User talk:Jhoshen1/Sandbox

Geometric proof for $\triangle X'A'B' \sim \triangle XAB$ 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 $XZY''$ such that

Next triangle $\triangle BXZ'' $ is constructed as follows

This congruency yields:

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

$\triangle X'A'B' \sim \triangle XAB$