and 
are limiting parallel to each other, the
convention is to pretend that they go through an ideal point
and denote them simply by the segments 
and 
. The
figure consisting of these rays and the segment PQ is then called a
trilateral . Note that here we do not assume that PQ is
necessarily perpendicular to . Trilaterals are also
called singly asymptotic triangles . The angles of a trilateral are the angles 
and 
, and these are called its interior
angles.
Just as we have defined a trilateral--singly asymptotic triangle--we can define a DAT--doubly asymptotic triangle.
Given a point P and two non-opposite rays through P--
and
--let 
be the line that is limiting parallel to
and limiting parallel to in the opposite direction. Let
denote the ideal point in the direction of and let 
denote
the ideal point in the direction of . Then we can identify the line
with the segment 
and the segments ,
and form a doubly asymptotic triangle. The angle
is the fan angle of and P.
Many of the properties of trilaterals are analogous to those of triangles.
Lemma 14.1:[Crossbar Theorem for Trilaterals] A line which subdivides an angle of a trilateral intersects the opposite side.
Lemma 14.2:[Pasch's Theorem for Trilaterals]
A line which intersects a side
of the trilateral 
but does not pass through a vertex will
intersect another side, provided the line is not limiting parallel to either
or 
at .
Lemma 14.3: The sum of the interior angles of a trilateral is less than
.
Lemma 14.4:[Exterior Angle Theorem]
If 
is a trilateral, 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 .
Two trilaterals are congruent if the angles and segment of one are congruent, respectively, to the angles and segment of the other.
Lemma 14.5: If in the trilateral 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.