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A followup to my previous question.
In this proof they say $\angle R_{1}RR_{2}=\beta+\alpha-\gamma$. I understand why this is true, but I dont understand how the proof made that logical jump? In other words, based on what the proof had already shown (which isnt much) how can you make the jump to say $\angle R_{1}RR_{2}=\beta+\alpha-\gamma$
Figure for reference:
Also, why does $\angle RR_{2}R_{1}=\gamma^{+}$ imply $R_{1}$ is on $AR_{2}$?
In the proof, the relation $|\angle R_1RR_2| = \alpha + \beta - \gamma$ comes from canceling the "$+\pi/3$"s in the expression $\alpha^+ + \beta^+ - \gamma^{++}$ (where "$x^+$" abbreviates "$x+\pi/3$").
The goal here, of course, is to find the simplest expression for the angle measure. Reducing the number of "$+\pi/3$"s certainly helps, as does later replacing $\alpha + \beta$ with $\pi/3-\gamma$. The ultimate expression, $\pi/3 - 2\gamma$, is nicer-looking than the original form.
If that bit of angle-chasing bothers you, here's a conceptually-simpler, if algebraically-longer, route to the same result:
(Keep in mind that $R_1$ is the reflection of $P$ in $\overline{RB}$, and $R_2$ is the reflection of $Q$ in $\overline{RA}$. Also, $\alpha + \beta + \gamma = \pi/3$.)
Simply observe that $\angle R_1RR_2$ is the "overlap" caused by fitting a number of known angles about $R$. So, its measure is the amount by which the total of those known angles exceeds $2\pi$, and we can compute thusly: $$\begin{align} |\angle R_1RR_2| &= \left(\;|\angle R_1RB| + |\angle BRP| \;\right) + |\angle PRQ| + \left( \; |\angle QRA| + |\angle ARR_2| \;\right) - 2\pi \\ &= \left(\; 2 \alpha^+ \;\right) + \pi/3 + \left(\; 2\beta^+ \;\right) - 2\pi \\ &= 2(\alpha +\pi/3) + 2(\beta+\pi/3) + \pi/3 - 2\pi \\ &= 2(\alpha + \beta)-\pi/3 \\ &= 2(\pi/3 - \gamma) - \pi/3 \\ &= \pi/3 - 2\gamma \end{align}$$
Based on where the proof goes next, I might have opted to massage the measure into this form: $$|\angle R_1RR_2| = \pi - 2\gamma^+$$ This makes it immediately clear that $\angle R_1RR_2$ is the vertex angle of an isosceles triangle with base angles $\gamma^+$.
I'm not certain if this will help with the particular question but here is a proof I displayed as an illustrative example in my Mathematica Presentations application. In the actual presentation one can click to display the various steps, and the "Morley's Theorem Illustrated" panel allows the three vertices to be moved. The first picture also shows the control structure and the subsequent pictures show only the panel contents. The entire proof takes up much less display in a notebook. I believe having the freedom for multiple displays helps to clarify a difficult proof.
This treatment follows Coxeter who in turn was following Raoul Bricard, Nouvelles Annales de Mathématiques (5), 1, (1922), pp 254-258.
