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Proof 2
Given Triangle $\triangle A'B'C'$
 * [[File:Morleys-Theorem-Fig1x.png]]

Constructed Triangle $\triangle ABC$
 * [[File:Morleys-Theorem-Fig2n1.png]]

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 proof applies the Dijkstra's technique for the Morley's Theorem.

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


 * $\therefore \angle XAY = 180 \degrees - (60 \degrees + \beta + 60 \degrees + \gamma) = \alpha$

Construct $\triangle BXZ$ such that


 * $\therefore \angle XBZ = 180 \degrees - (60 \degrees + \alpha + 60 \degrees + \gamma ) = \beta$

Construct $\triangle CYZ$ such that


 * $\therefore \angle YCZ = 180 \degrees -( 60 \degrees + \beta + 60 \degrees + \alpha) = \gamma$

Construct $AB$, $BC$ and $AC$, the sides of $\triangle ABC$

$\angle AXB$ is calculated as follows

Using the Sine Rule, we have:

Substituting $BX$ and $AX$ from $(2)$ and $(3)$ into $(1)$, respectively, and noting that $XZ=XY$, yields

We shall show that for $(4)$ to hold, we must have $\delta = \alpha$ and $\theta = \beta$. In the range in which these angles lie, from $0 \degrees$ to $60 \degrees$, the $sine$ function is a strictly increasing function of its argument.

Because $ \angle AXB = 180 \degrees - \alpha -\beta =  180 \degrees - \delta -\theta $, we have
 * $ \alpha + \beta = \delta + \theta $

If we assume that $\alpha > \delta $ then $\beta < \theta$ and


 * $ \sin \alpha > \sin \delta$ and
 * $\sin \theta > \sin \beta  $
 * $\therefore \sin \alpha \sin \theta > \sin \delta \sin \beta $
 * $\leadsto \dfrac { \sin \alpha } { \sin \beta  }

> \dfrac {\sin \delta} {\sin \theta}$ which contradicts $(4)$.

Similarly, if we assume that $\alpha < \delta $ then $\beta > \theta$, we obtain
 * $\dfrac { \sin \alpha } { \sin \beta  }

< \dfrac {\sin \delta} {\sin \theta}$

Again we have a contradiction with $(4)$. Thus we conclude that $\delta = \alpha$ and $\theta = \beta$.

Consequently, $\angle BAX = \alpha $ and $\angle ABX = \beta $.

Given that $\angle B'A'X' = \alpha $ and $\angle A'B'X' = \beta $, we can establish the following triangle similarity.
 * $\triangle A'B'X' \sim \triangle ABX$


 * $\therefore (5) \;\; \dfrac { AB } { A'B' } = \dfrac { BX } { B'X' } = \dfrac {AX} { A'X' } $

In a similar fashion, we can be shown that $\angle CAY = \alpha $, $\angle ACY = \gamma $, $\angle CBZ = \beta $ and $\angle BCZ = \gamma $, which leads to the following triangle similarities:
 * $\triangle A'C'Y' \sim \triangle ACY$
 * $\therefore (6) \;\; \dfrac { AC } { A'C' } = \dfrac { CY } { C'Y' } = \dfrac {AY} { A'Y' } $


 * $\triangle B'C'Z' \sim \triangle BCZ$
 * $\therefore (7) \;\; \dfrac { BC } { B'C' } = \dfrac { CZ } { C'Z' } = \dfrac {BZ} { B'Z' } $

Because
 * $\angle ABC =\angle ABX + \angle XBZ + \angle ZBC = 3 \beta \;\;\;$ and
 * $\angle BAC =\angle BAX + \angle XAY + \angle CAY = 3 \alpha $

We have the following similarity
 * $ \triangle ABC \sim \triangle A'B'C' $
 * $\therefore (8) \;\; \dfrac { AB } { A'B' } = \dfrac { AC } { A'C' } = \dfrac { BC } { B'C' } $

Combining $(8)$ with $(5)$ and $(6)$ yields:
 * $\dfrac { AB } { A'B' } = \dfrac { AC } { A'C' } = \dfrac {AX} { A'X' } = \dfrac {AY} { A'Y' } $

and
 * $\angle XAY = \angle X'A'Y' = \alpha $
 * $\leadsto \triangle XAY \sim \triangle X'A'Y' $
 * $\leadsto (9) \;\; \dfrac { XY } { X'Y' } = \dfrac {AX} { A'X' } $

Combining $(8)$ with $(5)$ and $(7)$ yields:
 * $\dfrac { AB } { A'B' } = \dfrac { BC } { B'C' } = \dfrac { BX } { B'X' } = \dfrac {BZ} { B'Z' }$

and
 * $\angle XBZ = \angle X'B'Z' = \beta $

$\leadsto \triangle XBZ \sim \triangle X'B'Z' $
 * $\leadsto (10) \;\; \dfrac { XZ } { X'Z' } = \dfrac {BX} { B'X' } $

Combining $(8)$ with $(6)$ and $(7)$ yields:
 * $\dfrac { AC } { A'C' } = \dfrac { BC } { B'C' } = \dfrac { CY } { C'Y' } = \dfrac {CZ} { C'Z' }$

and
 * $\angle YCZ = \angle Y'C'Z' = \gamma $

$\leadsto \triangle YCZ \sim \triangle Y'C'Z' $
 * $\leadsto (11) \;\; \dfrac { YZ } { Y'Z' } = \dfrac {CZ} { C'Z' } $

However, by $(8)$, $(5)$ and $(7)$
 * $ (12) \;\; \dfrac { AB } { A'B' } = \dfrac {AX} { A'X' } = \dfrac {BX} { B'X' }  = \dfrac {CZ} { C'Z' }$

Combining $(9)$, $(10)$ and $(11)$ with $(12)$, yields
 * $ (13) \;\; \dfrac {XY} { X'Y' } = \dfrac {XZ} { X'Z' }  = \dfrac {YZ} { Y'Z' }$

Because by construction
 * $ XY = XZ = YZ $

equation $(13)$ implies that
 * $ X'Y' = X'Z' = Y'Z' $

Hence, $\triangle X'Y'Z'$ is an equilateral triangle, which proves the theorem.