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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 $overlineAB$ does not intersect $gamma$ at a right angle at $A$ and is not parallel to $gamma$ at $A$. Since the geodesic $overlineAB$ is not parallel to the horocycle, we can look at the interior of the horocycle. Since $overlineAB$ does not intersect $gamma$ in a right angle, there is one side, where you can add a new geodesic $overlineAC$ 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 $A$.