hyperbolic quadrilateral angle

Question

Consider a hyperbolic quadrilateral of $abcd$ in the hyperbolic plane $\mathbb{H}^2$ with the metric being the metric defined via the cayley map. Suppose $\angle b$, $\angle c$ ,$\angle d$ are all of angle $\frac{\pi}{2}$. Show that $ad$ is longer than $bc$.

I'm not sure how to start this, I know that angle $\angle a$ cannot be $\frac{\pi}{2}$, for otherwise we obtain a contradiction by Gauss Bonnet. But i'm not sure how else to proceed.

Answer 1

I'm going to assume you know a little bit about distance-preserving transformations of the hyperbolic plane.

Situate the quadrilateral so that the side $dc$ coincides with the imaginary axis in the upper-half plane, with $|d| > |c|$, the point $c$ coincides with $i$, and side $cd$ coincides with the unit circle.

Note that the dilation $\varphi: z\mapsto |d|z$ is a hyperbolic isometry that:

1. Carries $c$ to $d$
2. Carries the unit circle to the circle, centered at $0$, containing the side $ad$, because the angle at $d$ is $\frac{\pi}{2}$

Now - because the angle at $b$ is $\frac{\pi}{2}$, and the geodesic containing side $ab$ is tangent to the line connecting $b$ and $|d|b$, the side $bc$ will be carried by $\varphi$ to a subset of the side $ad$. Since the map $\varphi$ is an isometry, we must have $|bc| < |ad$.

I hope this picture helps illustrate.