Finding angles of hyperbolic triangles

Question

I am trying to learn about how to find the angles of hyperbolic triangles. Now below is a problem:

It has all the steps but I am not understanding the concept (the ones that are underlined in green and yellow)

If you notice, $\measuredangle BAC=\frac{\pi}{2}-\measuredangle OAP=\measuredangle OPA$ but I do not know why.

On the other hand, I understand the computation parts but it is meaningless if I dont get the concept.

Anyone can help out?

Answer 1

Note that $\measuredangle BAC$ is the angle centered at $A$ (and not, e.g. $B$). The geodesic $AC$ is orthogonal to the geodesic $AP$ (why?), implying the first relation. The second is similar, you only have to reflect the angle $\measuredangle 0BQ$ about the tangent to $BC$ in $B$.

The last equation in your question ("if you notice") is true, cause $OPA$ is a Euclidean rightangled triangle in which the angles sum up to $180°$. (This is part of the technique, to find helper triangles in which Euclidean geometry applies..)

Edit: it is, of course, essential for this reasoning, that the upper half plane is a conformal model of Hyperbolic space, i.e. the Euclidean angles are equal to the Hyperbolic ones.

Answer 2

$ AB,BC, CA$ are sides and parts of semicircles centered on x-axis with centers respectively at Infinity, Q and P... Angle between two lines also equals angle between their perpendiculars.

The semi-circles are geodesics in ( one of two) Poincare's hyperbolic geometry models.