Proving an equality for an equilateral triangle in the Poincare model 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.
 A: Edit: I don't have enough points here to leave comments yet, so the Laws of Cosines are at:
http://en.wikipedia.org/wiki/Law_of_cosines_%28hyperbolic%29
where you take the "distance scale" $k = 1.$
Let's see, basic trigonometry things you need include
$$1 + \tan^2 \theta = \sec^2 \theta,$$
$$  \tan 2 \theta = \frac{2 \tan \theta}{1 - \tan^2 \theta},       $$
and on and on. It might help if you identify the books or websites you are already reading on this material. And yes, the relationship of additive distance and multiplicative distance is as you describe, just remember that the cross ratio may come out negative so we take absolute value, so the multiplicative length is always 1 or larger.
ORIGINAL: With additive distance $D$ between points $A,B,$ so that your $a = e^D,$ one of the Laws of Cosines reads
$$ \cosh D = \frac{\cos \alpha + \cos^2 \alpha}{\sin^2 \alpha}. $$
$$ \frac{1}{2} \left( a + \frac{1}{a} \right) = \frac{\sec \alpha + 1}{\tan^2 \alpha} $$
by dividing top and bottom by $\cos^2 \alpha$
$$ \frac{1}{2} \left( \frac{a^2 +1}{a} \right) = \frac{\sec \alpha + 1}{\tan^2 \alpha} $$
$$  \frac{2a}{a^2 +1}  = \frac{\tan^2 \alpha}{\sec \alpha + 1}   $$
Multiply top and bottom by $\sec \alpha - 1,$
$$  \frac{2a}{a^2 +1}  = \frac{\tan^2 \alpha (\sec \alpha - 1)}{\sec^2 \alpha - 1} = \sec \alpha - 1.  $$
As $\alpha $ is acute, $\alpha / 2 < \pi / 4, $ and $t < 1.$
The identity 
$$  \tan \alpha = \frac{2 t}{1 - t^2}  $$
leads to
$$  \sec \alpha = \frac{1 + t^2}{1 - t^2},  $$ so
$$  \sec \alpha - 1 = \frac{2 t^2}{1 - t^2}.   $$
All together now,
$$  \frac{2a}{a^2 +1}  =  \sec \alpha - 1 = \frac{2 t^2}{1 - t^2}.  $$
