每日一题[1014]殊途同归

在$\triangle ABC$中，$AB=c$，$DC=kAD$，$\angle DBA=\alpha$，$\angle DBC=\beta$，则$BC=$_______．

正确答案是

分析与解　根据题意，有 $$\overrightarrow{BD}=\dfrac{k}{k+1}\overrightarrow {BA}+\dfrac 1{k+1}\overrightarrow {BC},$$ 两边分别与$\overrightarrow{BA}$和$\overrightarrow{BC}$作数量积，可得 $$\begin{aligned}BD\cdot BA\cdot \cos\alpha = \dfrac k{k+1}\cdot BA^2+\dfrac 1{k+1}\cdot BC\cdot BA\cdot\cos(\alpha+\beta),\\BD\cdot BC\cdot \cos\beta = \dfrac{k}{k+1}\cdot BA\cdot BC\cdot\cos(\alpha+\beta)+\dfrac 1{k+1}\cdot BC^2 ,\end{aligned}$$ 于是 $$\begin{aligned} BD\cos\alpha&=\dfrac{kc}{k+1}+\dfrac 1{k+1}BC\cos(\alpha+\beta),\\BD\cos\beta &=\dfrac{kc}{k+1}\cos(\alpha+\beta)+\dfrac 1{k+1}BC,\end{aligned}$$ 因此解得 $$BC=kc\cdot \dfrac{\cos\alpha\cos(\alpha+\beta)-\cos\beta}{\cos\beta\cos(\alpha+\beta)-\cos\alpha}.$$

另法　在$\triangle BAD$与$\triangle BCD$中分别应用正弦定理得 $$\begin{cases} \dfrac {AD}{\sin\alpha}=\dfrac c{\sin\angle BDA},\\\dfrac {DC}{\sin\beta}=\dfrac {BC}{\sin\angle BDC},\end{cases}$$ 两式相比得 $$\dfrac {AD}{DC}\cdot\dfrac {\sin\beta}{\sin\alpha}=\dfrac c{BC},$$ 于是得到$BC=\dfrac {kc\sin\alpha}{\sin\beta}$．