User talk:Jhoshen1/Sandbox

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

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