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Not sure about your formula at all, and your "midpoint" is not on the hyperbolic line $BC$ (remember the hyperbolic line is a half circle) https://en.wikipedia.org/wiki/Poincar%C3%A9_half-plane_model gives another formula: $$ \operatorname{dist} (\langle x_1, y_1 \rangle, \langle x_2, y_2 \rangle) = \operatorname{arcosh} \left( 1 + \frac{ {(x_2 - x_1)}^2 + ...


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Another point of view (which is the theme of John Stillwell's book "Four pillars of modern geometry"), is a synthetic derivation that starts with the upper half plane equipped with its group of rigid motions: namely, the group generated by the fractional linear action by $PSL_2(\mathbb{R})$ together with reflection across the upper $y$-axis. From this, ...


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Let's consider the hyperbolic (Minkowski) and complex (Euclidean) planes. In the Euclidean plane, you can describe any point by $k(\cos(\theta),\sin(\theta))$. Analogously, in the hyperbolic plane, you can describe points by $k(\cosh(u),\sinh(u))$. A unit hyperbolic vector (in the hyperbolic plane) is then $(\cosh(u),\sinh(u))$. The actual inner product ...


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By the angle of parrallelism https://en.wikipedia.org/wiki/Angle_of_parallelism When C is an ideal point $ AC = BC = \infty , \angle C = 0 $ Then by the angle of parallelism https://en.wikipedia.org/wiki/Angle_of_parallelism you get $\sin \Pi(AB) = \sin \angle BAC = \frac {1}{\cosh (AB) } = sech(AB)$. When $ \angle BAC$ is larger than $ \Pi(AB)$ there ...


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Here's one straightforward formulation of the parallel postulate: "Given a point in the space, and a co-line not incident to that point, there is a unique co-line incident to the point which does not meet the given co-line." By 'co-line' here I mean a $(n-1)$-dimensional subspace; in the case where the 'space' is the normal 2d plane this is just the ...


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Deriving additional points There is something you have not mentioned, and which might make things easier. The way I understand your question, you are essentially dealing with the situation where you have two points and want to find the center of the Euclidean circle which corresponds to the geodesic through these two points. In both cases, you can compute ...



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