Hyperbolic Triangles and maximizing

Question

Let $A$,$B$ and $C$ be the nodes of a hyperbolic triangle $\triangle$ in $\mathbb{H}^2$.

Suppose the angle $\alpha$ at $A$ has angle at least $\frac{\pi}{2}$. Show that side $BC$ has maximal length.

My attempt:

Let us maximise the angle $\alpha$ to $\frac{\pi}{2}$. In which case we may suppose $A=0$ and $BC$ is hypotenuse. Denoting the lengths of $BC$ by $c$, $AC$ by $a$ and $AB$ by $b$, we obtain:

$coshc=coshacoshb$

I'm not sure how to proceed. But my thoughts are to show that we have some sort of monotonic relationships between angle $\alpha$ and side $BC$.

May someone offer help, please?

Answer 1

You don't need to maximise the angle:

Note that angles of triangles are taken to be between $0$ and $\pi$ each.

So, if $\alpha \leq \frac{\pi}{2}$, then $cos \alpha \leq 0$. Since the metric is non-negative, this implies that

$\cosh(c) \geq \cosh(a) \cosh(b)$

(Here I am using the generalised pythagoreas theorem)

This allows us to conclude the maximality of the side in question: Because as $\cosh(x)\geq 1$ for all $x\in \mathbb{R}$, we must have $\cosh(c)\geq\max(\cosh(a),\cosh(b))$. Thus, $c\geq max(a,b)$.