I've been working a good while trying to establish an equality, but have made little success.
Suppose you're working in the Poincare disk model inside an ambient Euclidean plane. If an equilateral triangle in the Poincare model has sides equal to $AB$ and angles equal to $\alpha$, how can we show that $ \frac{2a}{1+a^2}=\frac{2t^2}{1-t^2} $ where $a=\mu(AB)$ is its multiplicative length, and $t=\tan(\alpha/2)$?
I made the following picture:
I suppose that $ABC$ is the equilateral triangle in the Poincare model, and I suppose that the usual midpoint (in the Euclidean sense) $D$ of $BC$ is at the origin of the Euclidean plane. The curve through $A$ and $B$ is the P-line passing through $A$ and $B$ in the Poincare model, and I came up with the ugly calculation that $ \mu(AB)=(AB,PQ)^{-1}=\frac{2(AP\cdot BQ\cdot AQ\cdot BP)}{AP^2BQ^2+AQ^2BP^2} $ where $(AB,PQ)$ is the cross-ratio of $AB$ and $PQ$, and $P$ is the point on the circumference of the Poincare model closer to $A$, and $Q$ closer to $B$. I took $\tan(\alpha/2)$ to be $DC/AD$, but I didn't see a way to plug in to get the desired equality. Perhaps I'm using the operations wrong? Thanks for any help.