You should review the Betweenness Axioms in our list of axioms. There are numerous Propositions proved in the text based on the Betweenness Axioms. We shall list most of them, but shall prove only a few.
Proposition 6.1: For any two points A and B,
Figure: Proposition 7.1
It seems clear from Figure 7.1 that every point P lying on the line
through A, B, and C must either belong to ray 
or to an
opposite ray 
. This statement seems similar to the second assertion
of Proposition 7.1, but it is actually much more complicated. You are
now discussing four points and not the three of Proposition 7.1. You
will prove this assertion in your homework, with the addition of another axiom.
Let us call the assertion
as the line separation property . This is something that you will prove, but knowing that it can be proven, we shall use it as we need.and a point P is collinear with A, B, and C, implies that
.
Recall the definitions of same side and opposite sides. Also,
recall the last Betweenness Axiom. It is called the plane
separation axiom .
Note that it indirectly guarantees that our geometry will
be only two-dimensional, for such an axiom will not hold in three-dimensional
space. The points A and C may lie in the same plane with yet the
point B need not. In this case the conditions would be satisfied, yet the
conclusion would not hold.
Definition: Let be any line. A side of the line
containing A is the set of all points that are on the same side of as
A. This is also called the half-plane bounded by 
containing A.
Proposition 6.2: Every line bounds exactly two half-planes and these half-planes have no point in common.
Proposition 6.3: Given 
and 
,
then 
and 
.
Figure: Proposition 7.3
Proof: Compare Figure 7.2. From Exercise 1
are four
distinct points on the line. From Incidence Axiom 3 there is at least
one point E not on the line 
. Line 
exists by
Incidence Axiom 1, and by hypothesis
. Thus, A and D are on opposite sides
of . We claim that A and B lie on the same side of
.
Assume not; i.e., assume that A and B lie on opposite sides of
. Then intersects 
in a point
between A and B. By Proposition 6.1 that point must be C. Thus,
we have that 
and which contradicts
Betweenness Axiom 3. Thus, A and B lie on the same side of
. Since A and D are on opposite sides of , as
we have proven earlier, by Betweenness Axiom 4 B and D lie on
opposite sides of . Hence the unique point of intersection of
the lines and 
lies between B and D. This
unique point is C, by an earlier step. Thus, .
This proves half of Proposition 7.3. The other half is left as an exercise.
Proposition 6.4:[Line Separation Property]
If 
and is the line through A, B, and C, then for
every point P incident with , P is incident with either ray
or the opposite ray .
The next theorem is very interesting.
Theorem 6.1:[Pasch's Theorem]
If 
is any
triangle and
is any line intersecting side AB in a point between A and B, then
also intersects either side AC or side BC. If C is not incident with
, then does not intersect both AC and BC.
Proof: Compare Figure 7.3. Either 
or 
.
If then we are
done. Thus, let us assume that . By hypothesis 
and 
. Thus, A and B are on opposite sides of
. Then, since there are only two sides to any line, either C is on the
same side of as A or it is on the opposite side. If A and C are
on the same side of , then B and C are on opposite sides of .
Thus, 
. Likewise, if A and C are on opposite
sides, then 
, and we are done.
The following propositions are left to you to prove.
Proposition 6.5: Given , then 
and B is the only point common to the segments AB and BC.
Proposition 6.6: Given , then B is the only
point common to the rays 
and 
, and 
.
Definition: Given an angle 
, we shall say that a point D is in
the interior of the angle if D is on the same side of
as C and on the same side of 
as B.
Note that this implies that the interior of an angle is the intersection of two half planes.
Proposition 6.7: Let be any angle and let
D be any point incident with line 
. D is in the interior of
if and only if 
.
Proposition 6.8: If D is in the interior of angle , then:
except A;
is in the interior of ;
then B is in the interior of 
.
Definition: Ray is between rays and if
and are not opposite rays and D is in the
interior of .
Theorem 6.2:[Crossbar Theorem]
If is between and , then intersects the segment BC.
Proof: We are given a ray between rays and
. Let us use proof by contradiction and assume that 
.
Let 
be the opposite ray to . If 
, then
and by Proposition 7.7 we have that P lies in the
interior of . However, this contradicts (ii) of
Proposition 7.8, which says that no point on the opposite ray can be
interior to the angle. Thus, we have that 
. Now,
this means that 
by Proposition 7.1,
part (ii). It follows that B and C are on the same side of the line
.
Let E be a point on the line so that . Then,
C and E are on opposite sides of , and by Betweenness
Axiom 4, B and E are on opposite sides of . Applying
Proposition 7.8, part (iii) to this situation, we know that B
is in the interior of , which means that B and E are on the
same side of the line . We now have a contradiction.
Thus, we have that 
.
Lastly, we need to consider the incidence of rays and triangles.
Definition: The interior of a triangle is the intersection of the interiors of the three angles. A point is in the exterior of the triangle if it is not in the interior or on any of the three sides.
Proposition 6.9:
emanating from a point exterior to
intersects side AB in a point between A and B, then also
intersects AC or BC.
, then it
intersects on of the sides, and if it does not pass through a vertex, it
intersects only one side.
