Talk:Morley's Trisector Theorem/Dijkstra's Proof

$\vartriangle$   $\triangle$   $\bigtriangleup$   $\thicksim$  $\sim$  $\cong$ $A {'}$ $A'$

An implicit assumption has been made that $\triangle XYZ$ is an equilateral triangle.

It has not been stated as such or proven as such.

It is possible to construct $\triangle XYZ$ as an equilateral triangle.

However in this case it would be necessary to prove that $\angle AXB = 180 \degrees - \paren {\alpha + \beta}$.


 * Furthermore the following statements are not proven

$\angle AXY = 60 \degrees + \beta $

$ \angle AYX = 60 \degrees + \gamma $

$ \angle BXZ = 60 \degrees + \alpha $

$ \angle BZY = 60 \degrees + \gamma $

There are additional problems with the proof.
 * I think that this is an inappropriate or incomplete proof for the Morley theorem.
 * I don't have a clue how to fix it --Jhoshen1 (talk) 04:39, 8 February 2021‎ (UTC)


 * You bring up valid points. However I believe that the proof goes as follows:


 * 1. Draw the equilateral triangle $\triangle XYZ$.


 * 2. Draw the triangles $\triangle AXY, \triangle BXZ, \triangle CYZ$ with the angles $\angle XAY = \alpha, \angle AXY = 60 \degrees + \beta, \angle AYX = 60 \degrees + \gamma$, etc.


 * 3. $\angle AXB = 360 \degrees - 60 \degrees - (60 \degrees + \alpha) - (60 \degrees + \beta) = 180 \degrees - \alpha - \beta$


 * 4. Show that $\angle XAB = \alpha, \angle XBA = \beta$


 * 5. Argue by symmetry, each vertex of $\triangle ABC$ is trisected.


 * Since most of the initial steps are by construction, no proof is required. If this is indeed how the proof goes, we just need to fix the typos and the wordings in the current proof. --RandomUndergrad (talk) 05:31, 8 February 2021 (UTC)


 * I think I have fixed the typos. They probably arose through the labels used in the original not matching those used in my transcription of it. Please check it for accuracy.


 * Incidentally, Jhoshen1: Please sign your posts. --prime mover (talk) 06:39, 8 February 2021 (UTC)