MATH4221 Euclidean and Non-Euclidean Geometries (2012 Spring)
© Henry Cheng @ HKUST
Tutorial Note 5
Mar 7, 2012 (Week 6)
MATH4221 Euclidean and Non-Euclidean Geometries
Tutorial Note 5
Topics covered in week 5:
8. Hilbert’s Axioms (1): Betweenness
8. Hilbert’s Axioms (1):
Betweenness
What you need to know:
 The fourth undefined term: Betweenness
 The four betweenness axioms



Definitions of segments, rays and angles
Definition of two points being on the same side of a line vs on opposite sides of a line
Pasch’s Theorem

Crossbar Theorem
On top of the axioms of incidence geometry we dealt with in the past three weeks, we
introduce another notion called betweenness. We now add “betweenness” as our fourth
undefined term, apart from “point”, “line” and “incidence”. Here are four new axioms:
(B1) (Between)
If 𝐴 ∗ 𝐵 ∗ 𝐶 , then (i) 𝐶 ∗ 𝐵 ∗ 𝐴 and (ii) 𝐴 , 𝐵 and 𝐶 are three
distinct collinear points.
(B2) (In-and-out)
For any two distinct points 𝐴 and 𝐶, there exist two (distinct) points
𝐵 and 𝐷 such that 𝐴 ∗ 𝐵 ∗ 𝐶 and 𝐴 ∗ 𝐶 ∗ 𝐷.
(B3) (Trichotomy)
If 𝐴, 𝐵 and 𝐶 are distinct and collinear, then one and only one
among 𝐵 ∗ 𝐴 ∗ 𝐶, 𝐴 ∗ 𝐵 ∗ 𝐶 and 𝐴 ∗ 𝐶 ∗ 𝐵 holds.
(B4) (Plane Separation) Given any line 𝑙, “being on the same side of 𝒍” is an equivalence
relation on the set of all points not incident to 𝑙, with exactly two
equivalence classes.
𝑙
To interpret the Plane Separation Axiom (B4), we need to
define what it means for two points to be “on the same
side of 𝑙”. Two points 𝐴 and 𝐵 not incident to 𝒍 are
said to be on opposite sides of 𝒍 if there exists 𝑋
incident to 𝑙 such that 𝐴 ∗ 𝑋 ∗ 𝐵. Otherwise 𝐴 and 𝐵
are said to be on the same side of 𝒍.
𝐴
𝑋
𝐵
𝐶
𝐴 and 𝐵 are on opposite sides of 𝑙;
𝐴 and 𝐶 are on the same side of 𝑙
Page 1 of 11
MATH4221 Euclidean and Non-Euclidean Geometries (2012 Spring)
© Henry Cheng @ HKUST
Example 8.1
Tutorial Note 5
Mar 7, 2012 (Week 6)
(Greenberg 3.1a, b) Suppose that 𝐴 ∗ 𝐵 ∗ 𝐶 and 𝐴 ∗ 𝐶 ∗ 𝐷.
Show that 𝐴,
𝐵, 𝐶 and 𝐷 are four distinct points that are collinear.
Proof:
We first show that the four points are distinct.
Since 𝐴 ∗ 𝐵 ∗ 𝐶, by Axiom (B1) we know that 𝐴, 𝐵 and 𝐶 are distinct.
Since 𝐴 ∗ 𝐶 ∗ 𝐷, by Axiom (B1) we know that 𝐴, 𝐶 and 𝐷 are also distinct.
It now remains to show that 𝐵 ≠ 𝐷. We suppose on the contrary that 𝐵 = 𝐷.
But this gives 𝐴 ∗ 𝐵 ∗ 𝐶 and 𝐴 ∗ 𝐶 ∗ 𝐵, which is a contradiction to Axiom (B3).
Therefore 𝐴, 𝐵, 𝐶 and 𝐷 are distinct points.
Now we show that the four points are collinear.
Since 𝐴 ∗ 𝐵 ∗ 𝐶, by Axiom (B1) we know that 𝐴, 𝐵 and 𝐶 are incident to some line 𝑙1.
Since 𝐴 ∗ 𝐶 ∗ 𝐷, by Axiom (B1) again, we know that 𝐴, 𝐶 and 𝐷 are incident to some line 𝑙2 .
Since 𝑙1 and 𝑙2 are both incident to both 𝐴 and 𝐶, by Axiom (I1), 𝑙1 = 𝑙2.
Therefore 𝐴, 𝐵, 𝐶 and 𝐷 are collinear.
∎
Please also get familiar with the definitions of a segment 𝐴𝐵 and a ray ⃗⃗⃗⃗⃗
𝐴𝐵 (the end points
⃗⃗⃗⃗⃗ and 𝐴𝐶
⃗⃗⃗⃗⃗ are called opposite rays if 𝐵 ∗ 𝐴 ∗ 𝐶.
are included in both definitions). Two rays 𝐴𝐵
An opposite ray to a given ray always exists by Axiom (B2).
Example 8.2
Let 𝑙 be a line, 𝐴 be a point incident to 𝑙, and 𝐵 be a point
not incident to 𝑙. Show that every point of the ray ⃗⃗⃗⃗⃗
𝐴𝐵 , except 𝐴, is on the
same side of 𝑙 as 𝐵.
(Greenberg 3.9)
Proof:
⃡⃗⃗⃗⃗ exists by Axiom (I1).
Clearly 𝐴 and 𝐵 are distinct points, so the line 𝐴𝐵
We proceed to prove by contradiction.
⃗⃗⃗⃗⃗ distinct from 𝐴, such that 𝐶 is
Suppose on the contrary that there exists a point 𝐶 on 𝐴𝐵
not on the same side of 𝑙 as 𝐵.
Since 𝐵 is on the same side of 𝑙 as itself, 𝐵 and 𝐶 must be distinct. We consider the two
cases where (i) 𝑪 is incident to 𝒍 or (ii) 𝐶 and 𝐵 are on opposite sides of 𝑙.
(i) If 𝐶 is incident to 𝑙, then 𝐴 and 𝐶 are distinct
⃡⃗⃗⃗⃗ = 𝑙 by Axiom (I1).
points incident to 𝑙, so 𝐴𝐶
But 𝐶 was a point on the ray ⃗⃗⃗⃗⃗
𝐴𝐵 distinct from
⃡⃗⃗⃗⃗ = 𝐴𝐶
⃡⃗⃗⃗⃗ by Axiom (I1).
both 𝐴 and 𝐵, so 𝐴𝐵
Thus ⃡⃗⃗⃗⃗
𝐴𝐵 = 𝑙, and this contradicts that 𝐵 is not
incident to 𝑙.
Page 2 of 11
Don’t miss out case (i)!!
𝐵
𝐶
𝐴
𝑙
The diagrams must
be inconsistent,
since we’re proving
by contradiction!
MATH4221 Euclidean and Non-Euclidean Geometries (2012 Spring)
© Henry Cheng @ HKUST
Tutorial Note 5
Mar 7, 2012 (Week 6)
(ii) If 𝐶 and 𝐵 are on opposite sides of 𝑙, then there exists a
𝑙
𝐵
point 𝐷 incident to 𝑙 such that 𝐵 ∗ 𝐷 ∗ 𝐶.
⃡⃗⃗⃗⃗ .
By Axiom (B1) we immediately see that 𝐷 is incident to 𝐵𝐶
Since 𝐶 was a point on the ray ⃗⃗⃗⃗⃗
𝐴𝐵 distinct from 𝐴 and 𝐵,
we have
⃡⃗⃗⃗⃗ by Axiom (I1), so 𝐷 is incident to ⃡⃗⃗⃗⃗
 ⃡⃗⃗⃗⃗
𝐴𝐵 = 𝐵𝐶
𝐴𝐵 , and
𝐷=𝐴
𝐶
𝑙
𝐵
𝐷
𝐴
 Either 𝐴 ∗ 𝐵 ∗ 𝐶 or 𝐴 ∗ 𝐶 ∗ 𝐵, so 𝐴 and 𝐷 are distinct
𝐶
by Axiom (B3).
⃡⃗⃗⃗⃗ and 𝑙, so 𝐴𝐵
⃡⃗⃗⃗⃗ = 𝑙 by Axiom (I1).
Now 𝐴 and 𝐷 are distinct points both incident to 𝐴𝐵
This again contradicts that 𝐵 is not incident to 𝑙.
∎
𝑃
2
We have actually eliminated the projective geometry ℙ from our
consideration once we introduce the notion of betweenness, because
2
there is no concept of “betweenness” in ℙ – just consider any three
2
2
𝑆2
𝑄
𝑅′
𝑅
𝑄′
distinct “points” on a “line” in ℙ (a great circle of 𝑆 ). We have
𝑃′
also eliminated all the finite geometries from our consideration,
because with the notion of “betweenness”, it can be shown (by mathematical induction) that
every segment is incident to infinitely many points. Try this out as an exercise! (Problem 2)
Example 8.3
(Greenberg 3.8) In incidence geometry with betweenness, show that every
line is incident to at least five distinct points.
Proof:
Let 𝑙 be a line. Then by Axiom (I2), there exist two distinct points 𝐵 and 𝐷 incident to 𝑙.
By Axiom (B2), there exist three points 𝐴, 𝐶 and 𝐸 incident to 𝑙 such that 𝐴 ∗ 𝐵 ∗ 𝐷,
𝐵 ∗ 𝐶 ∗ 𝐷 and 𝐵 ∗ 𝐷 ∗ 𝐸.
These three statements 𝐴 ∗ 𝐵 ∗ 𝐷, 𝐵 ∗ 𝐶 ∗ 𝐷 and 𝐵 ∗ 𝐷 ∗ 𝐸 immediately imply that each of
the points 𝐴, 𝐶 and 𝐸 are distinct from 𝐵 and 𝐷.
It now remains to show that 𝐴, 𝐶 and 𝐸 are distinct points. But this follows immediately
since otherwise Axiom (B3) is violated.
Therefore 𝐴, 𝐵, 𝐶, 𝐷 and 𝐸 are five distinct points incident to 𝑙.
∎
Given 𝐴 ∗ 𝐵 ∗ 𝐶, the fact that 𝐴𝐶 = 𝐵𝐴 ∪ 𝐵𝐶 and 𝐵𝐴 ∩ 𝐵𝐶 = {𝐵} is known as the Segment
Separation Theorem. It means that any point on the segment 𝐴𝐶 distinct from 𝐵 must lie
on exactly one of the smaller segments 𝐵𝐴 and 𝐵𝐶 (Assignment 2 Q1).
⃗⃗⃗⃗⃗ and ⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗ = {𝐵} is known
This theorem also generalizes to the line ⃡⃗⃗⃗⃗
𝐴𝐶 : ⃡⃗⃗⃗⃗
𝐴𝐶 = ⃗⃗⃗⃗⃗
𝐵𝐴 ∪ 𝐵𝐶
𝐵𝐴 ∩ 𝐵𝐶
as the Line Separation Theorem. It means that any point on ⃡⃗⃗⃗⃗
𝐴𝐶 distinct from 𝐵 must lie on
⃗⃗⃗⃗⃗ (i.e. 𝐵 “separates” the line ⃡⃗⃗⃗⃗
exactly one of the opposite rays ⃗⃗⃗⃗⃗
𝐵𝐴 and 𝐵𝐶
𝐴𝐶 ). You can refer
to Exercise Problem 1 for this generalization.
Page 3 of 11
MATH4221 Euclidean and Non-Euclidean Geometries (2012 Spring)
© Henry Cheng @ HKUST
Tutorial Note 5
Mar 7, 2012 (Week 6)
Besides the concepts of segments and rays, we can also define the notion of angles in incidence
geometry with betweenness. An angle is an equivalence class of the equivalence relation ~,
which is defined on the set of all ordered triples of non-collinear points as
⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗⃗⃗′ )
either (𝐵𝐴
𝐵𝐴′ and ⃗⃗⃗⃗⃗
𝐵𝐶 = 𝐵𝐶
(𝐴, 𝐵, 𝐶) ~ (𝐴′ , 𝐵 ′ , 𝐶 ′ ) ⇔ 𝐵 = 𝐵 ′ and (
).
⃗⃗⃗⃗⃗ = 𝐵𝐶
⃗⃗⃗⃗⃗⃗⃗′ and 𝐵𝐶
⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗⃗⃗⃗
or (𝐵𝐴
𝐵𝐴′ )
We also say that a point 𝐷 is in the interior of an angle ∠𝐵𝐴𝐶 if
the points 𝐶 and 𝐷 are on the same side of ⃡⃗⃗⃗⃗
𝐴𝐵 and the points 𝐵
and 𝐷 are on the same side of ⃡⃗⃗⃗⃗
𝐴𝐶 . In this case we also say that
⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗
𝐴𝐵 ∗ 𝐴𝐷 ∗ 𝐴𝐶 . This definition looks easier to understand if you
𝐵
𝐴
𝐷
𝐶
visualize it using the diagram on the right.
The following are two theorems regarding triangles in the subject of
𝐶
incidence geometry with betweenness.
(i) Pasch’s Theorem: For any triangle Δ𝐴𝐵𝐶, if a line 𝑙 meets
𝐴𝐵\{𝐴, 𝐵}, then it must also meet side 𝐴𝐶 or side 𝐵𝐶; and if it
meets both sides, then it must meet them at 𝐶.
(ii) Crossbar Theorem: If ⃗⃗⃗⃗⃗
𝐴𝐵 ∗ ⃗⃗⃗⃗⃗
𝐴𝐷 ∗ ⃗⃗⃗⃗⃗
𝐴𝐶 , then ⃗⃗⃗⃗⃗
𝐴𝐷 meets 𝐵𝐶 .
𝐴
(In other words, any ray emanating from a vertex of a triangle
toward the interior of the triangle must meet the opposite side.)
𝐵
𝐵
𝑙1
𝑙3
𝑙2
Pasch’s Theorem
𝐶
𝐷
Remark: A triangle refers just to three non-collinear points, and a “side”
of a triangle always refers to a segment. The interior of a
triangle is the intersection of the interiors of its three angles.
Example 8.4
𝐴
Crossbar Theorem
(Prenowitz & Jordan 12.II.1, 2) Let Δ𝐴𝐵𝐶 be a triangle. Prove that
(a) If 𝐵 ∗ 𝐷 ∗ 𝐶 and 𝐴 ∗ 𝐸 ∗ 𝐶, then 𝐴𝐷 meets 𝐵𝐸.
(b) If 𝐵 ∗ 𝐷 ∗ 𝐶 and 𝐴 ∗ 𝐸 ∗ 𝐷, then there exists 𝐹 such that 𝐴 ∗ 𝐹 ∗ 𝐶 and
𝐵 ∗ 𝐸 ∗ 𝐹.
𝐴
Proof:
⃡⃗⃗⃗⃗ , we have ⃗⃗⃗⃗⃗
(a) Since 𝐵 ∗ 𝐷 ∗ 𝐶 and 𝐴 is not incident to 𝐵𝐶
𝐴𝐵 ∗ ⃗⃗⃗⃗⃗
𝐴𝐷 ∗ ⃗⃗⃗⃗⃗
𝐴𝐶 .
⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗ ∗ 𝐴𝐷
⃗⃗⃗⃗⃗ ∗ ⃗⃗⃗⃗⃗
But since 𝐴 ∗ 𝐸 ∗ 𝐶, we have 𝐴𝐶
𝐴𝐸 and so 𝐴𝐵
𝐴𝐸 .
By Crossbar Theorem, we have ⃗⃗⃗⃗⃗
𝐴𝐷 meets 𝐵𝐸 at some point 𝐹.
𝐸
𝐹
𝐵
𝐷
𝐶
⃗⃗⃗⃗⃗ .
On the other hand, since 𝐴 ∗ 𝐸 ∗ 𝐶 and 𝐵 is not incident to ⃡⃗⃗⃗⃗
𝐴𝐶 , we have ⃗⃗⃗⃗⃗
𝐵𝐴 ∗ ⃗⃗⃗⃗⃗
𝐵𝐸 ∗ 𝐵𝐶
⃗⃗⃗⃗⃗ = 𝐵𝐷
⃗⃗⃗⃗⃗ ∗ 𝐵𝐹
⃗⃗⃗⃗⃗⃗ and 𝐵𝐸
⃗⃗⃗⃗⃗ = 𝐵𝐹
⃗⃗⃗⃗⃗ , so 𝐵𝐴
⃗⃗⃗⃗⃗ ∗ 𝐵𝐷
⃗⃗⃗⃗⃗⃗ .
Since 𝐵 ∗ 𝐷 ∗ 𝐶 and 𝐵 ∗ 𝐹 ∗ 𝐸, we have 𝐵𝐶
Since 𝐴, 𝐹 and 𝐷 are collinear, this implies that 𝐴 ∗ 𝐹 ∗ 𝐷, so 𝐴𝐷 meets 𝐵𝐸 at 𝐹. ∎
Page 4 of 11
MATH4221 Euclidean and Non-Euclidean Geometries (2012 Spring)
© Henry Cheng @ HKUST
Tutorial Note 5
Mar 7, 2012 (Week 6)
⃗⃗⃗⃗⃗ ∗ 𝐵𝐸
⃗⃗⃗⃗⃗ ∗ 𝐵𝐷
⃗⃗⃗⃗⃗⃗ .
(b) Since 𝐴 ∗ 𝐸 ∗ 𝐷 and 𝐵 is not incident to ⃡⃗⃗⃗⃗
𝐴𝐷, we have 𝐵𝐴
⃗⃗⃗⃗⃗ and so ⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗ .
But since 𝐵 ∗ 𝐷 ∗ 𝐶, we have ⃗⃗⃗⃗⃗⃗
𝐵𝐷 = 𝐵𝐶
𝐵𝐴 ∗ ⃗⃗⃗⃗⃗
𝐵𝐸 ∗ 𝐵𝐶
By Crossbar Theorem, we have ⃗⃗⃗⃗⃗
𝐵𝐸 meets 𝐴𝐶 at some point 𝐹.
In other words, we have 𝐴 ∗ 𝐹 ∗ 𝐶.
𝐴
𝐹
𝐸
𝐵
𝐶
𝐷
⃡⃗⃗⃗⃗ , we have ⃗⃗⃗⃗⃗
On the other hand, since 𝐵 ∗ 𝐷 ∗ 𝐶 and 𝐴 is not incident to 𝐵𝐶
𝐴𝐵 ∗ ⃗⃗⃗⃗⃗
𝐴𝐷 ∗ ⃗⃗⃗⃗⃗
𝐴𝐶 .
⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗ = 𝐴𝐹
⃗⃗⃗⃗⃗ , so 𝐴𝐵
⃗⃗⃗⃗⃗ ∗ ⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗ .
Since 𝐴 ∗ 𝐸 ∗ 𝐷 and 𝐴 ∗ 𝐹 ∗ 𝐶, we have 𝐴𝐷
𝐴𝐸 and 𝐴𝐶
𝐴𝐸 ∗ 𝐴𝐹
Since 𝐵, 𝐸 and 𝐹 are collinear, this implies that 𝐵 ∗ 𝐸 ∗ 𝐹.
∎
Example 8.5
(Greenberg 3.12a) Let Δ𝐴𝐵𝐶 be an arbitrary triangle. Prove that if 𝑟 is a
ray that emanates from an exterior point of Δ𝐴𝐵𝐶 and meets side 𝐴𝐵 (at a
point between 𝐴 and 𝐵), then 𝑟 also meets side 𝐴𝐶 or side 𝐵𝐶.
Proof:
Let 𝑂 be the exterior point of Δ𝐴𝐵𝐶 where 𝑟 emanates from, and
𝐷 be the point where 𝑟 meets 𝐴𝐵.
𝐶
𝐴
𝐵
𝐷
Then by hypothesis, 𝑂, 𝐴, 𝐵 and 𝐷 are all distinct, 𝑟 = ⃗⃗⃗⃗⃗⃗
𝑂𝐷, and
𝑂
𝐴 ∗ 𝐷 ∗ 𝐵.
Now the line ⃡⃗⃗⃗⃗
𝑂𝐷 meets the side 𝐴𝐵, so by Pasch’s Theorem, ⃡⃗⃗⃗⃗
𝑂𝐷 also meets the side 𝐴𝐶 or
the side 𝐵𝐶 (at a point that is distinct from both 𝐴 and 𝐵).
Without loss of generality, we suppose that ⃡⃗⃗⃗⃗
𝑂𝐷 meets 𝐴𝐶 at some 𝐸, where either 𝑬 = 𝑪
or 𝑨 ∗ 𝑬 ∗ 𝑪.
Want to show:
⃗⃗⃗⃗⃗⃗ = 𝑶𝑫
⃗⃗⃗⃗⃗⃗ , from which we can
It now remains to show that 𝑶𝑬
immediately conclude that 𝑟 = ⃗⃗⃗⃗⃗⃗
𝑂𝐷 meets 𝐴𝐶 at 𝐸.
⃗⃗⃗⃗⃗⃗ and 𝑂𝐸
⃗⃗⃗⃗⃗ are opposite rays, so 𝐷 ∗ 𝑂 ∗ 𝐸.
Suppose not, then 𝑂𝐷
It can then be shown that 𝑶 is in the interior of 𝚫𝑨𝑩𝑪, which
gives a contradiction:
𝐶
𝐸
𝐴
𝐵
𝐷
𝑂
Suppose not:
 Since 𝐷 ∗ 𝑂 ∗ 𝐸, it follows that 𝑂 is in the interior of ∠𝐷𝐴𝐸.
But we have ⃗⃗⃗⃗⃗
𝐴𝐷 = ⃗⃗⃗⃗⃗
𝐴𝐵 (since 𝐴 ∗ 𝐷 ∗ 𝐵) and ⃗⃗⃗⃗⃗
𝐴𝐸 = ⃗⃗⃗⃗⃗
𝐴𝐶
(no matter 𝐸 = 𝐶 or 𝐴 ∗ 𝐸 ∗ 𝐶), so ∠𝐷𝐴𝐸 = ∠𝐵𝐴𝐶.
This implies that 𝑶 is in the interior of ∠𝑩𝑨𝑪.
𝐶
𝐸
𝐴
𝐵
𝐷
𝑂
𝐶
 Since 𝐷 ∗ 𝑂 ∗ 𝐸, it follows that 𝑂 is in the interior of ∠𝐷𝐵𝐸.
⃗⃗⃗⃗⃗ (since 𝐴 ∗ 𝐷 ∗ 𝐵), so ∠𝐷𝐵𝐸 = ∠𝐴𝐵𝐸.
⃗⃗⃗⃗⃗⃗ = 𝐵𝐴
But we have 𝐵𝐷
No matter 𝐸 = 𝐶 or 𝐴 ∗ 𝐸 ∗ 𝐶, we can conclude that 𝑶 is in
the interior of ∠𝑨𝑩𝑪.
Page 5 of 11
𝐸
𝐴
𝑂
𝐷
𝐵
MATH4221 Euclidean and Non-Euclidean Geometries (2012 Spring)
© Henry Cheng @ HKUST
Tutorial Note 5
Mar 7, 2012 (Week 6)
 Finally since 𝐴 ∗ 𝐷 ∗ 𝐵, 𝐷 is in the interior of ∠𝐴𝐶𝐵.
⃗⃗⃗⃗⃗ = ⃗⃗⃗⃗⃗
(i) If 𝐸 = 𝐶, then 𝐷 ∗ 𝑂 ∗ 𝐶 would imply that 𝐶𝑂
𝐶𝐷. Thus 𝑶 is in the interior of
∠𝑨𝑪𝑩.
(ii) If 𝐴 ∗ 𝐸 ∗ 𝐶, then 𝐶 and 𝐸 are distinct, so 𝐷 ∗ 𝑂 ∗ 𝐸 would imply that 𝑂 is in the
⃗⃗⃗⃗⃗ = 𝐶𝐴
⃗⃗⃗⃗⃗ and 𝐷 is in the interior of ∠𝐴𝐶𝐵, 𝑶 is still in
interior of ∠𝐸𝐶𝐷. But since 𝐶𝐸
the interior of ∠𝑨𝑪𝑩.
Therefore 𝑂 is in the interior of Δ𝐴𝐵𝐶, which is a contradiction. So ⃗⃗⃗⃗⃗
𝑂𝐸 = ⃗⃗⃗⃗⃗⃗
𝑂𝐷 indeed.
∎
Given an angle and a point in its interior, be careful not to assume that the point must lie on
some segment with end points on the two sides of the angle (as in the figure below)!
Example 8.6
(Greenberg 3.19)
Recall the Beltrami-Klein model ℳ𝐵𝐾 we introduced in
Example 4.2, in which we interpret
 “a point” as a usual point in the open unit disk 𝐷(0; 1);
 “a line” as a usual chord of 𝐷(0; 1); and
 “incidence” as belonging.
Apart from the incidence axioms, we can easily check that the betweenness
axioms also hold in this model. Now draw a diagram to show that given any
angle in this model, there exist points in the interior of the angle which do not
lie on any line that intersects both sides of the angle.
Solution:
Given any arbitrary angle ∠𝐴𝐵𝐶 in this model, the usual rays
⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗ meet 𝜕𝐷 at two distinct usual points 𝑋 and 𝑌
𝐵𝐴 and 𝐵𝐶
respectively. The usual chord 𝑋𝑌 divides 𝐷 into two disjoint
segments, one of them contains all the three points 𝐴, 𝐵 and
𝐶, while the other does not. Now every point 𝑃 in the segment
not containing the points 𝐴, 𝐵 and 𝐶 is not incident to any line
⃗⃗⃗⃗⃗ and 𝐵𝐶
⃗⃗⃗⃗⃗ .
that intersects both rays 𝐵𝐴
Page 6 of 11
𝑋
𝐷
𝑃
𝐴
𝐵
𝑌
𝐶
MATH4221 Euclidean and Non-Euclidean Geometries (2012 Spring)
© Henry Cheng @ HKUST
Tutorial Note 5
Mar 7, 2012 (Week 6)
Exercise
1. (Greenberg 3.4) In this problem, we try to prove the Line Separation Theorem assuming that
we have already proved the Segment Separation Theorem as in Assignment 2 Q1. It is now
given that 𝐴 ∗ 𝐵 ∗ 𝐶.
(a) If 𝑃 is a point collinear with 𝐴, 𝐵 and 𝐶, show that not(𝐴 ∗ 𝐵 ∗ 𝑃) ⇒ not(𝐴 ∗ 𝐶 ∗ 𝑃).
⃗⃗⃗⃗⃗ ⊂ ⃗⃗⃗⃗⃗
(b) Deduce that ⃗⃗⃗⃗⃗
𝐵𝐴 ⊂ ⃗⃗⃗⃗⃗
𝐶𝐴 and that 𝐵𝐶
𝐴𝐶 .
⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗
(c) Hence, prove that 𝐵𝐴 ∩ 𝐵𝐶 = {𝐵}.
2. (Greenberg 3.8) We have shown in Example 8.3 that every line is incident to at least five points
in incidence geometry with betweenness. Now show using mathematical induction that every
segment (and thus every line) is in fact incident to infinitely many points.
3. (Noronha 1.2.7) Let 𝐶 and 𝐷 be points on opposite sides of a line ⃡⃗⃗⃗⃗
𝐴𝐵 . Show that either
𝐵 is an interior point of ∠𝐶𝐴𝐷 or 𝐴 is an interior point of ∠𝐶𝐵𝐷.
4. (Greenberg 3.12b, 13) Let Δ𝐴𝐵𝐶 be an arbitrary triangle. Prove that
(a) Any ray 𝑟 emanating from an interior point of Δ𝐴𝐵𝐶 must intersect at least one of the
three sides 𝐴𝐵, 𝐴𝐶 or 𝐵𝐶. Moreover, if 𝑟 passes through none of 𝐴, 𝐵 and 𝐶, then it
intersects at most one side.
Hint: Let 𝑂 be the point where 𝑟 emanates from. First show that the line ⃡⃗⃗⃗⃗
𝐴𝑂 meets
the side 𝐵𝐶 at some point 𝐸 with 𝐴 ∗ 𝑂 ∗ 𝐸.
(b) Any line 𝑙 cannot be contained in the interior of Δ𝐴𝐵𝐶.
5. (Greenberg 3.14) Investigate the betweenness axioms (B1), (B2), (B3) and Theorem 2.7 in the
Lecture Note. Which of these statements are still true if all the collinear points in the form
⃗⃗⃗⃗⃗ ∗ 𝑂𝐶
⃗⃗⃗⃗⃗ ?
𝐴 ∗ 𝐵 ∗ 𝐶 are replaced by coterminal rays in the form ⃗⃗⃗⃗⃗
𝑂𝐴 ∗ 𝑂𝐵
Remark: Two rays are said to be coterminal if they emanate from the same point.
6. (Prenowitz & Jordan 12.II.3, 4) Let Δ𝐴𝐵𝐶 be a triangle. Prove that
(a) If 𝐴 ∗ 𝑋 ∗ 𝐵, 𝐴 ∗ 𝑌 ∗ 𝐶 and 𝐵 ∗ 𝑍 ∗ 𝐶, then 𝐴𝑍 meets 𝑋𝑌.
(b) If 𝐴 ∗ 𝑋 ∗ 𝐵 , 𝐴 ∗ 𝑌 ∗ 𝐶 and 𝑋 ∗ 𝑍 ∗ 𝑌 , then there exists 𝑊 such that 𝐵 ∗ 𝑊 ∗ 𝐶 and
𝐴 ∗ 𝑍 ∗ 𝑊.
Hint:
The results from Problem 5 may be helpful.
Page 7 of 11