TITLE: Trouble with Morley's theorem proof.
QUESTION [5 upvotes]: I am reading through this proof of Morley's theorem, and I am confused about one part. Why does $\angle ARR_{2}=\beta+\frac{\pi}{3}$? Also, why does the diagram put $\beta+\frac{\pi}{3}$ on $\angle ARQ$ when they say it should be on $\angle ARR_{2}$?
Here is the figure for reference:

Where $^{+}$ is shorthand for $+\frac{\pi}{3}$

REPLY [2 votes]: okay, first of all we see from $\Delta BRP$ that $\beta + \alpha^+ + \gamma^+ = \pi$. So if you compare this to $\Delta ARQ$, you'll get the $\beta^+$ in question.  
To answer the first question,  note that $\Delta ARQ$ and $\Delta ARR_2$ share two equa angle measures.