Is it possible that a right triangle is circumscribed by a horocycle? Or, is this statement a theorem in the hyperbolic gemetry?
For any horocycle $\gamma$, there are no three distinct ordinary points A,B,C on $\gamma$ forming a right triangle $\Delta$ABC
I'm pretty sure that this is a true statement but I do not know how to prove it.(or my guess might be wrong)
There are lots of right-angled triangles with vertices on a horocycle. A right-angled triangle has one right angle, the others are smaller. There are no hyperbolic triangles with three right angles.
Claim: Let $\gamma$ be a horocycle and $A,B$ two distinct points on it. Then there exists a point $C$ on the horocycle such that $\Delta ABC$ is a hyperbolic triangle with a right angle at $A$.
Proof: Let us work in the Poincaré disk model. A horocycle is a circle that touches the boundary of the disk. Let $A,B$ be two points on a horocycle $\gamma$. We note that the geodesic $\overline{AB}$ does not intersect $\gamma$ at a right angle at $A$ and is not parallel to $\gamma$ at $A$. Since the geodesic $\overline{AB}$ is not parallel to the horocycle, we can look at the interior of the horocycle. Since $\overline{AB}$ does not intersect $\gamma$ in a right angle, there is one side, where you can add a new geodesic $\overline{AC}$ that also points inside the horocycle. At some point this geodesic will leave the horocycle (it will not go to the boundary point where $\gamma$ touches the disk, since it does not intersect $\gamma$ at a right angle at $A$.) Let us call that point $C$. There is a geodesic connecting $C$ and $B$, completing the triangle. By construction the triangle $\Delta ABC$ has a right angle at $A$.