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For a right-angled geodesic triangle in curvature minus one, with legs $a,b$ and hypotenuse $c$, the displayed identity replaces the Euclidean theorem. Normalize the right vertex to $i$ in the upper half-plane, one endpoint to $iy$ with $y>1$, and the other to $s+it$ with $s^2+t^2=1$ and $t>0$. The Hyperbolic distance in the Poincare half-plane gives $\cosh a=(y+y^{-1})/2$, $\cosh b=1/t$ and $\cosh c=(1+y^2)/(2yt)$, proving the identity. For small lengths its expansion gives $c^2=a^2+b^2$ to leading order.

Write $B=\mathrm{diag}(\lambda ,{\lambda }^{-1})$ and $q={\lambda }^2>0$, so its Möbius transformation is $z\mapsto qz$, with $q\ne 1$. For $z=u+iv$, the Hyperbolic distance in the Poincare half-plane satisfiesAt $z=i$, the final parenthesis is $1$. Since $\cosh$ is strictly increasing on nonnegative arguments, every $z$ off the imaginary axis has