User talk:Jhoshen1/Sandbox

Geometric proof for $\triangle X'A'B' \sim \triangle XAB$
 * [[File:Morleys-Theorem-Auxiliary-Triangles.png]]

We consider 3 different cases for the geometric proof
 * (1) $\gamma < 30 \degrees $
 * (2) $\gamma > 30 \degrees $
 * (3) $\gamma = 30 \degrees $

For the first case, we construct triangle $\triangle AYX$ such that $\triangle AYX \cong \triangle AYX $.

Following that, we an construct isosceles triangle $YXZ''$ such that

Next triangle $\triangle BXZ'' $ is constructed where

This congruency yields:

Given that

we prove the desired triangle similarity:

For the case (2), where $\gamma > 30 \degrees $, the isosceles $XZY$ is external to triangles $\triangle AYX$ and $\triangle BXZ'' $. In case (3), where $\gamma = 30 \degrees $, the legs of the isosceles $XZY''$ degenerates into a single line segment. In either case (2) or (3), the proof for $\triangle  X'A'B'  \sim \triangle XAB$ is very similar to the proof for case (1).

In a similar fashion, we can prove that $\triangle Z'B'C' \sim \triangle ZBC$ and that $\triangle  Y'A'C' \sim \triangle YAC$