Basic stategy Suppose that ABC and PQR are htriangles. Then there is a hyperbolic transformation t which maps A to P, B to B' on the hline PQ, on the same side of P as Q, and C to C' on the same side of the hline as R.
Note that, as t preserves angle and hyperbolic distance,



(SAA) condition If htriangles ABC and PQR have


Proof Let t be the transformation implied by the Basic Strategy. Then B' lies on the hline PQ. By the Basic Strategy, d(A,B) = d(P,B'). By (2), d(A,B) = d(P,Q), so d(P,B') = d(P,Q), and hence B' = Q. By the Basic Strategy, <BAC = <QPC'. By (1), <BAC = <QPR, so C' lies on PR. By the Basic Strategy, <ACB = <PC'B'= <PC'Q (as B' = Q). By (3), <ACB = <PRQ, so <PC'Q = <PRQ. If C' = R, we are done, as t maps ABC to PQR.
Otherwise, we have a picture like that on the right, where
