To construct the Poincaré line joining two ideal points, or the so-called
line of enclosure , we have the following result.
The previous description for the construction of the unique line between two ordinary points follows from the fifth item above. We have already discussed the measure of segments, as well.
How do you copy a given angle at a given point, A, on a given ray,
?
then the ray is a diameter
of and the angle is copied as in Euclidean geometry, using another
diameter of .
, the center of , then we must find the unique circle
through A orthogonal to and tangent to the given Euclidean
line 
which forms the Euclidean angle with the tangent to at
A.
orthogonal to , by the above facts, must
pass through 
, the inverse of A with respect to . The
center, C, of lies on the perpendicular bisector of AA'. Let m
be this bisector.
to be tangent to at A, we must have 
perpendicular to . Thus, C lies on the unique perpendicular to
at A, say n. 
.
Lemma 16.1: If 
, then 
where r
is the radius of and e is the number so that 
.
Proof: If P and Q are the ends of the diameter through OB then
which is what was to be proven.
This will give us enough information to prove that all of the axioms of
neutral geometry are satisfied, except for the SAS axiom. It requires a
bit of work. Assume that we are given two p-triangles 
and
inside so that 
, 
and 
. We need to show that these triangles are congruent in the
Poincaré model.
First, we show that we may assume that A=X=O, where O is the center of
. Let be the circle through A and B orthogonal to
and let 
be the circle through A and C orthogonal to
. Then, since both circles are orthogonal to it follows that
.
Let 
and let 
be the circle of radius s
centered at A'. Then
Thus, 
, so that r is ``perpendicular'' to s by the
converse of the Pythagorean theorem. This means then that and
are orthogonal circles. By the construction of this circle
, O is the inverse of A in . By the last item
above, inversion in maps onto a p-congruent
triangle 
. Likewise, can be mapped by
inversion onto a p-congruent triangle 
.
By Lemma 17.2 and the SAS axiom for Euclidean geometry, we have that
d(O,B)=d(O,Y), d(O,C)=d(O,Z) and 
. Thus, the
Euclidean triangles, 
and 
, are congruent by
SAS. Therefore, there is a rotation about the center of and
possibly a reflection through a diameter which sends to
. This will map onto itself and the orthogonal circle
through B and C to the orthogonal circle through Y and Z. This
preserves the p-length, and hence the triangles are p-congruent.
Theorem 16.2: Two triangles in the Poincaré model of hyperbolic geometry
are p-congruent if and only if they can be mapped onto each other by a
succession of inversions in circles orthogonal to and/or reflections
in diameters of .
Figure 17.12: SAS in the Poincaré Model
At this point, I would rather think of measuring angles in radian measure, not
degree measure. In this case we know that the sum of the angles of any
triangle in hyperbolic geometry is less than 
.
Also, let 
denote
the radian measure of the angle of parallelism corresponding to the hyperbolic
distance d. We can define the standard trigonometric functions, not as
before--using right triangles--but in a standard way. Define
In this way we have avoided the problem of the lack of similarity in
triangles, the premise upon which all of real Euclidean trigonometry is based.
What we have done is to define these functions analytically, in terms of
a power series expansion. These functions are defined for all real numbers x
and satisfy the usual properties of the trigonometric functions.
Theorem 16.3: In the Poincaré model of hyperbolic geometry the angle of
parallelism satisfies the equation
Proof By the definition of the angle of parallelism, 
for
some point P to its foot Q in some p-line . Now, is half
of the radian measure of the fan angle at P, or is the radian measure of
, where 
is the limiting parallel ray to
through P.
We may choose to be a diameter of and Q=O, the center of
, so that P lies on a diameter of perpendicular to .
The limiting parallel ray through P is the arc of a circle so that
is orthogonal to ,
is tangent to at 
.
Figure 17.13: Angle of Parallelism
The tangent line to at P must meet at a point R inside
. This point R is the pole of 
from . By
Lemma 17.1 
.
Let us denote 
. Then in 
,
or
Now, 
.
Applying Lemma 17.2 we have
Using the
identity 
it follows that
Simplifying this it becomes
Also, we can write this as
.