geometryhyperbolic-geometrytrigonometry
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?
First off, the other two answers here are wrong because you can prove the statement using the same flaw if P was outside the triangle. The proof using the same flaw but done in the outside of the triangle is here: [https://www.youtube.com/watch?v=Yajonhixy4g][1]
The video clearly does this proof with P on the outside. If P were on the segments of the triangle, then the proof will still hold because triangle AEP will be congruent to triangle AFP by SAA (shared sides, bisected angles, 90 degree angles). You know that P and D are the same points because the definition of P is where the perpendicular bisector meets the bisector of the opposite angle, and provided the case where P's on BC, it can only possibly be a triangle if P and D are on the same place. D is the bisector of BC, so so is P. EP = FP because we proved that AEP = AFP. BP = CP because P bisects and is between BC. Since EBP and FCP are right triangles and the Hypotenuse and one of the legs are congruent, we have triangle EBP is congruent to triangle FCP. From here, basic elementary algebra can be use to prove AB = AC.
So, what is wrong with this proof? The answer is betweenness. Every single one of these proofs rely on a certain order of points your drawing above should be:
We still have AE=AF, PE=PF, and PB=PC, and it's still true that BE=FC, but AB=AC is not true. This is because F is still between A and C, but E is not between A and B. Since E is not between A and B, SAA does not follow, and neither does SAS for this particular case.
Best Answer
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.