“奔驰”定理的3种证明，并应用于三角形5心计算：内心、外心、重心、垂心、旁心

首先介绍奔驰定理的几种证明方法，然后应用于计算三角形的5心：内心、外心、重心、垂心、旁心。

已知P是△ABC内一点，求证：

S_{\triangleBPC}\cdot\overrightarrow{PA}+S_{\triangleAPC}\cdot\overrightarrow{PB}+S_{\triangleAPB}\cdot\overrightarrow{PC}=\overrightarrow{0}\\

延长AP交BC于Q，易知

\begin{aligned}\frac{|PA|}{|PQ|}&=\frac{S_{\triangleAPB}}{S_{\triangleQPB}}=\frac{S_{\triangleAPC}}{S_{\triangleQPC}}\\&=\frac{S_{\triangleAPB}+S_{\triangleAPC}}{S_{\triangleQPB}+S_{\triangleQPC}}\\&=\frac{S_{\triangleAPB}+S_{\triangleAPC}}{S_{\triangleCPB}}\\\therefore\quad\frac{|PA|}{|AQ|}&=\frac{S_{\triangleAPB}+S_{\triangleAPC}}{S_{\triangleABC}}\end{aligned}\\

所以

\overrightarrow{AP}=\frac{S_{\triangleAPB}+S_{\triangleCPA}}{S_{\triangleABC}}\cdot\overrightarrow{AQ}\\

根据共线向量，可得

\overrightarrow{AQ}=\frac{S_{\triangleAPB}}{S_{\triangleAPB}+S_{\triangleCPA}}\cdot\overrightarrow{AC}+\frac{S_{\triangleCPA}}{S_{\triangleAPB}+S_{\triangleCPA}}\cdot\overrightarrow{AB}\\

所以

\begin{aligned}\overrightarrow{AP}&=\frac{S_{\triangleAPB}}{S_{\triangleABC}}\cdot\overrightarrow{AC}+\frac{S_{\triangleCPA}}{S_{\triangleABC}}\cdot\overrightarrow{AB}\\-\overrightarrow{PA}&=\frac{S_{\triangleAPB}}{S_{\triangleABC}}\cdot(\overrightarrow{PC}-\overrightarrow{PA})+\frac{S_{\triangleCPA}}{S_{\triangleABC}}\cdot(\overrightarrow{PB}-\overrightarrow{PA})\\-S_{\triangleABC}\cdot\overrightarrow{PA}&=S_{\triangleAPB}\cdot(\overrightarrow{PC}-\overrightarrow{PA})+S_{\triangleCPA}\cdot(\overrightarrow{PB}-\overrightarrow{PA})\\-S_{\triangleABC}\cdot\overrightarrow{PA}&=S_{\triangleAPB}\cdot\overrightarrow{PC}+S_{\triangleCPA}\cdot\overrightarrow{PB}-(S_{\triangleAPB}+S_{\triangleCPA})\cdot\overrightarrow{PA}\\(S_{\triangleAPB}+S_{\triangleCPA})\cdot\overrightarrow{PA}-S_{\triangleABC}\cdot\overrightarrow{PA}&=S_{\triangleAPB}\cdot\overrightarrow{PC}+S_{\triangleCPA}\cdot\overrightarrow{PB}\\-S_{\triangleCPB}\cdot\overrightarrow{PA}&=S_{\triangleAPB}\cdot\overrightarrow{PC}+S_{\triangleCPA}\cdot\overrightarrow{PB}\\\end{aligned}\\

所以

S_{\triangleBPC}\cdot\overrightarrow{PA}+S_{\triangleAPC}\cdot\overrightarrow{PB}+S_{\triangleAPB}\cdot\overrightarrow{PC}=\overrightarrow{0}\\

设\angleAPB=\alpha,\angleAPC=\beta,PA=x,PB=y,PC=z，待证等式左侧与向量\overrightarrow{PA}作数量积

\begin{aligned}&(S_{A}\overrightarrow{PA}+S_{B}\overrightarrow{PB}+S_{C}\overrightarrow{PC})\cdot\overrightarrow{PA}\\=&\frac{1}{2}yz\sin[2\pi-(\alpha+\beta)]\cdotx^{2}+\frac{1}{2}zx\sin\beta\cdotxy\cos\alpha+\frac{1}{2}xy\sin\alpha\cdotzx\cos\beta\\=&\frac{1}{2}x^{2}yz[-\sin(\alpha+\beta)+\cos\alpha\sin\beta+\sin\alpha\cos\beta]\\=&0\end{aligned}\\

同理，可得

(S_{A}\overrightarrow{PA}+S_{B}\overrightarrow{PB}+S_{C}\overrightarrow{PC})\cdot\overrightarrow{PB}=0\\(S_{A}\overrightarrow{PA}+S_{B}\overrightarrow{PB}+S_{C}\overrightarrow{PC})\cdot\overrightarrow{PC}=0\\

而向量\overrightarrow{PA},\overrightarrow{PB},\overrightarrow{PC}不共线，所以

S_A\overrightarrow{PA}+S_B\overrightarrow{PB}+S_C\overrightarrow{PC}=\overrightarrow{0}\\

记

\overrightarrow{s}=S_{\triangleBPC}\cdot\overrightarrow{PA}+S_{\triangleAPC}\cdot\overrightarrow{PB}+S_{\triangleAPB}\cdot\overrightarrow{PC}\\

所以

\overrightarrow{s}\times\overrightarrow{PA}=S_{\triangleBPC}\cdot\overrightarrow{PA}\times\overrightarrow{PA}+S_{\triangleAPC}\cdot\overrightarrow{PB}\times\overrightarrow{PA}+S_{\triangleAPB}\cdot\overrightarrow{PC}\times\overrightarrow{PA}\\

因为

\overrightarrow{PA}\times\overrightarrow{PA}=\overrightarrow{0}\\|S_{\triangleAPC}\cdot\overrightarrow{PB}\times\overrightarrow{PA}|=S_{\triangleAPC}\cdot|\overrightarrow{PB}|\cdot|\overrightarrow{PA}|\cdot\sin\angleBPA=S_{\triangleAPC}\cdotS_{\triangleAPB}\\|S_{\triangleAPB}\cdot\overrightarrow{PC}\times\overrightarrow{PA}|=S_{\triangleAPB}\cdot|\overrightarrow{PC}|\cdot|\overrightarrow{PA}|\cdot\sin\angleCPA=S_{\triangleAPB}\cdotS_{\triangleAPC}\\\therefore|S_{\triangleAPC}\cdot\overrightarrow{PB}\times\overrightarrow{PA}|=|S_{\triangleAPB}\cdot\overrightarrow{PC}\times\overrightarrow{PA}|\\

注意两向量S_{\triangleAPC}\cdot\overrightarrow{PB}\times\overrightarrow{PA},S_{\triangleAPB}\cdot\overrightarrow{PC}\times\overrightarrow{PA}的方向相反，所以

\overrightarrow{s}\times\overrightarrow{PA}=\overrightarrow{0}\\

同理

\overrightarrow{s}\times\overrightarrow{PB}=\overrightarrow{0}\\\overrightarrow{s}\times\overrightarrow{PC}=\overrightarrow{0}\\

不难确定

S_{\triangleBPC}\cdot\overrightarrow{PA}+S_{\triangleAPC}\cdot\overrightarrow{PB}+S_{\triangleAPB}\cdot\overrightarrow{PC}=\overrightarrow{0}\\

上述定理中的点P具有任意性，所以当P为