Axioms:
Incidence Axioms
I-1:
Each two distinct points determine a line.
I-2:
Three noncollinear points determine a plane.
I-3:
If two points lie in a plane, then the line determined by those two points lies in
that plane.
I-4:
If two planes meet, their intersection is a line.
I-5:
Space consists of at least four noncoplanar points, and contains three noncollinear
points. Each plane contains at least three noncollinear points, and each line
contains at least two distinct points.
Metric Axioms:
D-1:
Each pair of points (A,B) is associated with a unique real number AB$0, called the
distance from A to B.
D-2:
For all points A and B, AB $ 0 with equality only when A=B.
D-3:
For all points A and B, AB=BA.
Definition: For any three points A, B, and C, we say that B is between A and C, and we write AB-C, iff A, B, and C are collinear points and AC = AB + BC.
Definition: If A, B, C, and D are distinct collinear points, then A-B-C-D represents the composite
of A-B-C, A-C-D, A-B-D, and B-C-D.
Theorem 1: If A-B-C then C-B-A, and neither A-C-B nor B-A-C.
Outline of proof: CB+BA = BC+AB = AB +BC = AC = CA. If AB+BC =AC, and AC
+CB = AB, then AB + BC +CB = AB, or in other words, AB + 2BC = AB, so BC=0, a
contradiction. Similar proof for B-A-C.
Theorem 2: If A-B-C, B-C-D, and A-B-D hold, then A-B-C-D.
Outline of Proof: We need to prove A-C-D. Now AC + CD = AB + BC + CD = AB +BD
= AD.
Note: In general, A-B-C-D is not the same as AB+BC+CD=AD. Suppose A, B, C and D are
collinear, with AB=2, BC=3, CD=4, AC=6, BD=8, AD=9. H er e,
AB + BC + CD = 2 + 3 + 4 = 9 = AD, but none of A-B-C, B-C-D, A-B-D, or A-C-D holds.
Definitions:
Segment
Ray
= {A,B} c {X: A-X-B}
= {A,B} c {X: A-X-B} c {Y: A-B-Y}
Angle pABC =
for A,B,C noncollinear.
The points A and B are called the endpoints of
origin of
, the point A is called the endpoint or
. The point B of pABC is the vertex and rays
The extension of
of B) or the line
is either the ray
and
are its sides.
(in the direction of B) or the ray
(in both directions). The extension of the ray
(in the direction
is the line
.
Some easy to prove facts:
1.
2.
Proofs of these are almost immediate from the definitions.
Our text also notes that
It would be an interesting and useful exercise to try to prove this fact from Axioms I-1 through I5 and D-1 through D-3. In fact, it cannot be done. Let’s go back to our last model:
Suppose A, B, C and D are collinear, with AB=2, BC=3, CD=4, AC=6, BD=8, AD=9.
Clear ly, bot h B a n d D a r e on line
set
B-C nor A-C-D holds.
, bu t t h ey would n ’t en d u p in t h e
, since neither A-
The moral of the story is that D-1 – D-3 are not particularly strong. Lots of funny things can go
wrong. In some developments, extra axioms are thrown in to specifically make sure that this
definition of line holds, e.g.
“D-4" :
whenever A-C-B.
If we start with a betweenness condition on our line, this makes our above “definition” of line
work. Is this enough to AB+BC+CD=AD equivalent to A-B-C-D? (It would be interesting to try
to prove this one way or the other.)
What we really need is a much more powerful axiom, that makes distances work nicely and
makes lines behave essentially like number lines. Here it is:
D-4 (The Ruler Postulate): The points of each line l may be assigned to the real numbers x, -4 <
x < 4, called coordinates, in such a manner that:
(1)
(2)
(3)
(4)
each point on l is assigned to a unique coordinate
no two points are assigned to the same coordinate
any two points on l may be assigned to zero and a positive coordinate, respectively
if points A and B on l have coordinates a and b, then AB = *a-b*.
We denote that the coordinate of A is a by writing, A[a].
Some implications of the Ruler Postulate:
1.
2.
3.
There are now as many points on a line as there are real numbers.
Distances must now correspond to ordering of real numbers.
We have it in our power to create points using distances from other points.
We can now prove a number of useful facts relating coordinates to betweeness:
Theorem 3:
If A[a], B[b], and C[c] are three collinear points with their coordinates, then A-B-C iff a<b<c or
c<b<a.
Outline of proof:
7): Given a<b<c or c<b<a. If a<b<c, let AB = b-a, BC =c-b, AC = c-a. Then
AB + BC = b - a + c - b = c - a = AC
Therefore, A-B-C. If c<b<a, let AB = b-a, BC = b-c, AC = a-c, and the same results
follow.
6): Given A-B-C. If either b<a<c or c<a<b, then part 1 of our proof would give B-A-C,
a contradiction. Similarly, if b<c<a or a<c<b, we would have B-C-A, a contradiction.
The only possibilities left are a<b<c or c<b<a.
This theorem is a basic tool. In particular we can now prove:
Theorem: If A, B, C, and D are collinear, and AB + BC + CD = AD, then A-B-C-D.
Outline of proof: Using the ruler postulate, let the coordinate of A be 0, and the
coordinate of B be b>0. Let the coordinates of C and D be c and d, respectively. There
are only 6 ways of choosing positions for c and d, and 2 ways of ordering them in these
positions, for a total of 12 possibilities. These are:
Order
AB
BC
CD
AB+BC+CD
AD
Implication
0<b<c<d
b
c-b
d-c
d
d
d=d
0<b<d<c
b
c-b
c-d
2c-d
d
c=d
0<c<d<b
b
b-c
d-c
2b-2c+d
d
b=c
0<c<b<d
b
b-c
d-c
2b-2c+d
d
b=c
0<d<b<c
b
c-b
c-d
2c-d
d
c=d
0<d<c<b
b
b-c
c-d
2b-d
d
b=d
d<c<0<b
b
b-c
c-d
2b-d
-d
b=0
c<d<0<b
b
b-c
d-c
2b-2c+d
-d
b=c-d<0
d<0<c<b
b
b-c
c-d
2b-d
-d
b=0
c<0<d<b
b
b-c
d-c
2b-2c+d
d
b=c
c<0<b<d
b
b-c
d-c
2b-2c+d
d
b=c
d<0<b<c
b
c-b
c-d
2c-d
-d
c=0
The only of these orderings that doesn’t lead to a contradiction is the first, namely, 0<b<c<d.
With this ordering,
AB + BC = b + c-b = c = AC so A-B-C
AB + BD = b + d-b = d = AD so A-B-D
AC + CD = c + d-c = d = AD so A-C-D
BC + CD = c-b + d-c =d-b = BD so B-C-D.
Thus, A-B-C-D.
with A … C, then
Lemma: If A and B are two distinct points and
Outline of proof: By ruler postulate let the coordinates of A, B, and C be 0, b>0 and c,
respectively. Since either A-C-B or A-B-C, c>0. If X is any point on ray
and x is its
coordinate, then x$ 0. Now since b, c, and x are real numbers, either 0#x#c (so A-X-C
and X is on ray
) or 0<c<x (so A-C-X and again X is on
arbitrary point X from
must also be on
.
). Either way, the
Theorem 4: If C0
and A… C, then
=
.
Proof: Exercise; use Lemma.
Theorem 5
Using the ruler postulate, we are able to create points, segments, and etc. with properties
dependent on length and distance. For example we can:
“Copy” one segment onto another.
Find midpoints, thus bisecting segments.
“Double” segments.
“Triple” segments
Find a segment that is 31/57 the length of a given segment.
Etc. etc. ad nauseam.