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 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.

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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, $$\frac{2a}{a^2 +1} = \sec \alpha - 1 = \frac{2 t^2}{1 - t^2}.$$

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Thanks for your answer, Will. I do have a few questions though. You say that if $AB$ has additive distance $D$, then $\mu(AB)=e^D$? Does this mean that the multiplicative distance given by the cross-ratio is always just $e$ raised to the additive distance? I'd never heard that before, but it seems quite nice. Also, do you have a reference for the first identity you used for $\cosh D$? I've never seen it, and couldn't find it on google. Also, I didn't quite follow the identities for $\tan\alpha$ and $\sec\alpha$ in terms of $t$ as I've never seen them before. Does it follow from...con't –  yunone May 3 '11 at 5:19
this section on the Weierstrass substitution? If so, I'll try to look deeper into that. Thanks again. –  yunone May 3 '11 at 5:19
Thanks for adding in the edit. Aside from the derivation of those identities, I followed all your calculations, thanks! –  yunone May 3 '11 at 6:10
I see, it lets me comment, at least on my own answer. I incorrectly assumed you were reading Hartshorne, as I was not sure otherwise how you came to prefer the multiplicative distance.  For ordinary triangles (no infinite sides) there is a Law of Sines and two Laws of Cosines. See if you can find those and write them out carefully in your preferred notation. Note that I do not use the upper half plane model at all, just the Laws. –  Will Jagy May 3 '11 at 6:20
I actually am reading Hartshorne's Geometry: Euclid and Beyond, but I hadn't seen those ideas in his text. I'll take your advice and try to translate those laws into the multiplicative notation. –  yunone May 3 '11 at 6:26