奔驰定理

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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$S_{\triangle PBC}\cdot \overrightarrow{PA}+S_{\triangle PCA}\cdot \overrightarrow{PB}+S_{\triangle PAB}\cdot \overrightarrow{PC}=0$