Suppose that ABC and PQR are h-triangles.
Then there is a hyperbolic transformation t which maps
A to P,
B to B' on the h-line PQ, on the same side of P as Q, and
C to C' on the same side of the h-line as R.
Note that, as t preserves angle and hyperbolic distance,
If h-triangles ABC and PQR have
Let t be the transformation implied by the Basic Strategy.
Then B' lies on the h-line PQ.
By the Basic Strategy, d(A,B) = d(P,B').
By (1), d(A,B) = d(P,Q), so d(P,B') = d(P,Q), and hence B' = Q.
By the Basic Strategy and (2), (3)
(4) d(P,C') = d(P,R), and
Let Kp be the h-circle, with h-centre P, through R, and
By (4) and (5), C' lies on Kp and Kq
Thus t maps ABC to PQR, i.e. the h-triangles are h-congruent.