本帖最后由 isee 于 2018-4-17 17:28 编辑
只能说,论坛里的朋友水准真的很高,不仅秒了,且看透了问题的实际。
考场硬算。
另解
$$\vv {BD}=x\vv{BA}+y\vv{BC}\iff \vv {CD}=x\vv{BA}+(y-1)\vv{BC}.$$
两边分别点乘$\vv{BA},\vv{BC}$可得到
$$\left\{\begin{aligned} \vv{CD}\cdot {BA}=xBA^2,\\\vv{CD}\cdot \vv{BC}=(y-1)BC^2 \end{aligned}\right.$$
设$\angle ACB=\alpha$,则$\angle ACD=\pi-\alpha$,方便书写,记$BC=a,AB=c,\frac{CD}{AC}=m$,
$$\left\{\begin{aligned} x=CD\sin\alpha=m,\\y=m+1\end{aligned}\right.\Rightarrow x-y=-1.$$ |