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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:

Construct $\triangle BXZ$ such that:

Construct $\triangle CYZ$ such that:

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

$\angle AXB$ is calculated as follows

To proceed, it is necessary to prove that $ \triangle X'A'B' \sim \triangle XAB$. We shall provide two alternate proofs for this preposition; a trigonometric proof and a geometric proof.

Trigonometric proof for $ \triangle X'A'B' \sim \triangle XAB$

Applying the Sine Rule for $\triangle XBZ$ and $\triangle XAY$, we have:

Dividing $(1)$ by $(2)$ and noting that $XZ = XY$, we obtain:

Applying the Sine Rule to $\triangle A'B'X' $, we get

Combining $(3)$ and $(4)$, yields:

For $\triangle A'B'X' $, we have:

and we have already shown that:

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

In a similar fashion, we can obtain the following triangle similarities:

These similarities lead to: $\angle CAY = \alpha $, $\angle ACY = \gamma $, $\angle CBZ = \beta $ and $\angle BCZ = \gamma $.

--- End of the Trigonometric proof for $ \triangle X'A'B' \sim \triangle XAB$ ---

Geometric proof for $\triangle X'A'B' \sim \triangle XAB$
 * Morleys-Theorem-Auxiliary-Triangles.png

We consider 3 different cases for the geometric proof
 * [1] $\gamma < 30 \degrees $
 * [2] $\gamma > 30 \degrees $
 * [3] $\gamma = 30 \degrees $

For the first case, we construct triangle $\triangle AYX$ such that $\triangle AYX \cong \triangle AYX $.

Following that, we an construct isosceles triangle $YXZ''$ such that:

Next triangle $\triangle BXZ'' $ is constructed where

We shall now prove that that $\triangle ABX \cong \triangle ABX$ and that $\triangle ABX \sim \triangle A'B'X'$. we note that:

Consequently,

and given that:

we have:

which proves $\triangle X'A'B' \sim \triangle XAB$ similarity

For the case [2], where $\gamma > 30 \degrees $, the isosceles $XZY$ is external to triangles $\triangle AYX$ and $\triangle BXZ'' $. In case [3], where $\gamma = 30 \degrees $, the legs of the isosceles $XZY''$ degenerates into a single line segment. 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 $\triangle Z'B'C' \sim \triangle ZBC$ and that $\triangle  Y'A'C' \sim \triangle YAC$

--- End of the geometric proof for''' $\triangle X'A'B' \sim \triangle XAB$ ---

Because


 * and

we have the following similarity

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:

Furthermore,

In a similar fashion, we can also prove the following triangle similarities:

which yield the following:

By construction:

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