Firstly, sorry if I put this is the wrong place. I'm not really sure where geometry goes.

Anyway, i'm trying to prove the SSS criterion for triangles in \mathbb{H}^2.

This is what i've tried:

Let PQR be a triangle in \mathbb{H}^2.

Arrange a Lorentz motion taking P to (0,0,1) and Q to (\sinh s , 0 , \cosh s ) on the line y=0 where s=d(P,Q).

Let r_1=d(P,R) and r_2=d(Q,R).

Then if we draw a circle of radius r_1 around P, and a circle of radius r_2 around Q, where they intersect should be R (they will intersect twice, but this doesn't matter since each of these triangles have the same side length).

Suppose that the circles do not intersect and R does not exist. Then this contradicts the fact that a Lorentz motion preserves hyperbolic lines.

Hence R must exist and must have the side lengths as the triangle before the motion.

Therefore, given any two triangles with the same sides, the triangles can be mapped to the same triangle with two points of (0,0,1) and (\sinh s, 0 , \cosh s). Hence they are congruent.

Is this right? I'm a little shaky on hyperbolic space.