A well-known result of the Euclidean planimetry says that the heights of any triangle have a common point called the orthocentre of the triangle. This result is not true in neutral geometry (i.e., geometry without the parallel and continuity axioms), because the heights of an obtuse triangle need not intersect. Nonetheless any two heights of an acute triangle (= a triangle whose all angles are acute) do intersect, by the Pasch axiom.

Problem. Is it true that the heights of any acute triangle have a common point (in neutral geometry)?

Remark. The answer to the problem is affirmative if the acute triangle is a midpoint triangle of another triangle, see Proposition 9.7 in these lecture notes of Polly Knight. But in the Hyperbolic Geometry an acute triangle (even equilateral) needs not be a midpoint triangle of another triangle. In that case what does happen with the intersecting points of the heights of an acute triangle? Do they coincide? Since this is rather a basic question in neutral geometry, I hope that an answer is known (to specialists). In the footnote 32 to Remark after Proposition 9.7, Polly Knight writes that in the Klein's model the heights of an acute triangle indeed have a common point. But neutral geometry has many other models, in particular, non-Archimedean models and even models in which the sum of a triangle is larger than $\pi$. In the mentioned footnote 32 Polly Knight writes that “a proof in neutral geometry is certainly valid a real bottle of wine”. So, is such a proof (worth a “real bottle of wine”) known?