# 莫雷角三分線定理

## 證明

### 引理

$\sin3\theta \equiv 4\sin\theta\sin(60^\circ+\theta)\sin(120^\circ+\theta)$

### 引理證明

$\sin3\theta = 3\sin\theta-4\sin^3\theta$
$= \sin\theta(3-4\sin^2\theta)= \sin\theta(3\cos^2\theta-\sin^2\theta)$
$= \sin\theta(\sqrt{3}\cos\theta+\sin\theta)(\sqrt{3}\cos\theta-\sin\theta)$
$= 4\sin\theta(\tfrac{\sqrt{3}}{2}\cos\theta+\tfrac{1}{2}\sin\theta)(\tfrac{\sqrt{3}}{2}\cos\theta-\tfrac{1}{2}\sin\theta)$
$= 4\sin\theta\sin(60^\circ+\theta)\sin(120^\circ+\theta)$

### 定理證明

$\triangle ABC$中：

$\alpha$$\angle A$的三等分角
$\beta$$\angle B$的三等分角
$\gamma$$\angle C$的三等分角

$\angle BXC = 120^\circ+\alpha$
$\angle CYA = 120^\circ+\beta$
$\angle AZB = 120^\circ+\gamma$

$\sin (120^\circ+\beta) = \frac{\overline{AC}\sin \gamma}{\overline{AY}}$
$\sin (120^\circ+\gamma) = \frac{\overline{AB}\sin \beta}{\overline{AZ}}$

$\angle EXF = \alpha$
$\angle XEF = 60^\circ+\beta \Rightarrow \sin (60^\circ+\beta) = \tfrac{\overline{XD}}{\overline{XE}}$
$\angle XFE = 60^\circ+\gamma \Rightarrow \sin (60^\circ+\gamma) = \tfrac{\overline{XD}}{\overline{XF}}$

$\overline{AB}\sin 3\beta = \overline{AC}\sin 3\gamma$

$\overline{AB}4\sin\beta\sin(60^\circ+\beta)\sin(120^\circ+\beta) = \overline{AC}4\sin\gamma\sin(60^\circ+\gamma)\sin(120^\circ+\gamma)$
$\overline{AB}\sin\beta\frac{\overline{XD}}{\overline{XE}}\frac{\overline{AC}\sin \gamma}{\overline{AY}} = \overline{AC}\sin\gamma\frac{\overline{XD}}{\overline{XF}}\frac{\overline{AB}\sin \beta}{\overline{AZ}}$

$\frac{\overline{XE}}{\overline{XF}} = \frac{\overline{AZ}}{\overline{AY}} \Rightarrow \triangle XEF \approx \triangle AZY$

$\angle AZY = \angle XEF = 60^\circ+\beta$
$\angle AYZ = \angle XFE = 60^\circ+\gamma$

$\angle BZX = 60^\circ+\alpha$
$\angle CYX = 60^\circ+\alpha$