User talk:Jhoshen1/Sandbox

The proof given here for Morley's theorem is very elegant. However, it is incomplete. It starts with an equilateral triangle and proves that the intersecting lines trisect the vertices of the triangle, which is the converse Morley's theorem. In the following writeup I will complete the proof. I will also modify the existing proof to clarify some issues and make it more rigorous. And if there is an agreement on my approach, I will modify the existing proof accordingly. (However, I need help with figures)

Please note $\triangle A'B'C'$ is the given triangle and $\triangle ABC$ is the constructed triangle

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 * Morleys-Theorem-2.png


 * Morleys-Theorem-Dijkstra-Proof.png

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


 * $\therefore \angle YCZ = 180 \degrees -( 60 \degrees + \beta + 60 \degrees + \alpha) = \gamma$

Hence, $XY = YZ = XZ = X'Y'$ and therefore $\triangle XYZ $ is an equilateral triangle.

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 $, 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 $

In a similar fashion, it can be shown that $\angle CAY = \alpha $, $\angle CBZ = \beta $, $\angle ACY = \gamma $ and $\angle BCZ = \gamma $.

We shall now prove that $\triangle X'Y'Z' $ is an equilateral triangle. We shall start with proving that $\triangle ABX \cong \triangle A'B'X'$


 * $\therefore \triangle ABX \cong \triangle A'B'X'\;\;\; $ angle-side-angle
 * $\leadsto BX = B'X' $
 * $\leadsto AB = A'B' $

Next we prove that $\triangle ABC \cong \triangle A'B'C' $
 * $AB = A'B' $
 * $\angle CAB = \angle C'A'B' = 3 \alpha$
 * $\angle ABC = \angle A'B'C' = 3 \beta $


 * $\therefore \triangle ABC \cong \triangle A'B'C'\;\;\; $ angle-side-angle
 * $\leadsto AC = A'C' $
 * $\leadsto BC = B'C' $

We now prove that $\triangle B'C'Z' \cong \triangle BCZ$


 * $\therefore B'C'Z' \cong \triangle BCZ\;\;\;$ angle-side-angle
 * $\leadsto B'Z' = BZ $
 * $\leadsto C'Z' = CZ $

Similarly, we prove that $\triangle A'C'Y' \cong \triangle ACY$

$\therefore \triangle A'C'Y' \cong \triangle ACY\;\;\;$ angle-side-angle
 * $\leadsto A'Y' = AY $
 * $\leadsto C'Y' = CY $

By proving that $C'Y' = CY $, $B'Z' = BZ $, $C'Z' = CZ $ and $B'X' = BX $, we can prove that $\triangle C'Y'Z' \cong \triangle CYZ$ and $\triangle B'X'Z' \cong \triangle BXZ$

For $\triangle C'Y'Z'$, we have


 * $\therefore \triangle C'Y'Z' \cong \triangle CYZ\;\;\;$ side-angle-side
 * $\leadsto Y'Z' = YZ $

For $\triangle B'X'Z'$, we have


 * $\therefore \triangle B'X'Z' \cong \triangle BXZ\;\;\;$ side-angle-side
 * $\leadsto X'Z' = XZ $

Because $ Y'Z' = YZ = XY = X'Y' $ and $ X'Z' = XZ = XY = X'Y' $, we have
 * $ Y'Z' = X'Z' = X'Y' $

which proves that $\triangle X'Y'Z$ is an equilateral triangle