18 三角形分邊線長公式

屬性

$b^{2}=x^{2}+d^{2}-2dx\cos D\Rightarrow x^{2}=b^{2}-d^{2}-2dx\cos D$
$a^{2}=x^{2}+c^{2}-2cx\cos {D}'=x^{2}+c^{2}-2cx\cos \left( \pi -D \right)$
$=x^{2}+c^{2}+2cx\cos D\Rightarrow x^{2}=a^{2}-c^{2}+2cx\cos D$
$\left\{ \begin{align} & x^{2}=b^{2}-d^{2}-2dx\cos D \\ & x^{2}=a^{2}-c^{2}+2cx\cos D \\ \end{align} \right.\Rightarrow \left\{ \begin{align} & cx^{2}=c\left( b^{2}-d^{2} \right)-2cdx\cos D \\ & dx^{2}=d\left( a^{2}-c^{2} \right)+2cdx\cos D \\ \end{align} \right.$
$\Rightarrow \left( c+d \right)x^{2}=c\left( b^{2}-d^{2} \right)+d\left( a^{2}-c^{2} \right)\Rightarrow x^{2}=\frac{c\left( b^{2}-d^{2} \right)+d\left( a^{2}-c^{2} \right)}{\left( c+d \right)}$