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Proof 2
Given Triangle $\triangle A'B'C'$
 * [[File:Morleys-Theorem-Fig1xxxx.png]]

Constructed Triangle $\triangle ABC$
 * [[File:Morleys-Theorem-Fig2xx.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.

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 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' $ (the given triangle), we get

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

For $\triangle A'B'X' $, we have
 * $\angle A'X'B' = 180 \degrees -  \beta  - \alpha $

and we have already shown that
 * $\angle AXB = 180 \degrees -  \beta -\alpha$
 * $\leadsto \angle AXB = \angle A'X'B'$
 * $\therefore \triangle A'B'X' \sim \triangle ABX ;\;\;\;\;\;\;$ Side-Angle-Side

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

In a similar fashion, it can be shown that $\angle CAY = \alpha $, $\angle ACY = \gamma $, $\angle CBZ = \beta $ and $\angle BCZ = \gamma $, which leads to the following triangle similarities:
 * $\triangle A'C'Y' \sim \triangle ACY$


 * $\triangle B'C'Z' \sim \triangle BCZ$

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' \;\;\;\;\;\;\;$ Angle-Angle

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
 * $\dfrac {AX} { A'X' } = \dfrac {AB} { A'B' } $
 * $\dfrac {AY} { A'Y' } = \dfrac {AC} { A'C' } = \dfrac {AB} { A'B' } $


 * $\leadsto \dfrac {AX} { A'X' } = \dfrac {AY} { A'Y' } $

Furthermore,
 * $\angle XAY = \angle X'A'Y' = \alpha $
 * $\therefore \triangle XAY \sim \triangle X'A'Y' \;\;\;\;\;\;\;$ Side-Angle-Side
 * $\leadsto \dfrac { XY } { X'Y' } = \dfrac {AX} { A'X' } = \dfrac {AB} { A'B' } $

In a similar fashion, we can also prove the following triangle similarities
 * $\triangle XBZ \sim \triangle X'B'Z' $
 * $\triangle YCZ \sim \triangle Y'C'Z' $

which together with $ \triangle ABC \sim \triangle A'B'C' $ yield the following
 * $\dfrac {XZ} { X'Z' } = \dfrac { BX } { B'X' } = \dfrac {AB} { A'B' }$
 * $ \dfrac {YZ} { Y'Z' } = \dfrac { CY } { C'Y' } = \dfrac {AC} { A'C' } = \dfrac {AB} { A'B' }$
 * $\therefore \dfrac {XY} { X'Y' } = \dfrac {XZ} { X'Z' } = \dfrac {YZ} { Y'Z' } $

By construction
 * $ XY = XZ = YZ $
 * $\therefore \dfrac {XY} { X'Y' } = \dfrac {XY} { X'Z' } = \dfrac {XY} { Y'Z' } $


 * $\leadsto X'Y' = X'Z' = Y'Z' $

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