Suppose we have a right triangle ABC in the Poincare half plane such that angle C = $\frac \pi2$. Is it possible to construct this triangle with the property that $d(A, B)^2 + d(A, C)^2 = d(B, C)^2$ where $d(P_1, P_2)$ is the hyperbolic distance from point $P_1$ to $P_2$?
I'm leaning towards no (purely intuition), but I can't quite figure out exactly why.
Thanks for any help in advance.
As Blue wrote in a comment, the counterpart to the Pythagorean theorem in hyperbolic geometry (with Gaussian curvature $-1$) is
$$\cosh a\cdot\cosh b=\cosh c$$
The Taylor series of $\cosh$ is $\cosh x = 1+\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots$. For small lengths you may ignore all terms of degree $4$ and above, and you end up with something that converges to the regular Euclidean law. Using big-Oh notation for the error terms you'd get
$$\cosh a\cdot\cosh b=\left(1+\frac{a^2}2+O(a^4)\right)\left(1+\frac{b^2}2+O(b^4)\right)=\\ 1+\frac{a^2}2+\frac{b^2}2+O(a^4)+O(a^2b^2)+O(b^4)=1+\frac{c^2}2+O(c^4)=\cosh c$$
I find this consideration useful in understanding how the product in hyperbolic geometry can correspond to the sum in Euclidean geometry.