What is hyperbolic geometry? It is an example of one of many non-euclidean geometries. It is often called Bolyai-Lobachevskiian geometry after two of its discovers János Bolyai and Nikolai Ivanovich Lobachevsky . Bolyai first announced his discoveries in a 26 page appendix to a book by his father, the Tentamen, in 1831. You can read about what happened to him and his work in the text. Another of the great mathematicians who seems to have preceeded Bolyai in his work is Carl Fredrich Gauss . He seems to have done some work in the area dating from 1792, but never published it. The first to publish a complete account of non-euclidean geometry was Lobachevsky in 1829. It was first published in Russian and was not widely read. In 1840 he published a treatise in German.
We shall add one more axiom to the list of axioms for Neutral Geometry.
HYPERBOLIC AXIOM: In hyperbolic geometry there exists a
line 
and a point P not on such that at least two distinct
lines parallel to pass through P.
We shall denote the set of all points in the plane by 
, and
call this the hyperbolic plane.
Lemma 12.1: There exists a triangle whose angle sum is less than
.
Proof: Let be a line and P a point not on such that two
parallels to pass through P. We can construct one of these parallels
as previously done using perpendiculars. Let Q be the foot of the
perpendicular to through P. Let m be the perpendicular to
through P. Then m and are non-intersecting. Let n
be another line through P which does not intersect . This line exists
by the Hyperbolic Axiom. Let 
be a ray of n lying between
and a ray 
of m.
CLAIM: There is a point 
on the same side of
as X and Y so that 
.
PROOF OF CLAIM. The idea is to construct a sequence of angles
so that 
. We will then apply
Archimedes Axiom for real numbers to complete the proof.
There is a point 
so that 
. Then 
is isosceles and 
. Also, there is a point
so that 
and 
. Then
is isosceles and 
. Since
is exterior to it follows that
so then
. Continuing with this construction, we
find a point 
so that 
and
Applying the
Archimedean axiom we see that for any positive real number, for example
,
there is a point so that R is on the same side of
as X and Y and 
. Thus, we have proved our
claim.
Now, 
lies in the interior of 
, for if not then
is in the interior of 
. By the Crossbar Theorem
it follows that 
which implies that n and
are not non-intersecting--a contradiction. Thus, 
. Then,
Therefore, 
and 
.
The Hyperbolic Axiom only hypothesizes the existence of one line and one point not on that line for which there are two non-intersecting lines. With the above theorem we can now prove a much stronger theorem.
Theorem 12.1:[Universal Hyperbolic Theorem] In for
every line and for every point P not on there pass
through P at least two distinct lines, neither of which intersect .
Proof: Drop a perpendicular to and construct a
line m through P perpendicular to . Let R be any other
point on , and construct a perpendicular t to through R. Now,
let S be the foot of the perpendicular to t through P. Now,
does not intersect since both are perpendicular to t.
At the same time 
. Assume that 
, then 
is a rectangle. By Theorem 11.5, if one rectangle exists all
triangles have defect 0. We have a contradiction to Lemma 13.1.
Thus, , and we are done.