Basic stategy 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,
|
|
|||
(SAS) condition If h-triangles ABC and PQR have
|
||||
Proof 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, <BAC = <QPC'. By (2), <BAC = <QPR, so C' lies on QR. By the Basic Strategy, d(A,C) = d(P,C'). By (3), d(A,C) = d(P,R), so d(P,C') = d(P,R), and hence C' = R. Thus t maps ABC to PQR, so the h-triangles are h-congruent. |
||||