TODO: 微博图床已挂, 本篇缺图, 需要找回

在 \(\triangle ABC\) 中, \(\angle A = 4 \angle C\), \(\angle B = 2 \angle C\), 试证: \((BC+CA) \cdot AB = BC \cdot CA\).

Qer: zhangboxin 20171218

三角证法

记 \(\alpha = \angle C = \dfrac\pi7\), 则 \(7\alpha = \pi\). 运用正弦定理, 待证等式等价于
\[
(\sin 2\alpha + \sin 4\alpha) \cdot \sin\alpha = \sin 2\alpha \cdot \sin 4\alpha
\]
而
\[
\begin{aligned}
&(\sin 2\alpha + \sin 4\alpha) \cdot \sin\alpha \\
&= [ \sin (3\alpha - \alpha) + \sin (3\alpha + \alpha) ] \cdot \sin\alpha \\
&= 2\sin 3\alpha \cos \alpha \sin\alpha \\
&= \sin 3\alpha \sin2\alpha \\
&= \sin (\pi-4\alpha) \sin2\alpha \\
&= \sin 2\alpha \cdot \sin 4\alpha
\end{aligned}
\]
得证.

几何证法

还是记 \(\alpha = \angle C\). 如图, 在 \(BA\) 延长线上取点 \(D\) 两点, 满足 \(AD = AC\), 则
\[
\begin{gathered}
\angle ADC = \angle ACD = \dfrac12 \angle CAB = \angle ABC = 2\alpha\\
CD = BC
\end{gathered}
\]
又在 \(AD​\) 延长线上取点 \(E​\), 满足 \(DE = CD = BC​\), 则
\[
\angle DEC = \angle DCE = \dfrac12 \angle ADC = \angle ACB = \alpha
\]

由 \(\angle DEC = \angle ACB\) 知
\[
\begin{aligned}
BC^2 &= BA \cdot BE \\
&= AB \cdot (AB + AD + AE) \\
&= AB \cdot (AB + BC + CA)
\end{aligned}
\]
故
\[
(BC + CA) \cdot AB = BC^2 - AB^2
\]
又过点 \(B\) 作 \(AC\) 的平行线, 交 \(\triangle ABC\) 外接圆于 \(F\), 则四边形 \(ABFC\) 是等腰梯形, 且不难得到 \(CF = AB\), \(\angle CBF = \alpha\), \(\angle BCF = \angle BFC = 3\alpha\), \(BF = BC = AF\). 在圆内接四边形 \(ABFC\) 中运用托勒密定理得
\[
AB \cdot CF + AC \cdot BF = BC \cdot AF
\]
即
\[
AB^2 + AC \cdot BC = BC^2
\]
故
\[
BC^2 - AB^2 = AC \cdot BC
\]
因此
\[
(BC + CA) \cdot AB = AC \cdot BC
\]