For convenience we'll work in the hyperbolic upper half plane $H$. We are given a hyperbolic quadrilateral $Q$ with vertices $a,b,c,d$ and geodesic segment edges $[ a,b ]$ $[ b,c ]$ $[ c,d ]$ $[ d,a ]$. We are also given that $\angle abc=\angle bad=\frac{\pi}{2}$ and $\rho_H([a,d])=\rho_H([b,c])$ where $\rho_H([x,y])$ is the hyperbolic length of the geodesic segment from $x$ to $y$. Can we conclude that $\rho_H([a,b])<\rho_H([c,d])$?

It's clear that by a transformation by an isometry, we can put the segment $[a,b]$ on the imaginary axis and then the segments $[a,d]$ and $[b,c]$ will be segments of half circles with centre at $0$ and of equal length, so the points $c$ and $d$ lie on the same 'ray' extending from the origin. This clearly implies that $\alpha=\angle bcd=\angle cda$ and, by the Gauss-Bonnet Theorem, we may also deduce that $0<Area_H(Q)=2\pi-(\frac{pi}{2}+\frac{pi}{2}+\alpha+\alpha)$ and so conclude that $\alpha<\frac{\pi}{2}$ (although this can be fairly easily seen graphically).

I feel like the proof that $\rho_H([a,b])<\rho_H([c,d])$ should be obvious and hinges on the fact that $0<\alpha<\frac{\pi}{2}$, but playing around with hyperbolic trigonometric identities has just led to a rather large mess.

I should point out that this was the beginning of an attempt to prove something which I have now proved using a different method. I am, however, curious to see if this method would have worked.

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Don't you mean "$\rho_H([c,d]) > \rho_H([a,b])$"? – Blue Apr 21 '13 at 17:48
I do indeed. Thanks. – Dan Rust Apr 21 '13 at 17:51

I'll use upper-case letters for points, and lower-case letters for distances.

Edit. As @Daniel mentions in his own answer, the $|AD|=|BC|$ condition is unnecessary. I've revised my answer to be more general.

We have quadrilateral $\square ABCD$ with right angles at $A$ and $B$. Write $p :=|AB|$, $q =|CD|$, $r = |AD|$, $s=|BC|$ for the edge lengths; finally, let $t=|BD|$ be the length of a diagonal.

In right triangle $\triangle ABD$, the Law of Cosines and Law of Sines give us $$\cosh t = \cosh p \cosh r \qquad\qquad \sinh r = \sinh t \sin \angle ABD$$ In $\triangle CBD$, we have \begin{align} \cosh q &= \cosh s \cosh t - \sinh s \sinh t \cos \angle CBD \\ &= \cosh s (\cosh p \cosh r) - \sinh s ( \sinh t \sin \angle ABD ) \\ &= \cosh p \cosh r \cosh s - \sinh r \sinh s\\ &= ( 2 \sinh^2 \frac{p}{2}+1) \cosh r \cosh s - \sinh r \sinh s \\ &= 2 \sinh^2\frac{p}{2} \cosh r \cosh s + \cosh(r-s) \end{align} so that \begin{align} 2\sinh^2 \frac{q}{2} + 1 &= 2 \sinh^2\frac{p}{2} \cosh r \cosh s + \cosh(r-s) \\[6pt] 2\sinh^2 \frac{q}{2} &= 2 \sinh^2\frac{p}{2} \cosh r \cosh s + \cosh(r-s) - 1 \\[6pt] 2\sinh^2 \frac{q}{2} &= 2 \sinh^2\frac{p}{2} \cosh r \cosh s + 2\sinh^2\frac{r-s}{2} \\[6pt] \sinh^2 \frac{q}{2} &= \sinh^2\frac{p}{2} \cosh r \cosh s + \sinh^2\frac{r-s}{2} \end{align}

Clearly, then, $\sinh\frac{q}{2} \ge \sinh\frac{p}{2}$, so that $q\ge p$, with equality if and only if $r=s=0$.

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This is great. Thanks. Didn't even need to use $\alpha <\frac{\pi}{2}$. – Dan Rust Apr 21 '13 at 18:02

I'll add an answer to my own question in case anyone would like to see an alternative proof. This is based on the method I eventually ended up using, which I alluded to in the question. It has the disadvantage of not being as elementary as Blue's proof as it makes use of a rather general lemma about pairs of non-intersecting geodesics. The lemma is stated by the following:

Lemma If $l_1$ and $l_2$ are two geodesics in $H$ which do not intersect, then there exists a unique pair $z\in l_1$ and $w\in l_2$ such that $\rho_H(z,w)=\min\{\rho_H(x,y)|\:x\in l_1, y\in l_2\}$ and further, the unique geodesic $l$ passing through $z$ and $w$ is the unique geodesic which meets both $l_1$ and $l_2$ at right angle

Using the construction mentioned in the question, $Q$ can be moved isometrically so that $a$ and $b$ lie on the imaginary axis and and the segments $[a,d]$ and $[b,c]$ are segments of semicircles $l_1$ and $l_2$ centered at the origin.

Given that $l_1$ and $l_2$ do not intersect, we may use the above lemma and, given that the imaginary axis intersects $l_1$ and $l_2$ at right angles at $a$ and $b$, it follows that $a$ and $b$ must be the unique minimising pair of elements. We also know that $c\in l_2$ and $d\in l_1$ and $c\neq b$, $d\neq a$ and so can conclude that $\rho_H(a,b)<\rho_H(c,d)$.

This has the added bonus that we do not need the hypothesis that $\rho_H(a,d)=\rho_H(b,c)$, only that one is non-zero.

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