Theorem: Morley's Trisector Theorem

Let $\triangle ABC$ be a triangle.
Let the internal angles of $\triangle ABC$ be trisected.
Let the points where these angle trisectors first intersect be $D$, $E$ and $F$.
:500px
Then $\triangle EDF$ is equilateral.

Proof:
By comparing the '''given triangle''' $\triangle A'B'C'$ with the '''constructed triangle'''  $\triangle ABC $, we shall prove that $\triangle X'Y'Z' \sim \triangle XYZ$ where $\triangle XYZ$ is an equilateral triangle.
'''The Given Triangle''' $\triangle A'B'C'$
:File:Morleys-Theorem-Fig1xxxx.png
'''The Constructed Triangle''' $\triangle ABC$
:File:Morleys-Theorem-Fig2xx.png
We begin by constructing $\triangle XYZ$, an equilateral triangle such that:
:$XY = YZ = XZ$
Noting that $\alpha + \beta + \gamma = 60 \degrees$, we construct $\triangle AXY$ such that: 
{{begin-eqn}}
{{eqn | l = \angle AXY
      | r = 60 \degrees + \beta
}}
{{eqn | l = \angle XZB
      | r = 60 \degrees + \gamma
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \therefore \angle XAY
      | r = 180 \degrees - (60 \degrees + \beta + 60 \degrees + \gamma)
}}
{{eqn | r = 60 \degrees - \beta - 60 \degrees
}}
{{eqn | r = \alpha
}}
{{end-eqn}}
Construct $\triangle BXZ$ such that: 
{{begin-eqn}}
{{eqn | l = \angle ZXB
      | r = 60 \degrees + \alpha
}}
{{eqn | l = \angle XZB
      | r = 60 \degrees + \gamma
}}
{{eqn | l = \therefore \angle XBZ
      | r = \beta
}}
{{end-eqn}}
Construct $\triangle CYZ$ such that: 
{{begin-eqn}}
{{eqn | l = \angle BXZ
      | r = 60 \degrees + \beta
}}
{{eqn | l = \angle AYX
      | r = 60 \degrees + \alpha
}}
{{eqn | l = \therefore \angle YCZ
      | r = \gamma
}}
{{end-eqn}}
Construct $AB$, $BC$ and $AC$, the sides of $\triangle ABC$.
$\angle AXB$ is calculated as follows:
{{begin-eqn}}
{{eqn | l = \angle AXB
      | r = 360 \degrees - ( \angle AXY +\angle YXZ + \angle BXZ )
}}
{{eqn | l = 
      | r = 360 \degrees - ((60 \degrees + \beta) +  60 \degrees + (60 \degrees + \alpha))
}}
{{eqn | l = 
      | r = 360 \degrees - (180 \degrees + \beta  + \alpha)
}}
{{eqn | l = 
      | r = 180 \degrees -  \beta  - \alpha
}}
{{end-eqn}}
To proceed, it is necessary to prove that $ \triangle A'X'B' \sim \triangle AXB$.
We shall provide two alternate proofs for this preposition; a trigonometric proof and a geometric proof.
'''Trigonometric proof for''' $\triangle A'X'B' \sim \triangle AXB$ 
Applying the Sine Rule for $\triangle XBZ$ and $\triangle XAY$, we have:
{{begin-eqn}}
{{eqn | n = 1
      | l = BX
      | r = XZ \map \sin {\angle XZB} / \sin \beta = XZ \map \sin {60 \degrees + \gamma} / \sin \beta
}}
{{eqn | n =2
      | l = AX
      | r = XY \map \sin {\angle AYX } /  \sin \alpha = XY \map \sin {60 \degrees + \gamma } /  \sin \alpha
}}
{{end-eqn}}
Dividing $(1)$ by $(2)$ and noting that by construction $XZ = XY$, we obtain:
{{begin-eqn}}
{{eqn | n = 3
      | l = \dfrac {BX } { AX}
      | r = \dfrac {\sin \alpha} {\sin \beta}
}}
{{end-eqn}}
Applying the Sine Rule to $\triangle A'X'B' $, we get:
{{begin-eqn}}
{{eqn | n = 4
      | l = \dfrac {B'X' } { A'X'}
      | r = \dfrac {\sin \alpha} {\sin \beta}
}}
{{end-eqn}}
Combining $(3)$ and $(4)$, yields:
{{begin-eqn}}
{{eqn | l = \dfrac {BX } { AX}
      | r = \dfrac {B'X'} {A'X'}
}}
{{end-eqn}}
For $\triangle A'X'B' $, we have: 
{{begin-eqn}}
{{eqn | l = \angle A'X'B'
      | r = 180 \degrees -  \beta  - \alpha
}}
{{end-eqn}}
and we have already shown that: 
{{begin-eqn}}
{{eqn | l = \angle AXB
      | r = 180 \degrees -  \beta  - \alpha
}}
{{eqn | l = \leadsto \angle AXB
      | r = \angle A'X'B'
}}
{{eqn | o = \sim
      | l = \therefore \triangle A'X'B'
      | r = \triangle AXB
      | c = Triangles with One Equal Angle and Two Sides Proportional are Similar
}}
{{end-eqn}}
Consequently, $\angle BAX = \alpha $ and $\angle ABX = \beta $.
In a similar fashion, we can obtain the following triangle similarities:
{{begin-eqn}}
{{eqn | o = \sim
      | l = \triangle A'Y'C'
      | r = \triangle CYC
}}
{{eqn | o = \sim
      | l = \triangle B'Z'C'
      | r = \triangle BZC
}}
{{end-eqn}}
--- End of the '''Trigonometric Proof''' for $\triangle A'X'B' \sim \triangle AXB$ --- <br>  
'''Geometric proof for''' $\triangle A'X'B' \sim \triangle AXB$<br>
:600px
We consider 3 different cases for the geometric proof 
:[1] $\gamma < 30 \degrees $
:[2] $\gamma > 30 \degrees $ 
:[3] $\gamma = 30 \degrees $
The outline for proving  case [1] is given as follows: 
::Construct $\triangle A' 'Y' 'X' '$.
::Construct $\triangle B' 'X' 'Z' ' $. 
::Prove that $\triangle A' 'X' 'B' ' \cong \triangle AXB$.
::Prove that $\triangle  A'X'B' \sim \triangle AXB$. 
We construct triangle $\triangle A' 'Y' 'X' '$ such that  $\triangle A' 'Y' 'X' ' \cong \triangle AYX $.
{{begin-eqn }}
{{eqn | l = \therefore \angle X' 'Y' 'A' '
      | r = \angle XYA
}}
{{eqn | r =  60 \degrees + \gamma
      | c = by the construction of $\triangle XYA$
}}
{{end-eqn}}
Next, we  construct triangle $\triangle B' 'X' 'Z' ' $ as follows:
{{begin-eqn }}
{{eqn | l = \angle Z' 'X' 'Y' '
      | r =  60 \degrees - 2 \gamma
      | c = construct $\angle Z' 'X' 'Y' '$
}}
{{eqn | l = \leadsto \angle X' 'Z' 'B' ' '
      | r = 180 \degrees - \angle X' 'Y' 'A' ' - \angle Z' 'X' 'Y' ' '
}}
{{ eqn | r = 180 \degrees - (60 \degrees + \gamma) -  (60 \degrees - 2 \gamma)
}}
{{ eqn | r = 60 \degrees + \gamma
}}
{{ eqn | r = \angle XZB
       | c = by the construction of $\triangle XZB$
}}
{{eqn | l = \therefore X' 'Z' ' 
      | r = X' 'Y' '
      | c = $\triangle Y' 'X' 'Z' '$  is an isosceles by Triangle with Two Equal Angles is Isosceles
}}
{{eqn | r = XY
      | c = by $\triangle A' 'Y' 'X' ' \cong \triangle AYX $
}}
{{eqn | r = XZ
      | c = by equilateral $\triangle XYZ$  construction
}}
{{eqn | l = \angle B' 'X' 'Z' '
      | r = \angle BXZ
      | c = construct $\angle B' 'X' 'Z' '$
}}
{{ eqn | r = 60 \degrees +\alpha 
       | c = by the construction of $\triangle XZB$ 
}}
{{eqn | o = \cong 
      | l = \therefore \triangle B' 'X' 'Z' '
      | r = \triangle BXZ 
      | c = Triangle Angle-Side-Angle Congruence
}}
{{end-eqn}}
We shall now prove that that $\triangle A' 'X' 'B' ' \cong \triangle AXB$.
We note that:
{{begin-eqn}}
{{eqn | l = \angle X' 'A' 'Y' '
      | r = \angle XAY
      | c = by $\triangle A' 'Y' 'X' ' \cong \triangle AYX $
}}
{{eqn | r = \alpha
}}
{{eqn | l = \angle X' 'B' 'Z' '
      | r = \angle XBZ
      | c = by $\triangle B' 'X' 'Z' ' \cong \triangle BXZ$ 
}}
{{eqn | r = \beta
}}
{{eqn | l = \angle AXB
      | r = 180 \degrees - \angle XAY -\angle XBZ
}}
{{eqn | r = 180 \degrees - \alpha - \beta
}}
{{eqn | l = \angle A' 'X' 'B' '
      | r = 180 \degrees - \angle X' 'A' 'Y' ' - \angle X' 'B' 'Z' '
}}
{{eqn | r = 180 \degrees - \alpha - \beta
}}
{{eqn | r = \angle AXB
}}
{{end-eqn}}
{{begin-eqn }}
{{eqn | l = B' 'X' ' 
      | r = BX
      | c = by $\triangle B' 'X' 'Z' ' \cong \triangle BXZ$
}}
{{eqn | l =  A' 'X' '
      | r = AX
      | c = by $\triangle A' 'Y' 'X' ' \cong \triangle AYX $
}}
{{eqn | o = \cong
      | l = \therefore \triangle AXB
      | r = \triangle A' 'X' 'B' '
      | c = Triangle Side-Angle-Side Congruence
}}
{{end-eqn}}
Consequently,
{{begin-eqn }}
{{eqn | l = \angle XBA
      | r = \angle X' 'B' 'A' '
}}
{{eqn | r = \beta 
}}
{{eqn | l = \angle XAB
      | r = \angle X' 'A' 'B' '
}}
{{eqn | r = \alpha
}}
{{end-eqn}}
and given that:
{{begin-eqn }}
{{eqn | l = \angle  X'B'A'
      | r = \beta 
}}
{{eqn | l = \angle  X'A'B' 
      | r = \alpha 
}}
{{end-eqn}}
yields the desired result:
{{begin-eqn }}
{{eqn | o = \sim
      | l = \triangle  A'X'B' 
      | r = \triangle AXB
      | c = Triangles with Two Equal Angles are Similar
}}
{{end-eqn}}
For case [2], where $\gamma > 30 \degrees$, $\angle Z' 'X' 'Y' ' =  2 \gamma - 60 \degrees$ and
$\triangle X' 'Z' 'Y' '$ is external to $\triangle A' 'Y' 'X' '$ and $\triangle B' 'X' 'Z' '$.
In case [3], where $\gamma = 30 \degrees $, $\angle Z' 'X' 'Y' ' =  0 \degrees$ and $\triangle X' 'Z' 'Y' '$ degenerates into line segment $X' 'Z' '$. 
In either case, [2] or [3],
the proof for $\triangle  X'A'B' \sim \triangle XAB$ is very similar to the proof for case [1].
In a similar fashion, we can prove that:
:$\triangle  B'Z'C' \sim \triangle BZC$ and $\triangle  A'Y'C' \sim \triangle AYC$
and that:
:$\angle CAY = \alpha $, $\angle ACY = \gamma $, $\angle CBZ = \beta $ and $\angle BCZ = \gamma $.
--- End of the '''Geometric Proof''' for''' $\triangle  A'X'B' \sim \triangle AXB$ ---<br> 
Because
{{begin-eqn}}
{{eqn | l = \angle ABC 
      | r = \angle ABX + \angle XBZ + \angle ZBC
}}
{{eqn | r = 3 \beta
}}
{{end-eqn}}
:and
{{begin-eqn}}
{{eqn | l = \angle BAC
      | r = \angle BAX + \angle XAY + \angle CAY
}}
{{eqn | r = 3 \alpha
}}
{{end-eqn}}
we have the following similarity:
{{begin-eqn}}
{{eqn | o = \sim
      | l = \triangle ABC
      | r = \triangle A'B'C'
      | c = Triangles with Two Equal Angles are Similar
}}
{{end-eqn}}
Using $ \triangle ABC \sim \triangle A'B'C' $, $\triangle A'B'X' \sim \triangle ABX$ and $\triangle A'C'Y' \sim \triangle ACY$ triangle similarities, we observe that:
{{begin-eqn}}
{{eqn | l = \dfrac {AX} { A'X' }
      | r = \dfrac {AB} { A'B' }
      | c = by $\triangle A'B'X' \sim \triangle ABX$
}}
{{eqn | l = \dfrac {AY} { A'Y' }
      | r = \dfrac {AC} { A'C' } 
      | c = by $\triangle A'C'Y' \sim \triangle ACY$
}}
{{eqn | r = \dfrac {AB} { A'B' } 
      | c = by $ \triangle ABC \sim \triangle A'B'C' $
}}
{{eqn | l = \leadsto \dfrac {AX} { A'X' }
      | r = \dfrac {AY} { A'Y' } 
      | | c = the 2 ratios are equal to $\dfrac {AB} { A'B'}$
}}
{{end-eqn}}
Furthermore,
{{begin-eqn}}
{{eqn | l = \angle XAY 
      | r = \angle X'A'Y'
}}
{{eqn | r = \alpha
}}
{{eqn | o = \sim
      | l = \therefore \triangle XAY 
      | r = \triangle X'A'Y'
      | c = Triangles with One Equal Angle and Two Sides Proportional are Similar
}}
{{eqn | l =  \leadsto \dfrac { XY } { X'Y' }
      | r = \dfrac {AX} { A'X' }
}}
{{eqn | r = \dfrac {AB} { A'B' }
      | c = by $\triangle A'B'X' \sim \triangle ABX$
}}
{{end-eqn}}
In a similar fashion, we can also prove the following triangle similarities:
{{begin-eqn}}
{{eqn | o = \sim
      | l = \triangle XBZ
      | r = \triangle X'B'Z'
}}
{{eqn | o = \sim
      | l = \triangle YCZ
      | r = \triangle Y'C'Z'
}}
{{end-eqn}}
which yield the following:
{{begin-eqn}}
{{eqn | l = \dfrac {XZ} { X'Z' } 
      | r = \dfrac { BX } { B'X' }
      | c = by $\triangle XBZ \sim \triangle X'B'Z'$
}}
{{eqn | r = \dfrac {AB} { A'B' }
      | c = by $\triangle A'B'X' \sim \triangle ABX$ 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \dfrac {YZ} { Y'Z' } 
      | r = \dfrac { CY } { C'Y' }
      | c = by $\triangle YCZ \sim \triangle Y'C'Z'$
}}
{{eqn | r = \dfrac {AC} { A'C' } 
      | c = by $\triangle A'C'Y' \sim \triangle ACY$
}}
{{eqn | r = \dfrac {AB} { A'B'}
      | c = by $\triangle A'B'C' \sim \triangle ABC$ 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \therefore \dfrac {XY} { X'Y' }
      | r = \dfrac {XZ} { X'Z' } = \dfrac {YZ} { Y'Z' }
      | c = the 3 ratios are all equal to $\dfrac {AB} { A'B'}$
}}
{{end-eqn}}
By construction:
{{begin-eqn}}
{{eqn | l = XY  
      | r = XZ = YZ
}}
{{eqn | l = \therefore \dfrac {XY} { X'Y' }
      | r = \dfrac {XY} { X'Z' } = \dfrac {XY} { Y'Z' }
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \leadsto X'Y'  
      | r =  X'Z' = Y'Z'
}}
{{end-eqn}}
Hence, $\triangle X'Y'Z'$ is an equilateral triangle, which proves the theorem.
{{qed}}