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.