User talk:Jhoshen1/Sandbox

In the last segment of this proof, we used trigonometry. We shall now provide an alternate purely geometrical proof for this section. We consider 3 different cases where (1) $\gamma < 30 \degrees $, (2) $\gamma > 30 \degrees $, and $\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
 * $ XZ = XY = XY  $
 * $ \angle XZY = \angle XYZ = \angle AYZ = 60 \degrees + \gamma $
 * $\therefore \angle ZXY = 180 \degrees - 2 \angle XYZ = 180 \degrees - 2 (60 \degrees + \gamma) = 60 \degrees - 2 \gamma$

Next triangle $\triangle BXZ'' $ is constructed as follows
 * $\angle BXZ'' = \angle BXZ = 60 \degrees +\alpha $
 * $ XZ = XZ = XY $
 * $ \angle XZB'' = \angle XZB = \angle 60 \degrees + \gamma $


 * $\therefore \triangle BXZ'' \cong \triangle BXZ \;\;\;\;\;$  Angle-Side-Angle

This congruency yields:


 * $\leadsto \angle XBZ'' = \angle XBZ = \beta $

From the construction of $\triangle AYX''$, we have
 * $\angle XAY'' = \alpha $
 * $\therefore \angle AXB = 180 \degrees - \angle XAY - XBZ'' = 180 \degrees - \alpha - \beta$

Also,
 * $\angle AXB = 180 \degrees - \angle XAY - XBZ = 180 \degrees - \alpha - \beta$


 * $\therefore \angle AXB'' = \angle AXB $

Also from the construction of $\triangle AYX''$, we have
 * $ AX = AX$

From the congruency for $\triangle BXZ''$, we get
 * $BX = BX$
 * $\therefore \triangle BXA \cong \triangle BXA'' \;\;\;\;\;$  Side-Angle-Side

This congruency yields:
 * $\angle XBA =\angle XBA'' =\beta $
 * and
 * $\angle XAB =\angle XAB'' =\alpha $

It is also given that
 * $\angle X'B'A'  = \beta$
 * $\angle X'A'B'  = \alpha $

Yilding the following triangle similarity
 * $\triangle X'A'B'  \sim \triangle XAB \;\;\;\;\;$   Angle-Angle

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.