Assume convex (hyperbolic) quadrilateral $\square ABCD$ has right angles at $A$, $B$, and $D$. Let $x = |AD|$ and $y =|BC|$. Prove $x < y$.
I know that the sum of the angles is less than $360^\circ$. Yet, other than the scalene inequality for triangles, I don't know how to best relate two sides together. I thought of trying by contradiction, but I didn't see any major contradiction that would happen if $y \geq x$.
On
Let $C^\prime$ be a point on $\overrightarrow{BC}$ such that $|AD|=|BC^\prime|$. Then (as per the Claim in my answer to your other question), $\square ABC^\prime D$ has congruent angles at $D$ and $C^\prime$. These congruent angles must each be smaller than a right angle; in particular, $\angle ADC^\prime < 90^\circ = \angle ADC$, so that $C^\prime$ is in the interior of $\angle ADC$ and thus lies somewhere between $B$ and $C$. $\square$
Assuming this is in hyperbolic geometry, this property can help you. Line AD is a unique line that joins two ultraparallel lines formed by the extensions of AB and DC and is perpendicular to both of them. It is also the closest distance between the two lines. That is, any other line between the AB and DC will be longer.