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 would need help with figures)


 * As the existing proof is a reproduction of an existing publication in the literature, it is preferred if it is kept as it is, unless you want to go to the original source and investigate what may need to be done to complete it according to whatever it says in there. I confess I wasn't paying much attention when I posted this one up, so I may have got all sorts of things wrong.


 * So if you do come up with a better way of working this proof, which essentially turns it into a different proof, the post it up as a completely separate proof rather than change this one. --prime mover (talk) 21:22, 9 February 2021 (UTC)

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 * Morleys-Theorem-Dijkstra-Proof.png

Construct $\triangle AXY$ such that


 * $\therefore \angle XAY = \alpha$

Construct $\triangle AXY$ such that


 * $\therefore \angle XBZ = \beta$

Construct $\triangle CYZ$ such that


 * $\therefore \angle YCZ = \gamma$
 * $\alpha + \beta + \gamma = 60 \degrees$
 * and $\triangle XYZ $ is an equilateral triangle.

Because $\angle AXB = 180 \degrees - \paren {\alpha + \beta}$, it follows that:
 * if $\angle BAX = \alpha + x$ then $\angle ABX = \beta - x$

Using the Sine Rule, we have:


 * $ \dfrac {\map \sin {\alpha + x} } {\map \sin {\beta - x} } = \dfrac {BX} {AX} $

Where
 * $ BX =  XZ \map \sin {60 \degrees + \gamma} / \sin \beta $

and
 * $ AX = XY \map \sin {60 \degrees + \gamma } / \sin \alpha $

Also given that $XZ=XY$, yields
 * $ \dfrac {\map \sin {\alpha + x} } {\map \sin {\beta - x} } = \dfrac {\sin \alpha} {\sin \beta} $

In the range in which these angles lie, the of the above is a strictly increasing function of $x$.

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 $.

Using triangle congruencies, we shall prove that $\triangle X'Y'Z' $ is an equilateral triangle.


 * $AX=A'X'\;\;\;$ by construction
 * $\angle BAX = \angle B'A'X' = \alpha $
 * $\angle BXA = \angle B'X'A' =  180 \degrees - \alpha - \beta$
 * $\therefore \triangle ABX \cong \triangle A'B'X'\;\;\; $ angle-side-angle
 * $\leadsto B'X' = BX $
 * $\leadsto A'B' = AB $


 * $\angle C'A'B = \angle CAB = 3 \alpha$
 * $\angle A'B'C = \angle ABC = 3 \beta $
 * $\therefore \triangle A'B'C' \cong \triangle ABC\;\;\; $ angle-side-angle
 * $\leadsto A'C' = AC $
 * $\leadsto B'C' = BC $