Morley's Trisector Theorem/Proof 2

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 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:

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

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

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

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

--- End of the Trigonometric Proof for $ \triangle A'X'B' \sim \triangle AXB$ ---

Geometric proof for $ \triangle A'X'B' \sim \triangle AXB$
 * 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 $

The outline for proving case [1] is given as follows:
 * Construct $\triangle AYX''$.
 * Construct $\triangle BXZ'' $.
 * Prove that $\triangle AXB'' \cong \triangle AXB$.
 * Prove that $\triangle A'X'B' \sim \triangle AXB$.

We construct triangle $\triangle AYX$ such that $\triangle AYX \cong \triangle AYX $.

Next, we construct triangle $\triangle BXZ'' $ as follows:

We shall now prove that that $\triangle AXB'' \cong \triangle AXB$. We note that:

Consequently,

and given that:

yields the desired result:

For case [2], where $\gamma > 30 \degrees $, $\angle ZXY'' =  2 \gamma - 60 \degrees $ and $\triangle XZY$ is external to $\triangle AYX$ and $\triangle BXZ'' $.

In case [3], where $\gamma = 30 \degrees $, $\angle ZXY = 0 \degrees $ and $\triangle XZY$ degenerates into line segment $XZ$.

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 that $\triangle  A'Y'C' \sim \triangle AYC$.

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

--- End of the Geometric Proof for''' $\triangle A'X'B' \sim \triangle AXB$ ---

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.