MATH 4510/5510: Brown
Betweenness-1
Notes on Betweenness1
Axiom 1 (B-1). If A ∗ B ∗ C, then A, B, and C are three distinct points all lying on the same line, and
C ∗ B ∗ A.
←→
Axiom 2 (B-2). Given any two distinct points B and D, there exist points A, C, and E lying on BD such
that A ∗ B ∗ D, B ∗ C ∗ D, and B ∗ D ∗ E.
Axiom 3 (B-3). If A, B, and C are three distinct points lying on the same line, then one and only one of
the points is between the other two.
Proposition 1. For any points A and B:
−−→ −−→
1. AB ∩ BA = AB,
−−→ −−→
←→
2. AB ∪ BA = {AB}.
Proof.
−−→ −−→
−−→
−−→
1. By the definition of segment and ray AB ⊂ AB and AB ⊂ BA, so AB ⊂ AB ∩ BA. Now suppose
−−→ −−→
C ∈ BA ∩ AB. If C = A or C = B, then C is an endpoint of AB. Otherwise A, B, and C are collinear
by definition of ray and B-1. Then by B-3, exactly one of the relations A ∗ B ∗ C, A ∗ C ∗ B, or C ∗ A ∗ B
−−→
−−→
holds. If A ∗ B ∗ C holds, then C is not on BA; if C ∗ A ∗ B holds, then C is not on AB. In either case
−−→ −−→
C does not belong to both rays. Hence A ∗ C ∗ B holds, so C is on AB. Therefore, AB ∩ BA = AB.
−−→
←→
−−→
←→
−−→ −−→
←→
←→
2. By the definition of ray AB ⊂ {AB} and BA ⊂ {AB}, so AB ∪ BA ⊂ {AB}. Now suppose C ∈ {AB}.
If C 6= A and C 6= B, then by Betweenness 3, either A ∗ C ∗ B, A ∗ B ∗ C, or C ∗ A ∗ B. If A ∗ C ∗ B,
−−→
−−→
−−→
−−→
then C is on both AB and BA. If A ∗ B ∗ C, the C is on AB. If C ∗ A ∗ B, then C is on BA. In all of
−−→ −−→
←→
−−→ −−→
−−→ −−→
←→
these cases, C ∈ AB ∪ BA. Hence {AB} ⊂ AB ∪ BA. Therefore, AB ∪ BA = {AB}.
−→ −−→
Line Separation Property: If C ∗ A ∗ B and P is collinear with A, B, and C, then P ∈ AC ∪ AB.
Definition. Let l be any line, A and B any points that do not lie on l. If A = B or if segment AB contains
no point lying on l, we say A and B are on the same side of l, whereas if A 6= B and segment AB does
intersect l, we say that A and B are on opposite sides of l.
Axiom 4 (B-4). For every line l, and for any three points A, B, and C not lying on l:
1. If A and B are on the same side of l and B and C are on the same side of l, then A and C are on the
same side of l.
2. If A and B are on opposite sides of l and B and C are on opposite sides of l, then A and C are on
the same side of l.
Corollary 1. If A and B are on opposite sides of l and B and C are on the same side of l, then A and C
are on opposite sides of l.
Proof. Suppose that A and B are on opposite sides of l and B and C are on the same side of l. Suppose
further that A and C are on the same side of l (RAA). Since A and C are on the same side of l and B and
C are on the same side of l, by B-4(i) A and B are on the same side of l. This is a contradiction, so A and
C are on opposite sides of l.
Define a side of a line l to be the set of all points that are on the same side of l as some particular point
A not lying on l. Denote this side by HA . If C is on the same side of l as A, then HC = HA . A half-plane
bounded by l is a side of l.
Proposition 2. Every line bounds exactly two half-planes and these half-planes have no point in common.
1 The statements of the propositions and many proofs are taken from the book Euclidean and Non-Euclidean Geometries by
M. Greenberg.
MATH 4510/5510: Brown
Betweenness-2
Proof. Since for every line l there exists a point not on l, there exists a point A not on l. By I-2 there is a
point O lying on l. By B-2, there exists a point B such that B ∗ O ∗ A. Then segment AB intersects l at O,
so by definition A and B are on opposite sides of l, so l has at least two sides. Let C be any point distinct
from A and B not lying on l. If C and B are not on the same side of l, then A and C are on the same side
of l, by B-4(ii). Then the set of points not on l is the set HA ∪ HB . Suppose C is on both sides of l (RAA).
Then A and B are on the same side of l by B-4(i). This is a contradiction, so HA ∩ HB = ∅.
Lemma 1. Given A ∗ B ∗ C and A ∗ C ∗ D. A, B, C, and D are four distinct collinear points.
Proof. By B-1, A, B, and C are three distinct collinear points, and A, C, and D are three distinct collinear
points. Suppose B = D (RAA). By B-3, if A, B, and C are three distinct collinear points only one of the
points is between the other two, but if B = D, then A ∗ B ∗ C and A ∗ C ∗ B. This is a contradiction, so
←→
←→
←→
B 6= D. By I-1 the line AC is unique. B is on AC and D is on AC, so B and D are on the same line.
Therefore, A, B, C, and D are four distinct collinear points.
Proposition 3. Given A ∗ B ∗ C and A ∗ C ∗ D. Then B ∗ C ∗ D and A ∗ B ∗ D.
Proof. By lemma 1, A, B, C, and D are four distinct collinear points. Since for every line, there is a point
←→
not on that line, there exists a point E not lying on the line through A, B, C, and D. Consider line l = EC
by I-1. Since AD meets l at C, A and D are on opposite sides of l. Suppose A and B are on opposite sides
←→
of l (RAA). Then l meets AB at a point between A and B by the definition of opposite sides. This point
must be C, since distinct nonparallel lines intersect at exactly one point. Then A ∗ C ∗ B and A ∗ B ∗ C,
but this contradicts Betweenness 3. Hence A and B are on the same side of l. By Corollary 1, B and D are
←→
on opposite sides of l. Then l mets BD at a point between B and D. This point must be C, since distinct
nonparallel lines intersect at exactly one point. Therefore, B ∗ C ∗ D.
←→
Consider line m = EB by Incidence 1. Since AC meets l at B, A and C are on opposite sides of l.
←→
Suppose C and D are on opposite sides of l (RAA). Then l meets CD at a point between C and D. This
point must be B, since distinct nonparallel lines intersect at exactly one point. Then C ∗ B ∗ D and B ∗ C ∗ D,
but this contradicts Betweenness 3. Hence C and D are on the same side of l. By Corollary 1, A and D are
←→
on opposite sides of l. Then l mets AD at a point between A and D. This point must be B, since distinct
nonparallel lines intersect at exactly one point. Therefore, A ∗ B ∗ D.
Corollary 2. Given A ∗ B ∗ C and B ∗ C ∗ D. Then A ∗ B ∗ D and A ∗ C ∗ D.
Proof. By lemma 1, A, B, C, and D are four distinct collinear points. Since for every line there is a point
←→
not on that line, there exists a point E not lying on the line through A, B, C, and D. Let l = EC by
Incidence 1. By the argument in the proof of proposition 3, A and B are on the same side of l. Since BD
meets l at C, B and D are on opposite sides of l. By corollary 1, A and D are on opposite sides of l. Then
←→
AD intersects l at a point between A and D. This point must be C, so A ∗ C ∗ D. Then by proposition 3
since A ∗ B ∗ C and A ∗ C ∗ D, we have A ∗ B ∗ D.
Proposition 4. If C ∗ A ∗ B and l is the line through A, B, and C, then for every point P lying on l, P
−−→
−→
either lies on AB or AC.
−−→
−−→
Proof. Either P lies on AB or it does not. If P does lie on AB, then we’re done, so suppose P does not lie
−→
−−→
−−→
on AB. Then P ∗ A ∗ B by the definition of AB and Betweenness 3. If P = C, then P lies on AC and we’re
done, so assume P 6= C. By Betweenness 3, either P ∗ C ∗ A, C ∗ P ∗ A, or C ∗ A ∗ P .
Suppose C ∗ A ∗ P (RAA). By Betweenness 3, either C ∗ P ∗ B, C ∗ B ∗ P or P ∗ C ∗ B.
Case 1: Suppose C ∗ P ∗ B. By the RAA hypothesis C ∗ A ∗ P . By Proposition 3, A ∗ P ∗ B. This contradicts
Betweenness 3 since we assumed above the P ∗ A ∗ B.
Case 2: Suppose C ∗ B ∗ P . We are given that C ∗ A ∗ B. By Proposition 3, A ∗ B ∗ P . This contradicts
Betweenness 3 since we assumed P ∗ A ∗ B.
Case 3: Suppose P ∗ C ∗ B. We are given that C ∗ A ∗ B. By Proposition 3, P ∗ C ∗ A. This contradicts
Betweenness 3 since we assumed C ∗ A ∗ P as the RAA hypothesis.
−→
Since all three cases lead to a contradiction, we have P ∗ C ∗ A or C ∗ P ∗ A, so P lies on AC.
Theorem 1 (Pasch’s Theorem). If A, B, and C are distinct noncollinear points and l is any line intersecting
AB in a point between A and B, then l also intersects either AC or BC. If C does not lie on l, then l does
not intersect both AC and BC.
MATH 4510/5510: Brown
Betweenness-3
Proof. Either C lies on l or it does not. If C lies on l then we are done, so assume C does not lie on l. By
hypothesis A and B do not lie on l and segment AB does intersect l by Betweenness 1. Then A and B lie
on opposite sides of l. Since C does not lie on l, either C is on the same side of l as A or C is on the same
side of l as B by Proposition 2.
Suppose C is on the same side of l as A. By the definition of same side AC does not intersect l. By
←→
corollary 1, C is on the opposite side of l as B. By the definition of opposite sides, BC intersects l at a point
between B and C, i.e. BC intersects l.
←→
Suppose C is on the opposite side of l as A. By the definition of opposite sides AC intersects l at a point
between A and C, i.e. AC intersects l. By Betweenness 4, B and C are on the same side of l, so BC does
not intersect l.
Proposition 5. Given A ∗ B ∗ C. Then AC = AB ∪ BC and B is the only point common to segments AB
and BC.
Proof. We prove the first part by showing AB ∪ BC ⊂ AC and AC ⊂ AB ∪ BC. Suppose D is any point
on segment AB. If D = A, then D is on AC. If D = B, then D is on AC since A ∗ B ∗ C, so suppose
D 6= B. Then A ∗ D ∗ B, by the definition of segment. By proposition 3, since A ∗ B ∗ C and A ∗ D ∗ B,
we have A ∗ D ∗ C, so D is on AC. Then AB ⊂ AC. By Betweenness 1, since A ∗ B ∗ C we also have
C ∗ B ∗ A. We repeat the above argument exactly with A and C interchanged, so see that BC ⊂ AC. Then
AB ∪ BC ⊂ AC.
Suppose P is any point on AC distinct from A, B, and C. Since for every line there is a point not on
←→
←→
that line, there exists a point Q not on AC. By Incidence 1, we have the line l = P Q. A and B are either on
←→
the same side of l or the opposite sides of l. Suppose A and B are on opposite sides of l. Then AB intersects
l at a point between A and B. This point must be P , so P is on AB.
Suppose A and B are on the same side of l. Suppose B and C are on the same side of l (RAA). By
←→
Betweenness 4 A and C are on the same side of l. Then AC does not intersect l at a point between A and
←→
C, but this is a contradiction since AC intersects l at P and A ∗ P ∗ C. Hence if A and B are on the same
←→
side of l, then B and C are on opposite sides of l. Then BC meets l at a point between B and C. This point
must be P , so P is on BC. Hence AC ⊂ AB ∪ BC. Therefore, AC = AB ∪ BC.
Suppose P 6= B and P ∈ AB ∩ BC (RAA). Since P ∈ AB ∩ BC, P is on both AB and BC. Since A
is not on BC and C is not on AB, we know that P 6= A and P 6= C. By hypothesis, A ∗ B ∗ C and by
definition of segment A ∗ P ∗ B. By proposition 3, we have P ∗ B ∗ C, but then P is not on BC. This is a
contradiction, so B is the only point common to both AB and BC.
−−→
−−→
−−→ −→
Proposition 6. Given A ∗ B ∗ C. Then B is the only point common to rays BA and BC, and AB = AC.
−−→
−−→
−−→ −−→
Proof. Suppose D 6= B is any point on both BA and BC, i.e. D ∈ BA ∩ BC (RAA). D 6= A since A is not
−−→
−−→
on BC, so if D is on BA, then D ∗ A ∗ B or A ∗ B ∗ C. By hypothesis, A ∗ B ∗ C, so if D ∗ A ∗ B, then
−−→
D ∗ B ∗ C by corollary 2. Then D is not on BC, so D must satisfy A ∗ D ∗ B, i.e. D is on AB. D 6= C,
−−→
−−→
since C is not on BA, so if D is on BC, then B ∗ D ∗ C or B ∗ C ∗ D. By hypothesis and Betweenness 1,
−−
→
C ∗ B ∗ A, so if D ∗ C ∗ B, then D ∗ B ∗ A by corollary 2. Then D is not on BA, so D must satisfy B ∗ D ∗ C,
i.e. D must be on BC. Hence D is common to both AB and BC. By proposition 5, if A ∗ B ∗ C, then B is
the only point common to both AB and BC. This is a contradiction, so D = B.
−→
−−→
Let P be any point on AB. If P = A, then P is on AC and if P = B, then since A ∗ B ∗ C, P is on
−→
−−→
AC, so assume P 6= A and P 6= B. Since P is on AB, either A ∗ P ∗ B or A ∗ B ∗ P . If A ∗ P ∗ B, then
−→
−→
by proposition 3, A ∗ P ∗ C, so P is on AC. Suppose A ∗ B ∗ P . Suppose P is not on AC (RAA). Then
P ∗ A ∗ C. By proposition 3, since P ∗ A ∗ C and P ∗ B ∗ A, we have B ∗ A ∗ C. By hypothesis A ∗ B ∗ C, so
−→
−−→ −→
this contradicts Betweenness 3, so P is on AC. Hence AB ⊂ AC.
−−→
−−→
−→
Now let P be any point on AC. If P = A, then P is on AB and if P = C, then P is on AB since
−→
A ∗ B ∗ C, so assume P 6= A and P 6= C. Since P is on AC, either A ∗ P ∗ C or A ∗ C ∗ P . If A ∗ C ∗ P ,
−−→
−−→
then by proposition 3, A ∗ B ∗ P , so P is on AB. Suppose A ∗ P ∗ C. Suppose P is not on AB (RAA). Then
P ∗ A ∗ B. By corollary 2, since C ∗ P ∗ A and P ∗ A ∗ B, we have C ∗ A ∗ B. By hypothesis A ∗ B ∗ C, so
−−→
−→ −−→
−−→ −→
this contradicts Betweenness 3, so P is on AB. Hence AC ⊂ AB. Therefore, AB = AC.
Corollary 3. Every ray has a unique opposite ray.
MATH 4510/5510: Brown
Betweenness-4
−−→
−→
Proof. Suppose E ∗ A ∗ B and E ∗ A ∗ C and B 6= C. Then rays AB and AC are defined to be opposite
−→
−→
−→
to AE. By proposition 4, since B is not on AE, B is on ray AC, so either A ∗ B ∗ C or A ∗ C ∗ B. By
−→
−−→ −→
proposition 6, AB = AC. Hence there is a unique ray opposite to AE.
Definition. Given an angle ∠CAB, define a point D to be in the interior of ∠CAB if D is on the same
←→
←→
side of AC as B and if D is on the same side of AB as C. (The interior of an angle is the the intersection
of two half-planes.)
−−→
Lemma 2. Given a line l, a point A on l, and a point B not on l. Then every point of AB, except A, is
on the same side of l as B.
−−→
Proof. Suppose C is a point on AB. Then C = A, C = B, A ∗ C ∗ B, or A ∗ B ∗ C by Betweenness 3 and the
←→
definition of ray. Suppose B and C are on opposite sides of l (RAA). Then l meets BC at a point between
B and C by the definition of opposite sides. This point must be A, since nonparallel lines meet at exactly
−−→
one point. Then C ∗ A ∗ B. This is a contradiction, so every point on AB is on the same side of l as B.
←→
Proposition 7. Given an angle ∠CAB and point D lying on BC. Then D is in the interior of ∠CAB if
and only if B ∗ D ∗ C.
←→
Proof. Suppose D is a point on BC distinct from both B and C. By Betweenness 3, either D ∗ B ∗ C,
←→
B ∗ C ∗ D, or B ∗ C ∗ D. First, suppose that D ∗ B ∗ C. Then AB meets segment CD at B, so C and D are
←→
on opposite sides of AB, by the definition of opposite sides. Hence D is not in the interior of ∠CAB. Next
←→
←→
suppose that B ∗ C ∗ D. Then AC meets segment BD at C, so B and D are on opposite sides of AC, by the
definition of opposite sides. Hence D is not in the interior of ∠CAB. Finally, suppose B ∗ D ∗ C. Then D
−−→
←→
−−→
is on BC by the definition of ray. By Lemma 2 D is on the same side of AB as C. Also, D is on CB by the
←→
definition of ray. By Lemma 2 D is on the same side of AC as B. Hence D is in the interior of ∠CAB.
Proposition 8. If D is in the interior of ∠CAB,
−−→
1. so is every other point on ray AD except A,
−−→
2. no point on the opposite ray to AD is in the interior of ∠CAB, and
3. if C ∗ A ∗ E, then B is in the interior of ∠DAE.
Proof.
←→
←→
1. By Lemma 2, since A is a point on AB and D is a point on the same side of AB as C, every point on
−−→
←→
←→
AD except A is on the same side of AB as C. Similarly, by Lemma 2, since A is a point on AC and
←→
−−→
←→
D is a point on the same side of AC as D, every point on AD except A is on the same side of AC as
−−→
B. Hence every point on AD except A is in the interior of ∠CAB.
−−→
2. Suppose F is any point except A on the opposite ray to AD. By the definition of opposite ray and
←→
Betweenness 3 F ∗ A ∗ D. Then segment F D meets AC at A. Then D and F are on opposite sides of
←→
←→
AC. Since D is in the interior of ∠CAB, D and B are on the same side of AC. By Corollary 1, B and
←→
F are on opposite sides of AC, so F is not in the interior of ∠CAB.
3. Since C ∗ A ∗ E, by Betweenness 1 C, A, and E are collinear. Since D is in the interior of ∠CAB, D
←→ ←→
and B are on the same side of AC = AE. To finish we need to show that B and E are on the same
←→
side of AD.
←→
←→
Since C ∗ A ∗ E, segment CE meets AB at A, so C and E are on opposite sides of AB by the definition
←→
of opposite sides. Since D is in the interior of ∠CAB, D and C are on the same side of AB. By
←→
−−→
Corollary 1, D and E are on opposite sides of AB. By lemma 2, every point on BE is on the same
←→
−−→
←→
side of AB as E, and every point of AD is on the same side of AB as D. By corollary 1, every point
−−→
←→
−−→
−−→
on EB is on the opposite side of AB as every point on AD. In particular, AD does not meet segment
EB.
←→ ←→
−−→
From part (2), B and F are on opposite sides of AC = AE. By lemma 2, every point on EB is on the
←→
−→
←→
same side of AE as B and every point on AF is on the same side of AE as F . By corollary 1, every
MATH 4510/5510: Brown
Betweenness-5
−−→
←→
−→
−→
point of EB is on the opposite side of AE as every point of AF . In particular, AF does not meet
−→ −−→
←→
←→
segment EB. By Proposition 1, AF ∪ AD = {AD}, so segment BE does not meet AD. Hence E and
←→
B are on the same side of AD.
−−→
−→
−−→ −−→
−→
Definition. Ray AD is between rays AC and AB if AB and AC are not opposite rays and D is interior
to ∠CAB.
−−→
−→
−−→
−−→
Theorem 2 (Crossbar Theorem). If AD is between AC and AB, then AD intersects segment BC.
−−→
−→
−−→
Proof. Suppose AD is between AC and AB. By Betweenness 2, there exists a point F , such that F ∗ A ∗ D.
←→
By the argument in the proof of Proposition 8(b), B and F are on opposite sides of AC. By Lemma 2 and
−→
←→
−−→
Corollary 1, every point on AF is on the opposite side of AC as every point on CB. In particular, segment
−→
BC does not meet AF .
←→
By Betweenness 2, there is a point E such that E ∗ A ∗ C. Then segment EC meets AD at A, so E and
←→
C are on opposite sides of AD. From Proposition 7, if C ∗ A ∗ E, then B is in the interior of ∠DAE, so B
←→
←→
is on the same side of AD as E. By Corollary 1, B and C are on opposite sides of AD. By Proposition 1,
−→ −−→
←→
←→
←→
AF ∪ AD = {AD}. Since B and C are on opposite sides of AD, BC intersects AD. We have seen that BC
−−→
−→
does not intersect AF , so BC must intersect AD.
Definition. The interior of a triangle is the intersection of the interiors of its three angles. Define a point
to be exterior to the triangle if it is not in the interior and does not lie on any side of the triangle.
Proposition 9.
1. If a ray r emanating from an exterior point of △ABC intersects side AB in a point between A and B,
then r also intersects side AC or side BC.
2. If a ray emanates from an interior point of △ABC, then it intersects one of the sides, and if it does
not pass through a vertex, it intersects only one side.
Proof.
1. Suppose r is any ray emanating from a point X exterior to △ABC that intersects AB at a point D
−−→
←→
between A and B, i.e., r = XD. By Pasch’s Theorem XD intersects side AC or side BC at a point E.
−−→
We want to show E is on r = XD. Without loss of generality suppose E is on AC. E 6= X because E
lies on AC and X is exterior to △ABC. If E = D, then E is on r, so suppose E 6= D. By Betwenness
−−→
3, either X ∗ D ∗ E, X ∗ E ∗ D, or E ∗ X ∗ D. If either X ∗ D ∗ E or X ∗ E ∗ D holds, then E is on XD.
Suppose E ∗ X ∗ D (RAA). We will show that X is in the interior △ABC, arriving at a contradiction.
←→
By proposition 7, X is in the interior of ∠DAE = ∠BAC. By lemma 2, E is the same side of AB
−−→
←→
as C. By proposition 1, X is on DE, so by lemma 2 X is on the same side of AB as E. Then by
←→
←→
Betweenness 4, X is on the same side of AB as C. By lemma 2, D is on the same side of BC as A
←→
←→
and E is on the same side of BC as A. Then by Betweenness 4, D and E are on the same side of BC.
←→
←→
Then DE does not intersect BC. Suppose D and X are on opposite sides of BC (RAA). Then XD
←→
intersects BC at a point P where D ∗ P ∗ X. By proposition 3, since D ∗ P ∗ X and D ∗ X ∗ E, we
←→
have D ∗ P ∗ E, so DE intersects BC as P . This is a contradiction since we have already shown that
←→
←→
D and E are on the same side of BC. Hence D and X are on the same side of BC. By Betweenness
←→
4, X and A are on the same side of BC. Thus X is in the interior of ∠ABC. By the same type of
argument we can show that X is in the interior of ∠ACB. Then X is in the interior of △ABC, but
this is a contradiction because by hypothesis, X is exterior to △ABC. Hence ∼ E ∗ X ∗ D, i.e. E is
−−→
on XD.
−−→
2. Suppose D is any point in the interior of △ABC. By the Crossbar theorem AD intersects side BC at a
point E. Since D is in the interior of △ABC, D is in the interior of ∠ACB = ∠ACE. By proposition
7, A ∗ D ∗ E. Let r be any ray emanating from D. Then r is part of a line l through D, so l intersects
side AE of triangles △AEB and △AEC. By Pasch’s theorem l also intersects side AB or side AE in
△AEB and l intersects side CE or side AC in △AEC. Without loss of generality suppose l intersects
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Betweenness-6
AB at F and EC at G. Since D is in the interior of △ABC, D is in the interior of ∠ABC = ∠F BG,
so by proposition 7 F ∗ D ∗ G. Then by proposition 4 and corollary 3, r contains either F or G, so r
intersects a side of △ABC.