叶卢庆的博客

利用向量积解一道平面几何题

下面这道题目来自彭翕成的博文:
图1

题目:如图1,$\triangle ABC$中,$X,Y,Z$分别是$BC,CA,AB$上的一点,$D,E,F$分别是$AX,BY,CZ$的中点,求证$S_{\triangle XYZ}=4S_{\triangle DEF}$.

不同于彭翕成的坐标法结合行列式法,下面我使用向量积将其证明.

证明:设$\overrightarrow{AZ}=\lambda_1\overrightarrow{AB}$,$\overrightarrow{BX}=\lambda_2\overrightarrow{BC}$,$\overrightarrow{CY}=\lambda_3\overrightarrow{CA}$.则\begin{align*}\overrightarrow{FD}\times \overrightarrow{FE}&=\frac{1}{2}(\overrightarrow{ZX}-\overrightarrow{AC})\times \frac{1}{2}(\overrightarrow{ZY}-\overrightarrow{BC})\\&=\frac{1}{4}(\overrightarrow{BX}-\overrightarrow{BZ}+\overrightarrow{BA}-\overrightarrow{BC})\times (\overrightarrow{AY}-\overrightarrow{AZ}-\overrightarrow{BC})\\&=\frac{1}{4}[\lambda_2\overrightarrow{BC}-(1-\lambda_1)\overrightarrow{BA}+\overrightarrow{BA}-\overrightarrow{BC}]\times [(1-\lambda_3)\overrightarrow{AC}+\lambda_1\overrightarrow{BA}-\overrightarrow{BC}]\\&=\frac{1}{4}[\lambda_1\overrightarrow{BA}-(1-\lambda_2)\overrightarrow{BC}]\times [(1-\lambda_3)(\overrightarrow{BC}-\overrightarrow{BA})+\lambda_1\overrightarrow{BA}-\overrightarrow{BC}]\\&=\frac{1}{4}[\lambda_1\overrightarrow{BA}-(1-\lambda_2)\overrightarrow{BC}]\times [(\lambda_3+\lambda_1-1)\overrightarrow{BA}-\lambda_3\overrightarrow{BC}]\\&=\frac{1}{4}(\lambda_1+\lambda_2+\lambda_3-\lambda_1\lambda_2-\lambda_2\lambda_3-\lambda_3\lambda_1-1)\overrightarrow{BA}\times \overrightarrow{BC}.\end{align*}而且\begin{align*} \overrightarrow{YZ}\times\overrightarrow{YX}&=(\overrightarrow{AZ}-\overrightarrow{AY})\times (\overrightarrow{YC}+\overrightarrow{CX})\\&=[-\lambda_1\overrightarrow{BA}-(1-\lambda_3)\overrightarrow{AC}]\times [\lambda_3\overrightarrow{AC}-(1-\lambda_2)\overrightarrow{BC}]\\&=[-\lambda_1\overrightarrow{BA}-(1-\lambda_3)(\overrightarrow{BC}-\overrightarrow{BA})]\times [\lambda_3(\overrightarrow{BC}-\overrightarrow{BA})-(1-\lambda_2)\overrightarrow{BC}] \\&=[(-\lambda_1-\lambda_3+1)\overrightarrow{BA}-(1-\lambda_3)\overrightarrow{BC}]\times [-\lambda_3\overrightarrow{BA}+(\lambda_2+\lambda_3-1)\overrightarrow{BC}] \\&=(\lambda_1+\lambda_2+\lambda_3-\lambda_1\lambda_2-\lambda_2\lambda_3-\lambda_3\lambda_1-1)\overrightarrow{BA}\times\overrightarrow{BC}.\end{align*}因此可得$$\overrightarrow{FD}\times \overrightarrow{FE}=\frac{1}{4}\overrightarrow{YZ}\times \overrightarrow{YX},$$于是,$S_{\triangle XYZ}=4S_{\triangle DEF}$.