The absolute ruler unit of hyperbolic geometry [closed]

I will answer the first question, since the 2nd question is unclear.

Now, to the question itself. You should read the continuation of the quoted passage, as it clarifies what is said in your quote:

This fact is a direct result of

Big Idea: In hyperbolic geometry, if two triangles are similar, then they are congruent.

For instance, if I want to communicate to a friend what segment I am planning to use as a unit, I can say:

Consider an equilateral triangle in the hyperbolic plane with angles equal $30^\circ$. Its edges are congruent to each other and will serve as the unit of measurement from now on.

Or, I can say: consider an isosceles triangle where one of the angles if the right angle and the other two have the angle $22.5^\circ$.

As long as your friend resides in the same hyperbolic plane as you do (same curvature), they will understand what you mean unambiguously and, thus, will use the same unit of measurement. That's all what the quoted passage is saying.

Edit. Both triangles I have mentioned are constructible in hyperbolic plane (using the hyperbolic compass and ruler). One uses the following criterion, due to Mordukhai-Boltovskoy:

1. A number $x$ is constructible, as a length, in hyperbolic geometry if and only if $\cosh(x)$ is constructible in Euclidean geometry.
2. A number $\alpha$ is constructible as an angle-measure in hyperbolic geometry if and only if it is constructible as an angle-measure in Euclidean geometry.

See for instance this answer. Or, you can find a self-contained (and rather long) proof of this theorem in the Bachelor's Thesis

Ruben de Vries, Compass and Straightedge Constructions in the Hyperbolic Plane, Utrecht University, 2021.

In fact, you do not need the full power of this result: In the proof of (1) they first establish the result for right-angled triangles, which clearly suffices in both (equilateral and right-angled) examples.

1. In the case of the equilateral triangle I mentioned, by the dual hyperbolic law of cosines its side-length $a$ satisfies: $$ \cosh (a)= (\cos(\pi/6) + \cos^2(\pi/6) )/\sin^2(\pi/6) = 2\sqrt{3} + 6 $$ and the right-hand-side is Euclidean-constructible.

2. The hypotenuse $c$ the right-angled triangle satisfies $$ \cosh(c)= 3+2\sqrt{2} $$ which is also Euclidean-constructible.

More generally, if $\alpha, \beta, \gamma$ are angles constructible in hyperbolic, equivalently, in Euclidean, geometry, then the side-lengths of the corresponding hyperbolic triangle are constructible in hyperbolic geometry (again, by the dual hyperbolic cosine law). Thus, in this situation, the entire triangle is constructible in hyperbolic plane.