This should really be a comment on Marco Radeschi's [answer][1] from Feb 22 involving the area formula for spherical triangles, but since I'm new here I don't have the reputation to leave comments yet.

In reply to Igor's comment (on Marco's answer) wondering about an analogous proof for the area formula of hyperbolic triangles:  there is one along similar lines, and you're rescued from non-compactness by the fact that asymptotic triangles have finite area.  In particular, the proof in the spherical case relies on the fact that the area of a double wedge with angle $\alpha$ is proportional to $\alpha$; in the hyperbolic case, you need to replace the double wedge with a doubly asymptotic triangle (one vertex in the hyperbolic plane and two vertices on the ideal boundary) and show that if the angle at the finite vertex is $\alpha$, then the area is proportional to $\pi - \alpha$.  That follows from similar arguments to those in the spherical case (show that the area function depends affinely on $\alpha$ and use what you know about the cases $\alpha=0,\pi$).

Once you have that, then everything follows from the picture [here][2], since you know the area of the triply asymptotic triangle and of the three (yellow, red, blue) doubly asymptotic triangles.  (Sorry, I can't even post images yet -- I guess I need to do some math to persuade the site I'm not a spambot.)

(That picture is slightly modified from p. 221 of [this book][3], which has the whole proof in more detail.)


  [1]: http://mathoverflow.net/questions/8846/proofs-without-words/16064#16064
  [2]: http://www.math.psu.edu/climenha/hyperbolic-triangle.jpg
  [3]: http://books.google.com/books?id=XFkc0Yn-TE8C&printsec=frontcover&dq=Lectures+on+Surfaces&ei=bMvuS86VK4-2zQTRufDxCg&cd=1#v=onepage&q&f=false