharpoonup }{\mathrm{OD}}+c\stackrel{\rightharpoonup }{\mathrm{OC}}+(a\stackrel{\rightharpoonup }{\mathrm{DA}}+b\stackrel{\rightharpoonup }{\mathrm{DB}})=\stackrel{\rightharpoonup }{0}$
因向量$((a+b)\stackrel{\rightharpoonup }{\mathrm{OD}}+c\stackrel{\rightharpoonup }{\mathrm{OC}})$与$(a\stackrel{\rightharpoonup }{\mathrm{DA}}+b\stackrel{\rightharpoonup }{\mathrm{DB}})$线性无关，所以上式要取得$\stackrel{\rightharpoonup }{0}$，只有
$(a\stackrel{\rightharpoonup }{\mathrm{DA}}+b\stackrel{\rightharpoonup }{\mathrm{DB}})=\stackrel{\rightharpoonup }{0}$，再由$\overline{DA}$与$\overline{BD}$共线，
可得$\overline{AD}/\overline{DB}=\overline{AC}/\overline{CB}=b/a$，即线段$\overline{CD}$是∠ACB的角平分线，同理可证另两条角平分线$\overline{AO}$和$\overline{BO}$，O为ΔABC的内心。另外，
$\overline{OC}/\overline{OD}=(a+b)/c$。
三、对ΔABC外心O来讲有
${\stackrel{\rightharpoonup }{\mathrm{OA}}}^2={\stackrel{\rightharpoonup }{\mathrm{OB}}}^2={\stackrel{\rightharpoonup }{\mathrm{OC}}}^2$，
证明：线段$\overline{OA}$，$\overline{OB}$，$\overline{OC}$为外接圆的半径，所以等长，向量${\stackrel{\rightharpoonup }{\mathrm{OA}}}^2$内积为长度的平方。
四、对ΔABC垂心O来讲有
$\stackrel{\rightharpoonup }{\mathrm{OA}}\cdot \stackrel{\rightharpoonup }{\mathrm{OB}}=\stackrel{\rightharpoonup }{\mathrm{OB}}\cdot \stackrel{\rightharpoonup }{\mathrm{OC}}=\stackrel{\rightharpoonup }{\mathrm{OC}}\cdot \stackrel{\rightharpoonup }{\mathrm{OA}}$
证明：因为线段$\overline{AB}\perp \overline{CO}$，所以
$\stackrel{\rightharpoonup }{\mathrm{OC}}\cdot \stackrel{\rightharpoonup }{\mathrm{AB}}=0$，因
$\stackrel{\rightharpoonup }{\mathrm{AB}}=\stackrel{\rightharpoonup }{\mathrm{AO}}-\stackrel{\rightharpoonup }{\mathrm{BO}}$，所以
$\stackrel{\rightharpoonup }{\mathrm{OC}}\cdot (\stackrel{\rightharpoonup }{\mathrm{AO}}-\stackrel{\rightharpoonup }{\mathrm{BO}})=0$，化简得
$\stackrel{\rightharpoonup }{\mathrm{OC}}\cdot \stackrel{\rightharpoonup }{\mathrm{AO}}=\stackrel{\rightharpoonup }{\mathrm{OC}}\cdot \stackrel{\rightharpoonup }{\mathrm{BO}}$，即
$\stackrel{\rightharpoonup }{\mathrm{OC}}\cdot \stackrel{\rightharpoonup }{\mathrm{OA}}=\stackrel{\rightharpoonup }{\mathrm{OB}}\cdot \stackrel{\rightharpoonup }{\mathrm{OC}}$，同理可证
$\stackrel{\rightharpoonup }{\mathrm{OA}}\cdot \stackrel{\rightharpoonup }{\mathrm{OB}}=\stackrel{\rightharpoonup }{\mathrm{OB}}\cdot \stackrel{\rightharpoonup }{\mathrm{OC}}$，即
$\stackrel{\rightharpoonup }{\mathrm{OA}}\cdot \stackrel{\rightharpoonup }{\mathrm{OB}}=\stackrel{\rightharpoonup }{\mathrm{OB}}\cdot \stackrel{\rightharpoonup }{\mathrm{OC}}=\stackrel{\rightharpoonup }{\mathrm{OC}}\cdot \stackrel{\rightharpoonup }{\mathrm{OA}}$。
反之也可证，当
$\stackrel{\rightharpoonup }{\mathrm{OA}}\cdot \stackrel{\rightharpoonup }{\mathrm{OB}}=\stackrel{\rightharpoonup }{\mathrm{OB}}\cdot \stackrel{\rightharpoonup }{\mathrm{OC}}=\stackrel{\rightharpoonup }{\mathrm{OC}}\cdot \stackrel{\rightharpoonup }{\mathrm{OA}}$时，O为ΔABC垂心。