User talk:Jhoshen1/Sandbox

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

Construct $\triangle AXY$ and $\triangle BXZ$


 * Morleys-Theorem-Dijkstra-Proof.png

where

Because $\angle AXB = 180 \degrees - \paren {\alpha + \beta}$, it follows that:
 * if $\angle BAX = \alpha + x$ then $\angle ABX = \beta - x$

Using the Sine Rule, we have:


 * $ \dfrac {\map \sin {\alpha + x} } {\map \sin {\beta - x} } = \dfrac {BX} {AX} $

Where
 * $ BX =  XZ \map \sin {60 \degrees + \gamma} / \sin \beta $

and
 * $ AX = XY \map \sin {60 \degrees + \gamma } / \sin \alpha $

Also given that $XZ=XY$, yields
 * $ \dfrac {\map \sin {\alpha + x} } {\map \sin {\beta - x} } = \dfrac {\sin \alpha} {\sin \beta} $

In the range in which these angles lie, the of the above is a strictly increasing function of $x$.

Thus we conclude that $x = 0$.

Consequently, $\angle BAX = \alpha $ and $\angle ABX = \beta $

In a similar fashion, it can be shown that $\angle CAY = \alpha $, $\angle CBZ = \beta $, $\angle ACY = \gamma $ and $\angle MCZ = \gamma $.