Definition: Let 
be a set of lines in the plane. A line k is
transversal of if
, and
for all 
.
Let 
be transversal to m and n at points A and B, respectively.
We say that each of the angles of intersection of and m and of
and n has a transversal side in and a non-transversal
side not contained in .
Definition: An angle of intersection of m and k and one of n and k are
alternate interior angles if their transversal sides are opposite
directed and intersecting, and if their non-transversal sides lie on opposite
sides of . Two of these angles are corresponding angles
if their
transversal sides have like directions and their non-transversal sides lie on
the same side of .
Definition: If k and are lines so that 
, we
shall call these lines non-intersecting.
We want to reserve the word parallel for later.
Theorem 9.1:[Alternate Interior Angle Theorem] If two lines cut by a transversal have a pair of congruent alternate interior angles, then the two lines are non-intersecting.
Figure 10.1: Alternate interior angles
Proof:
Let m and n be two lines cut by the transversal . Let the points of
intersection be B and B', respectively. Choose a point A on m on one
side of , and choose 
on the same side of as A.
Likewise, choose 
on the opposite side of from A. Choose
on the same side of as C. Hence, it is on the opposite side
of from A', by the Plane Separation Axiom.
We are given that 
. Assume that the lines m
and n are not non-intersecting; i.e., they have a nonempty
intersection. Let us denote this point of intersection by D. D is on one
side of , so by changing the labeling, if necessary, we may assume that
D lies on the same side of as C and C'. By Congruence Axiom
1 there is a unique point 
so that 
. Since,
(by Axiom C-2), we may apply the SAS Axiom to prove that
From the definition of congruent triangles, it follows that
. Now, the supplement of 
is
congruent to the supplement of 
, by Proposition 8.5.
The supplement of is 
and . Therefore, 
is congruent to the supplement of . Since the angles share a side, they are themselves supplementary. Thus,
and we have shown that 
or that 
is more
that one point, contradicting Proposition 6.1. Thus, m and n must
be non-intersecting.
Corollary 1: If m and n are distinct lines both perpendicular to the
line , then m and n are non-intersecting.
Proof: is the transversal to m and n. The alternate interior
angles are right angles. By Proposition 8.14 all right angles are
congruent, so the Alternate Interior Angle Theorem applies. m and n
are non-intersecting.
Corollary 2: If P is a point not on , then the perpendicular
dropped from P to is unique.
Proof: Assume that m is a perpendicular to through P,
intersecting at Q. If n is another perpendicular to through
P intersecting at R, then m and n are two distinct lines
perpendicular to . By the above corollary, they are non-intersecting,
but each contains P. Thus, the second line cannot be distinct, and the
perpendicular is unique.
The point at which this perpendicular intersects the line , is called
the foot of the perpendicular.
Corollary 3: If is any line and P is any point not on ,
there exists at least one line m through P which does not intersect
.
Proof: By Corollary 2 there is a unique line, m, through P
perpendicular to . By Proposition 8.7 there is a unique line,
n, through P perpendicular to m. By Corollary 1 and n are non-intersecting.
Note that while we have proved that there is a line through P which does not
intersect , we have not (and cannot) proved that it is unique.