定理:已知$P$为$\triangle ABC$内一点,则$S_{\triangle PBC}\cdot \overrightarrow{PA}+S_{\triangle PCA}\cdot \overrightarrow{PB}+S_{\triangle PAB}\cdot \overrightarrow{PC}=0$
证明:在$PA$、$PB$、$PC$上分别取点$A^{\prime}$、$B^{\prime}$、$C^{\prime}$使得$S_{\triangle PA^{\prime}B^{\prime}}=S_{\triangle PB^{\prime}C^{\prime}}=S_{\triangle PC^{\prime}A^{\prime}}$
记$\dfrac{PA^{\prime}}{PA}=a$,$\dfrac{PB^{\prime}}{PB}=b$,$\dfrac{PC^{\prime}}{PC}=c$
由共边定理易得$P$是$\triangle A^{\prime}B^{\prime}C^{\prime}$的重心,则
$\overrightarrow{PA^{\prime}}+\overrightarrow{PB^{\prime}}+\overrightarrow{PC^{\prime}}=0$
所以$a\overrightarrow{PA}+b\overrightarrow{PB}+c\overrightarrow{PC}=0$
由面积公式易得$\dfrac{S_{\triangle PA^{\prime}B^{\prime}}}{S_{\triangle PAB}}=\dfrac{PA^{\prime}\cdot PB^{\prime}}{PA\cdot PB}=ab$,则$S_{\triangle PA^{\prime}B^{\prime}}=abS_{\triangle PAB}$
同理$S_{\triangle PB^{\prime}C^{\prime}}=bcS_{\triangle PBC}$,$S_{\triangle PC^{\prime}A^{\prime}}=caS_{\triangle PCA}$
所以$abS_{\triangle PAB}=bcS_{\triangle PBC}=caS_{\triangle PCA}$
那么$S_{\triangle PAB}:S_{\triangle PBC}:S_{\triangle PCA}=c:a:b$
因此$S_{\triangle PBC}\cdot \overrightarrow{PA}+S_{\triangle PCA}\cdot \overrightarrow{PB}+S_{\triangle PAB}\cdot \overrightarrow{PC}=0$