If 
and 
are limiting parallel rays, the convention is to
pretend that they pass through an ideal point
.
We then
denote the rays by 
and 
. The figure consisting of these
rays and the segment PQ is called a singly asymptotic triangle . You
can have doubly asymptotic triangles, with 2 ideal points, and trebly
asymptotic triangles, with 3 ideal points.
Lemma 13.3:[Exterior Angle Theorem] If 
is a singly asymptotic triangle, the exterior angles at P and Q are greater
than their respective opposite interior angles.
Proof: Choose 
so that 
. We need to show
that 
.
There is a unique ray lying on the same side of 
as
so that 
. If 
is so
that 
, then 
.
Thus, we have that the alternate interior angles are congruent. Let M be the
midpoint of PQ and let A and B be the feet of M in 
and 
, respectively. By AAS 
so that 
and 
. Thus, A, M, and B
are collinear. Thus, and have a common
perpendicular. Thus, 
. Thus
lies between 
and 
, from which it follows
that .
Lemma 13.4: If in the singly asymptotic triangles 
and 
we have 
, then 
Proof: First, let us assume that 
but 
. We may suppose that 
. Then, there is a unique ray 
between 
and
so that 
. Since 
and
are limiting parallel, must intersect
in a point X. There is a point 
so that
. Then by SAS 
. Therefore,
. Then, 
and
are not limiting parallel, a contradiction.
Now assume that 
and 
. We
may then assume that AB>CD. There is a unique point 
so that
. Let 
be the limiting parallel ray to
from P. Thus, is also parallel to
. The first part of this proof now implies 
. But 
is
exterior to 
, by the Exterior Angle Theorem 
, a contradiction.
Corollary: 
if and only if 
.
We are now in a position to show that 
.
Let 
, 
, and
. Recall that 
and 
. By Lemma 14.4 
.
Choose a point 
so that 
.
Since m and n were not limiting parallel lines, we have that
and 
. Now, 
by Angle Subtraction. Since , 
, and , we
must have that 
. Thus, 
is exterior
to the singly asymptotic triangle 
. Therefore, 
. This means that 
lies between 
and
, which implies that lies between and
. But since , we have that
Since 
, it lies on the same side of 
as does
A'. Thus, A and J are on opposite sides of . Thus,
. But, 
since H lies on
the same side of 
as does J.
It is possible to construct the limiting parallel ray to a line through a
given point with a compass and straightedge. Let 
and let Q be
the foot of P in 
. Let m be perpendicular to at P
so that 
. Let 
, 
and let S be the
foot of R in m. Then 
is a Lambert quadrilateral and PS<QR.
QR<PR since PR is a hypotenuse. Take a circle centered at P of radius
QR. S is inside this circle and R is outside this circle. There is a
unique point 
where RS intersects the circle. One can show
that the ray 
is limiting parallel to !
