Morley's Trisector Theorem/Dijkstra's Proof

Dijkstra's Proof
Choose angles $\alpha$, $\beta$ and $\gamma$ all greater than $0$ such that $\alpha + \beta + \gamma = 60 \degrees$.

Draw an equilateral triangle $\triangle XYZ$.

Construct $\triangle AXY$ and $\triangle BXZ$ with the angles as indicated:


 * Morleys-Theorem-Dijkstra-Proof.png

We have that:

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 $3$ times, in $\angle AXB$, $\angle AXY$ and $\angle BYZ$, we have:


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

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

Thus we conclude that $x = 0$.

The result follows.