Most of the results are analogues of euclidean conditions. Indeed,
many of the proofs mirror wellknown euclidean proofs.
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.
justification of basic strategy
Note that, as t preserves angle and hyperbolic distance,
d(A,B) = d(P,B'), d(A,C) = d(P,C'), d(B,C) = d(B',C'), and
<ABC = <PB'C', <BAC = <B'PC', <ACB = <PC'B'.
Our aim is to prove conditions which are sufficient for the
htriangles to be hcongruent. In each case, we show that the
transformation t of the basic strategy actually maps B to Q,
and C to R, i.e. that B' = Q, and C' = R.


Our first result is often known as the two sides and included angle
condition for congruence.
(SAS) condition
If htriangles ABC and PQR have
(1) d(A,B) = d(P,Q), 
(2) <BAC = <QPR, and 
(3) d(A,C) = d(P,R), 
then the htriangles are hcongruent.
We use a common notation: (SAS) in the title refers to side, angle, side.
The order is significant, the angle is that enclosed by the pairs of sides.
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 (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 PR.
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 htriangles are hcongruent.
(ASA) condition
If htriangles ABC and PQR have
(1) <BAC = <QPR, 
(2) d(A,B) = d(P,Q), and 
(3) <ABC = <PQR, 
then the htriangles are hcongruent.
proof of (ASA) condition
In euclidean geometry, the sum of the angles of a triangle is constant.
Thus, if two triangles have two corresponding angles equal, then the
third angles must be equal. Thus (ASA) and (SAA) are equivalent in that
geometry. Here, however, we need a separate proof.
(SAA) condition
If htriangles ABC and PQR have
(1) d(A,B) = d(P,Q), 
(2) <BAC = <QPR, and 
(3) <ACB = <PRQ, 
then the htriangles are hcongruent.
proof of (SAA) condition
(SSS) condition
If htriangles ABC and PQR have
(1) d(A,B) = d(P,Q), 
(2) d(A,C) = d(P,R), and 
(3) d(B,C) = d(Q,R), 
then the htriangles are hcongruent.
proof of (SSS) condition
The final result is not true in euclidean geometry. There, two
triangles with the corresponding angles equal need only be similar.
(AAA) condition
If htriangles ABC and PQR have
(1) <ACB = <PRQ, 
(2) <ABC = <PQR, and 
(3) <BAC = <QPR, 
then the htriangles are hcongruent.
proof of (AAA) condition


We can apply these results to prove other theorems, some
familiar, some strange looking.
The Isosceles Triangle Theorem
Suppose that ABC is an htriangle.
(1) If d(A,B) = d(A,C) then <ABC = <ACB.
(2) If <ABC = <ACB then d(A,B) = d(A,C).
We choose a proof which illustrates the fact that the hcongruence
of two htriangles involves an implicit correspondence between the
vertices. Thus, even if ABC is hcongruent to PQR, it is not usually
hcongruent to PRQ. In the proof, we meet the possibilty that we
may have ABC hcongruent to ACB.
proof of the isosceles triangle theorem
Of course, an htriangle with either of these properties is called
an isosceles htriangle.
Note that such htriangles do exist. The figure on the right was
produced by choosing B and C on an hcircle with hcentre A,
so that d(A,B) = d(A,C).
We can also define equilateral htriangles as those
with three equal sides (and hence three equal angles).
In euclidean geometry, the angle sum is π, so an equilateral
euclidean triangle has each of its angles equal to π/3.
In hyperbolic geometry, each angle must be less than π/3.
By the (AAA) condition, there is only one equilateral htriangle
with angles equal to α for each α < π/3.
illustration of equilateral hyperbolic triangles

