Morley's Trisector Theorem/Proof 2

== Theorem ==
== 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 $\triangle A'B'C'$
:
The Constructed Triangle $\triangle ABC$
:
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 by construction $XZ = XY$, 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 $\triangle A'X'B' \sim \triangle AXB$
:
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 $.
Next, we construct triangle $\triangle B' 'X' 'Z' ' $ as follows:
We shall now prove that that $\triangle A' 'X' 'B' ' \cong \triangle AXB$.
 
We note that:
 
Consequently,
and given that:
yields the desired result:
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 $\triangle A'X'B' \sim \triangle AXB$ --- 
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.
== Historical Note ==
Category:Morley's Trisector Theorem