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.

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. 