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Proof 2
This proof of Morley's Theorem is provided because the first proof given here is incomplete. That proof was given for the converse of Morley's Theorem and not the theorem itself. In this proof, we shall follow the Dijkstra's technique and complete the proof.

Given Triangle $\triangle A'B'C'$


 * [[File:Morleys-Theorem-Fig1.jpg]]

Constructed Triangle $\triangle ABC$


 * [[File:Morleys-Theorem-Fig2.jpg]]

We shall prove that triangle $ \triangle X'Y'Z'$ is congruent to triangle $\triangle XYZ $ where $\triangle XYZ $ is an equilateral triangle. $ \triangle X'Y'Z'$ is embedded in the given triangle $\triangle A'B'C'$ whereas $ \triangle XYZ $ is embedded in the constructed triangle $\triangle ABC$.

Construct $\triangle XYZ$, 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 $\triangle ABC$

Because

it follows that:
 * if $\angle BAX = \alpha + x$ then $\angle ABX = \beta - x$

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 $x = 0$. 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.

Under the assumption $x > 0 $ and $\beta - x > 0 $
 * $\map \sin {\alpha + x} > \sin \alpha$
 * $\sin \beta > \map \sin {\beta - x} $
 * $\leadsto \map \sin {\alpha + x} \sin \beta > \sin \alpha \map \sin {\beta - x}$
 * $\leadsto \dfrac {\map \sin {\alpha + x} } {\map \sin {\beta - x} }

> \dfrac {\sin \alpha} {\sin \beta}$ which contradicts $(4)$. If we now assume $x < 0 $ and $\alpha + x > 0 $, we obtain
 * $\dfrac {\map \sin {\alpha + x} } {\map \sin {\beta - x} }

< \dfrac {\sin \alpha} {\sin \beta}$ Again we have a contradiction with $(4)$. Thus we conclude that $x = 0$.

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' } $

Equations $(9)$, $(10)$ and $\angle XBZ = \angle X'B'Z' = \beta $ establish the following triangle congruence:
 * $ \triangle X'B'Z' \cong \triangle XBZ\;\;\;$ Side-Angle-Side

Similarly $(11)$, $(12)$ and $\angle YCZ = \angle Y'C'Z' = \gamma$ yield:
 * $ \triangle Y'C'Z' \cong \triangle YCZ\;\;\;$ Side-Angle-Side

Thus
 * $X'Z' = XZ $

and
 * $Y'Z' = YZ$

Because by construction $XZ=YZ=XY=X'Y'$, we have
 * $X'Z' = XZ = XY = X'Y' $
 * $Y'Z' = YZ = XY = X'Y' $
 * $\therefore Y'Z' = X'Z' = X'Y' $

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