莫雷角三分线定理

引理

由三倍角公式及和差公式可得出：

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

引理证明

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

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

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

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

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

定理证明

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条角三分线分别为${\displaystyle {\overline {BX}}}$ 、${\displaystyle {\overline {XC}}}$ 、${\displaystyle {\overline {CY}}}$ 、${\displaystyle {\overline {YA}}}$ 、${\displaystyle {\overline {AZ}}}$ 、${\displaystyle {\overline {ZB}}}$ ，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${\displaystyle \alpha +\beta +\gamma =60^{\circ }}$ ，由此等式还可以得出以下三式：

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

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

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

由正弦定理可得出：

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

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

从这里可以得出${\displaystyle \triangle XEF}$ 的三个内角，计算出${\displaystyle \angle XEF}$ 和${\displaystyle \angle XFE}$ 的正弦值：

${\displaystyle \angle EXF=\alpha }$ 

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

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

我们知道：

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

从引理我们可以得出：

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

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

化简后得出：

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

因为${\displaystyle \triangle XEF}$ 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：

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

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

综合以上结果，可得出${\displaystyle \angle XZY=\angle XYZ=60^{\circ }}$ ，因此${\displaystyle \triangle XYZ}$ 是等边三角形

更一般的莫雷角三分线定理由Taylor和Marr于1914年发表，将6条角三分线顺时钟和逆时钟旋转120°，其交点共可得出27个不同的等边三角形。

• 三等分角
• 三角恒等式

• Johnson, R. A. Modern Geometry: An Elementary Treatise on the Geometry of the Triangle and the Circle. Boston, MA: Houghton Mifflin, pp. 253-256, 1929.
• Morley's Miracle — Several proofs of Morley's theorem （页面存档备份，存于互联网档案馆）
• Morleys Theorem （页面存档备份，存于互联网档案馆） MathWorld
• Morley's Trisection Theorem MathPages