Constructing a regular right angled hyperbolic hexagon I would like to construct a regular right angled hexagon in a klein model.
I'm having a hard time understanding why this method works, here is what my professor did in class. Any additional comments will be greatly appreciated. Thanks in advance!
First in a circle O, pick any six points, $A,B,C,D,E,F$, connect $AB$, $CD$, and $EF$.

Then, construct line $a, b, c, d, e, f$ where $a,b$ intersects, $c,d$ intersects, and $e,f$ intersects, call them points $AA, BB,$ and $CC$.

Lastly, connect $AA, BB, CC$ to construct lines $\alpha, \beta,$ and $\gamma$, the hexagon inscribed in the circle hexagon ($\theta1 - \theta6$) all have right angles therefore is a regular hexagon.

 A: You will not make it that way, what your professor showed you is how to make an equiangular hexagon (all angles are equal in measure) in a Beltrami Klein disk model , you can transform that into an equiangular hexagon in a Poincare disk model, but that is a later worry.
The first step is to get a regular hexagon in a Beltrami Klein disk model 


*

*devide the circle in 12 equal arcs and number the dividing points.
(for nice looks don't start at 12 0'clock but start halfway between 11 and 12) 

*draw segments between  the points 1 and 4, 2 and 11, 3 and 6, 
5 and 8, 7 and 10, 9 and 12
You now have a regular hexagon in a Beltrami Klein disk model. ( you can check that all angles are right hyperbolic angles by the method given by your professor) 
Then for every segment:


*

*draw a line tangent to the circle for each of the endpoints of the segment.

*draw a circle from the intersection of these lines , trough the endpoints of the segment  (this circle should go to each endpoint and be orthogonal to the circle)
Do this for every of the segments 1-4, 2-11, 3-6, 5-8, 7-10, 9-12 
Done !
