莫雷角三分線定理

由三倍角公式及和差公式可得出：

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

${\displaystyle \sin 3\theta =3\sin \theta -4\sin ^{3}\theta }$

${\displaystyle =\sin \theta (3-4\sin ^{2}\theta )=\sin \theta (3\cos ^{2}\theta -\sin ^{2}\theta )}$

${\displaystyle =\sin \theta ({\sqrt {3}}\cos \theta +\sin \theta )({\sqrt {3}}\cos \theta -\sin \theta )}$

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

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

莫雷角三分線定理證明

在${\displaystyle \triangle ABC}$中：

${\displaystyle \alpha }$是${\displaystyle \angle A}$的三等分角

${\displaystyle \beta }$是${\displaystyle \angle B}$的三等分角

${\displaystyle \gamma }$是${\displaystyle \angle C}$的三等分角

作6條角三分線分別為${\displaystyle {\overline {BX}}}$、${\displaystyle {\overline {XC}}}$、${\displaystyle {\overline {CY}}}$、${\displaystyle {\overline {YA}}}$、${\displaystyle {\overline {AZ}}}$、${\displaystyle {\overline {ZB}}}$，作${\displaystyle D}$、${\displaystyle E}$、${\displaystyle F}$在${\displaystyle {\overline {BC}}}$上，且${\displaystyle {\overline {BC}}\bot {\overline {XD}}}$、${\displaystyle \angle BXE=\angle CXF=60^{\circ }}$

容易得出${\displaystyle \alpha +\beta +\gamma =60^{\circ }}$，由此等式還可以得出以下三式：

${\displaystyle \angle BXC=120^{\circ }+\alpha }$

${\displaystyle \angle CYA=120^{\circ }+\beta }$

${\displaystyle \angle AZB=120^{\circ }+\gamma }$

由正弦定理可得出：

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

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

從這裡可以得出${\displaystyle \triangle XEF}$的三個內角，計算出${\displaystyle \angle XEF}$和${\displaystyle \angle XFE}$的正弦值：

${\displaystyle \angle EXF=\alpha }$

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

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

我們知道：

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

從引理我們可以得出：

${\displaystyle {\overline {AB}}4\sin \beta \sin(60^{\circ }+\beta )\sin(120^{\circ }+\beta )={\overline {AC}}4\sin \gamma \sin(60^{\circ }+\gamma )\sin(120^{\circ }+\gamma )}$

${\displaystyle {\overline {AB}}\sin \beta {\frac {\overline {XD}}{\overline {XE}}}{\frac {{\overline {AC}}\sin \gamma }{\overline {AY}}}={\overline {AC}}\sin \gamma {\frac {\overline {XD}}{\overline {XF}}}{\frac {{\overline {AB}}\sin \beta }{\overline {AZ}}}}$

化簡後得出：

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

因為${\displaystyle \triangle XEF}$和${\displaystyle \triangle AYZ}$相似，所以可得出：

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

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

同理可得出：

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

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

綜合以上結果，可得出${\displaystyle \angle XZY=\angle XYZ=60^{\circ }}$，因此${\displaystyle \triangle XYZ}$是等邊三角形