# 莫雷角三分線定理

## 證明

### 引理

${\displaystyle \sin 3\theta \equiv 4\sin \theta \sin(60^{\circ }+\theta )\sin(120^{\circ }+\theta )}$

### 引理證明

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

### 定理證明

${\displaystyle \triangle ABC}$中：

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

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

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

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

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

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

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

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

${\displaystyle \angle BZX=60^{\circ }+\alpha }$
${\displaystyle \angle CYX=60^{\circ }+\alpha }$