Download Tutorial Note 6

MATH4221 Euclidean and Non-Euclidean Geometries (2012 Spring)
© Henry Cheng @ HKUST
Tutorial Note 6
Mar 14, 2012 (Week 7)
MATH4221 Euclidean and Non-Euclidean Geometries
Tutorial Note 6
Topics covered in week 6:
9. Hilbert’s Axioms (2): Congruence
10. Hilbert’s Axioms (3): Continuity and Parallelism
9. Hilbert’s Axioms (2):
Congruence
What you need to know:
 The fifth and sixth undefined terms:
Segment congruence and angle congruence



The six congruence axioms
Definition of congruent triangles
Properties of isosceles triangles



Total orders of segments and of angles
ASA Theorem and SSS Theorem for congruence of triangles
Right angles (90°), feet of perpendicular
We now introduce the last two undefined terms in our theory, “segment congruence” and
“angle congruence”, together with six new axioms:
(C1) (Sgmt Congruence) Segment congruence is an equivalence relation.
(C2) (Angle Congruence) Angle congruence is an equivalence relation.
(C3) (Sgmt Relocation) Given a segment 𝐴𝐵 and any ray ⃗⃗⃗⃗⃗⃗
𝐴′ 𝐶 , there exists 𝐵 ′ ∈ ⃗⃗⃗⃗⃗⃗
𝐴′ 𝐶
(distinct from 𝐴′ ) such that 𝐴𝐵 ≅ 𝐴′ 𝐵 ′.
(C4) (Angle Relocation) Given an angle ∠𝐴𝐵𝐶 and any ray ⃗⃗⃗⃗⃗⃗⃗⃗
𝐴′ 𝐵 ′ , there exists 𝐶 ′ on each
given side of the line ⃡⃗⃗⃗⃗⃗⃗⃗
𝐴′ 𝐵 ′ such that ∠𝐴𝐵𝐶 ≅ ∠𝐴′ 𝐵 ′ 𝐶 ′ .
(C5) (Segment Addition) If 𝐴 ∗ 𝐵 ∗ 𝐶, 𝐴′ ∗ 𝐵 ′ ∗ 𝐶 ′ , 𝐴𝐵 ≅ 𝐴′ 𝐵 ′ , 𝐵𝐶 ≅ 𝐵 ′ 𝐶 ′ , then 𝐴𝐶 ≅ 𝐴′ 𝐶 ′ .
(C6) (SAS Axiom)
If 𝐴 , 𝐵 , 𝐶 are non-collinear, 𝐴′ , 𝐵 ′ , 𝐶 ′ are non-collinear,
∠𝐵𝐴𝐶 ≅ ∠𝐵 ′ 𝐴′ 𝐶 ′ , 𝐴𝐵 ≅ 𝐴′ 𝐵 ′, 𝐴𝐶 ≅ 𝐴′ 𝐶 ′ , then Δ𝐴𝐵𝐶 ≅ Δ𝐴′ 𝐵 ′ 𝐶 ′ .
Remark:
Be careful not to mix up the notions of equality of two angles and congruence of two
angles. Equal angles are congruent, but congruent angles may not be equal.
𝐴
𝐴
𝐵
𝐴′
𝐵=
𝐵′
𝐴′
𝐶′
𝐶
𝐶
𝐵′
∠𝐴𝐵𝐶 ≅ ∠𝐴′ 𝐵′ 𝐶 ′
∠𝐴𝐵𝐶 = ∠𝐴′ 𝐵′ 𝐶 ′
Page 1 of 12
𝐶′
MATH4221 Euclidean and Non-Euclidean Geometries (2012 Spring)
© Henry Cheng @ HKUST
Example 9.1
Tutorial Note 6
Mar 14, 2012 (Week 7)
(Vertically opposite angles) Suppose that 𝐵 ∗ 𝐴 ∗ 𝐸 and 𝐶 ∗ 𝐴 ∗ 𝐷, where
⃡⃗⃗⃗⃗
𝐵𝐸 and ⃡⃗⃗⃗⃗
𝐶𝐷 are distinct lines. Using only the congruence axioms, show that
∠𝐵𝐴𝐶 ≅ ∠𝐷𝐴𝐸.
Proof:
⃗⃗⃗⃗⃗ , 𝐷′ ∈ 𝐴𝐷
⃗⃗⃗⃗⃗ and 𝐸 ′ ∈ ⃗⃗⃗⃗⃗
By the Segment Relocation Axiom (C3), there exists 𝐶 ′ ∈ 𝐴𝐶
𝐴𝐸 such that
𝐴𝐶 ′ ≅ 𝐴𝐵, 𝐴𝐷′ ≅ 𝐴𝐵 and 𝐴𝐸 ′ ≅ 𝐴𝐵. Without loss of generality we relabel 𝐶 ′ , 𝐷′ and 𝐸 ′
as 𝐶, 𝐷 and 𝐸.
𝐷
𝐵
Now since 𝐴𝐵 ≅ 𝐴𝐷, 𝐴𝐷 ≅ 𝐴𝐵 and ∠𝐵𝐴𝐷 = ∠𝐷𝐴𝐵, we have
Δ𝐴𝐵𝐷 ≅ Δ𝐴𝐷𝐵 by SAS Axiom (C6). In particular, ∠𝐴𝐵𝐷 ≅ ∠𝐴𝐷𝐵.
𝐴
𝐸
Next, since 𝐶 ∗ 𝐴 ∗ 𝐷, 𝐸 ∗ 𝐴 ∗ 𝐵, 𝐶𝐴 ≅ 𝐸𝐴 and 𝐴𝐷 ≅ 𝐴𝐵, we
𝐶
have 𝐶𝐷 ≅ 𝐸𝐵 by Segment Addition Axiom (C5).
Together with 𝐵𝐷 = 𝐷𝐵 and ∠𝐶𝐷𝐵 = ∠𝐴𝐷𝐵 ≅ ∠𝐴𝐵𝐷 = ∠𝐸𝐵𝐷, we have Δ𝐵𝐶𝐷 ≅ Δ𝐷𝐸𝐵
by SAS Axiom (C6). In particular, ∠𝐵𝐶𝐷 ≅ ∠𝐷𝐸𝐵 and 𝐵𝐶 ≅ 𝐷𝐸.
Finally, since 𝐵𝐶 ≅ 𝐷𝐸 , 𝐴𝐶 ≅ 𝐴𝐸 and ∠𝐵𝐶𝐴 = ∠𝐵𝐶𝐷 ≅ ∠𝐷𝐸𝐵 = ∠𝐷𝐸𝐴 , we have
Δ𝐵𝐶𝐴 ≅ Δ𝐷𝐸𝐴 by SAS Axiom (C6). In particular, ∠𝐵𝐴𝐶 ≅ ∠𝐷𝐴𝐸.
Remark:
∎
This is actually an easy consequence of Lemma 2.34 in the Lecture Note, which states
that supplementary angles of congruent angles are congruent.
We define total orders on the sets of congruence classes of segments and angles. We write
 𝐴𝐵 < 𝐶𝐷 if there exists 𝐸 such that 𝐶 ∗ 𝐸 ∗ 𝐷 and 𝐴𝐵 ≅ 𝐶𝐸.
 ∠𝐴𝐵𝐶 < ∠𝐷𝐸𝐹 if there exists 𝐺 in the interior of ∠𝐷𝐸𝐹 such that ∠𝐴𝐵𝐶 ≅ ∠𝐷𝐸𝐺.
Example 9.2
Suppose that 𝐴 ∗ 𝐵 ∗ 𝐶 and 𝐴′ ∗ 𝐵 ′ ∗ 𝐶 ′ .
Show that if 𝐴𝐵 < 𝐴′ 𝐵 ′ and
𝐵𝐶 < 𝐵 ′ 𝐶 ′ , then 𝐴𝐶 < 𝐴′ 𝐶 ′ .
Proof:
Since 𝐴𝐵 < 𝐴′ 𝐵 ′ , by Segment Relocation Axiom (C3) there exists
a point 𝑃 such that 𝐴′ ∗ 𝑃 ∗ 𝐵 ′ and 𝐴𝐵 ≅ 𝐴′ 𝑃.
Since 𝐴′ ∗ 𝐵 ′ ∗ 𝐶 ′ and 𝐴′ ∗ 𝑃 ∗ 𝐵 ′, we have 𝐴′ ∗ 𝑃 ∗ 𝐶 ′ and
𝑃 ∗ 𝐵 ′ ∗ 𝐶 ′ , so 𝑩′ 𝑪′ < 𝑷𝑪′ .
Since 𝐵𝐶 < 𝐵 ′ 𝐶 ′ , we have 𝐵𝐶 < 𝑃𝐶 ′ .
So by Segment Relocation Axiom (C3), there exists 𝑄 such that
𝑃 ∗ 𝑄 ∗ 𝐶 ′ and 𝐵𝐶 ≅ 𝑃𝑄.
Since 𝐴′ ∗ 𝑃 ∗ 𝐶 ′ and 𝑃 ∗ 𝑄 ∗ 𝐶 ′ , we have 𝐴′ ∗ 𝑃 ∗ 𝑄 and
𝐴′ ∗ 𝑄 ∗ 𝐶 ′ .
Now we have 𝐴 ∗ 𝐵 ∗ 𝐶, 𝐴′ ∗ 𝑃 ∗ 𝑄, 𝐴𝐵 ≅ 𝐴′ 𝑃 and 𝐵𝐶 ≅ 𝑃𝑄,
so 𝐴𝐶 ≅ 𝐴′ 𝑄 by Segment Addition Axiom (C5).
Since 𝐴′ ∗ 𝑄 ∗ 𝐶 ′ and 𝐴𝐶 ≅ 𝐴′ 𝑄, we have 𝐴𝐶 < 𝐴′ 𝐶 ′ .
Page 2 of 12
𝐶
𝐴
𝐴′
𝐵
′
𝑃 𝐵
𝑄
𝐶′
It does not matter whether
𝐵′ ∗ 𝑄 ∗ 𝐶 ′ or 𝑄 ∗ 𝐵′ ∗ 𝐶 ′ .
∎
MATH4221 Euclidean and Non-Euclidean Geometries (2012 Spring)
© Henry Cheng @ HKUST
Remark:
Tutorial Note 6
Mar 14, 2012 (Week 7)
𝐶
As in the figure on the right, one can also construct
a similar proof using the existence of a point 𝑃 such
that 𝐴′ ∗ 𝑃 ∗ 𝐵 ′ and 𝐴𝐵 ≅ 𝑃𝐵 ′ , and a point 𝑄
such that 𝐵 ′ ∗ 𝑄 ∗ 𝐶 ′ and 𝐵𝐶 ≅ 𝐵 ′ 𝑄. Try it out!
𝐵
𝐴
𝐴′
𝑄
𝐵′
𝑃
𝐶′
ASA and SSS are valid criteria for congruence of triangles in the subject of incidence geometry
with betweenness and congruence, just as in high school.
Example 9.3
(Greenberg 3.28 adapted)
Prove that triangle is equiangular if and only if it is
equilateral.
Proof:
(⇒) Let Δ𝐴𝐵𝐶 be an equiangular triangle, i.e. ∠𝐵𝐴𝐶 ≅ ∠𝐴𝐵𝐶 and ∠𝐴𝐵𝐶 ≅ ∠𝐴𝐶𝐵.
Then using ∠𝐵𝐴𝐶 ≅ ∠𝐴𝐵𝐶, ∠𝐴𝐵𝐶 ≅ ∠𝐵𝐴𝐶 and 𝐴𝐵 = 𝐵𝐴, we have Δ𝐴𝐵𝐶 ≅ Δ𝐵𝐴𝐶
by ASA Theorem. In particular, 𝐴𝐶 ≅ 𝐵𝐶.
Next using ∠𝐴𝐵𝐶 ≅ ∠𝐴𝐶𝐵, ∠𝐴𝐶𝐵 ≅ ∠𝐴𝐵𝐶 and 𝐵𝐶 = 𝐶𝐵, we have Δ𝐴𝐵𝐶 ≅ Δ𝐴𝐶𝐵 by
ASA Theorem. In particular, 𝐴𝐵 ≅ 𝐴𝐶.
Now the three sides of Δ𝐴𝐵𝐶 are all congruent, so Δ𝐴𝐵𝐶 is equilateral.
(⇐) Let Δ𝐴𝐵𝐶 be an equilateral triangle, i.e. 𝐴𝐵 ≅ 𝐴𝐶 and 𝐴𝐶 ≅ 𝐵𝐶.
Then using 𝐴𝐶 ≅ 𝐵𝐶 , 𝐵𝐶 ≅ 𝐴𝐶 and 𝐴𝐵 = 𝐵𝐴 , we have Δ𝐴𝐵𝐶 ≅ Δ𝐵𝐴𝐶 by SSS
Theorem. In particular, ∠𝐴𝐵𝐶 ≅ ∠𝐵𝐴𝐶.
Next using 𝐴𝐵 ≅ 𝐴𝐶 , 𝐴𝐶 ≅ 𝐴𝐵 and 𝐵𝐶 = 𝐶𝐵 , we have Δ𝐴𝐵𝐶 ≅ Δ𝐴𝐶𝐵 by SSS
Theorem. In particular, ∠𝐴𝐵𝐶 ≅ ∠𝐴𝐶𝐵.
Now the three angles of Δ𝐴𝐵𝐶 are all congruent, so Δ𝐴𝐵𝐶 is equiangular.
∎
Circle is a new type of object that can be defined using congruence. Given two distinct points
𝑂 and 𝐴, a circle 𝑐 centered at 𝑂 with radius 𝑂𝐴 is defined as the point set {𝑃: 𝑂𝑃 ≅ 𝑂𝐴}.
The interior and exterior of 𝑐 are defined as the point sets {𝑃: 𝑂𝑃 < 𝑂𝐴} and {𝑃: 𝑂𝐴 < 𝑂𝑃}
respectively.
Example 9.4
(Angle subtended by equal chords)
Let 𝛾 be a circle with center 𝑂. Show
that if 𝐴, 𝐵, 𝐶 and 𝐷 are points on the circle 𝛾 such that 𝐴𝐵 ≅ 𝐶𝐷, then
∠𝐴𝑂𝐵 ≅ ∠𝐶𝑂𝐷.
Proof:
Since 𝐴, 𝐵, 𝐶 and 𝐷 are points on the circle 𝛾 and 𝑂 is the center
of 𝛾, we must have 𝑂𝐴 ≅ 𝑂𝐶 and 𝑂𝐵 ≅ 𝑂𝐷.
Since 𝐴𝐵 ≅ 𝐶𝐷, we have Δ𝐴𝑂𝐵 ≅ Δ𝐶𝑂𝐷 by SSS Theorem.
In particular, we have ∠𝐴𝑂𝐵 ≅ ∠𝐶𝑂𝐷.
Page 3 of 12
𝐵
𝐴
𝐶
𝑂
𝛾
𝐷
∎
MATH4221 Euclidean and Non-Euclidean Geometries (2012 Spring)
© Henry Cheng @ HKUST
10. Hilbert’s Axioms (3):
Tutorial Note 6
Mar 14, 2012 (Week 7)
Continuity and Parallelism
What you need to know:
 Dedekind’s Continuity Axiom
 Consequences of Dedekind’s Axiom:
 Segment-circle Continuity Theorem
 Circle-circle Continuity Theorem
 Archimedean Principle
 Hilbert’s Euclidean Parallel Postulate
 Euclidean Plane: Incidence + betweenness + congruence + continuity + parallelism
We will need only one continuity axiom. To state this axiom in
a concise manner, we first need the concept of Dedekind cut.
A Dedekind cut of a line / ray / segment is a partition into two
non-empty sets 𝑆 and 𝑇 such that no points in 𝑆 is between
two points in 𝑇 and no points in 𝑇 is between two points in 𝑆.
The diagram shows an example of a Dedekind cut of a line 𝑙.
𝑂
𝑙
𝑆
𝑇
𝑂
{𝑆, 𝑇} forms a Dedekind cut of 𝑙
The only continuity axiom we need is the following Dedekind’s Axiom:
(Dedekind’s Axiom)
For any line 𝑙 with a Dedekind cut {𝑆, 𝑇}, there exists a unique point 𝑂
incident to 𝑙 such that 𝑋 ∗ 𝑂 ∗ 𝑌 for each 𝑋 ∈ 𝑆\{𝑂} and 𝑌 ∈ 𝑇\{𝑂}.
It has been shown in the lecture that Dedekind’s Axiom also holds for any ray.
demonstrate that it also holds for any segment as well.
Example 10.1 (Greenberg 3.7a adapted)
Let’s
Show that Dedekind’s Axiom also holds for
segments.
Proof:
Let 𝐴𝐵 be any segment with a Dedekind cut {𝑆, 𝑇}. Our aim is to show that there exists a
unique point 𝑂 ∈ 𝐴𝐵 such that 𝑋 ∗ 𝑂 ∗ 𝑌 for each 𝑋 ∈ 𝑆\{𝑂} and 𝑌 ∈ 𝑇\{𝑂}.
(Existence)
Since {𝑆, 𝑇} is a Dedekind cut of 𝐴𝐵, it is a partition of 𝐴𝐵 in particular. So we have either
𝐵 ∈ 𝑆 or 𝐵 ∈ 𝑇. Without loss of generality, we may assume that 𝐵 ∈ 𝑆.
⃗⃗⃗⃗⃗ .
Next pick a point 𝐶 such that 𝐴 ∗ 𝐵 ∗ 𝐶 by Axiom (B2), and define the set 𝑆 ′ ≔ 𝑆 ∪ 𝐵𝐶
Then 𝑆 ′ and 𝑇 are both non-empty and {𝑆 ′ , 𝑇} forms a partition of the ray ⃗⃗⃗⃗⃗
𝐴𝐶 , because
′
⃗⃗⃗⃗⃗ = 𝐴𝐵 ∪ 𝐵𝐶
⃗⃗⃗⃗⃗ = (𝑆 ∪ 𝑇) ∪ 𝐵𝐶
⃗⃗⃗⃗⃗ = (𝑆 ∪ 𝐵𝐶
⃗⃗⃗⃗⃗ ) ∪ 𝑇 = 𝑆 ∪ 𝑇, and
 𝐴𝐶
⃗⃗⃗⃗⃗ ) ∩ 𝑇 = (𝑆 ∩ 𝑇) ∪ (𝐵𝐶
⃗⃗⃗⃗⃗ ∩ 𝑇) = ∅ ∪ ∅ = ∅.
 𝑆 ′ ∩ 𝑇 = (𝑆 ∪ 𝐵𝐶
Page 4 of 12
MATH4221 Euclidean and Non-Euclidean Geometries (2012 Spring)
© Henry Cheng @ HKUST
Tutorial Note 6
Mar 14, 2012 (Week 7)
⃗⃗⃗⃗⃗ , so that we can
We now claim that this partition {𝑆 ′ , 𝑇} is in fact a Dedekind cut of the ray 𝐴𝐶
apply Dedekind’s Axiom for rays. To prove this claim, we need to show that no point in 𝑆 ′ is
between two points in 𝑇 and that no point in 𝑇 is between two points in 𝑆 ′ .
 Suppose that there exist 𝑃, 𝑃′ ∈ 𝑇 and 𝑄 ∈ 𝑆 ′ such that 𝑃 ∗ 𝑄 ∗ 𝑃′ .
Then since 𝑃 and 𝑃′ both lie on the segment 𝐴𝐵, we have 𝑄 ∈ 𝑃𝑃′ ⊂ 𝐴𝐵.
Now 𝑄 ∈ 𝐴𝐵 ∩ 𝑆 ′ = 𝑆, which is a contradiction to the fact that {𝑆, 𝑇} is a Dedekind cut.
 Suppose that there exist 𝑃, 𝑃′ ∈ 𝑆 ′ and 𝑄 ∈ 𝑇 such that 𝑃 ∗ 𝑄 ∗ 𝑃′ . Since we have
⃗⃗⃗⃗⃗ , we consider the following three cases:
defined 𝑆 ′ ≔ 𝑆 ∪ 𝐵𝐶
(i) If 𝑃, 𝑃′ ∈ 𝑆, then we have a contradiction to the fact that {𝑆, 𝑇} is a Dedekind cut.
⃗⃗⃗⃗⃗ and 𝑃′ ∈ 𝑆, then we have 𝐵 ∗ 𝑄 ∗ 𝑃′ , which is also a contradiction to the
(ii) If 𝑃 ∈ 𝐵𝐶
fact that {𝑆, 𝑇} is a Dedekind cut.
⃗⃗⃗⃗⃗ , then 𝑄 ∈ 𝐵𝐶
⃗⃗⃗⃗⃗ , which is impossible since 𝐵𝐶
⃗⃗⃗⃗⃗ ∩ 𝑇 = ∅.
(iii) If 𝑃, 𝑃′ ∈ 𝐵𝐶
⃗⃗⃗⃗⃗ .
So {𝑆 ′ , 𝑇} is indeed a Dedekind cut of the ray 𝐴𝐶
⃗⃗⃗⃗⃗ , there exists a point 𝑂 ∈ 𝐴𝐶
⃗⃗⃗⃗⃗ such that
Now applying Dedekind’s Axiom to the ray 𝐴𝐶
𝑋 ∗ 𝑂 ∗ 𝑌 for each 𝑋 ∈ 𝑆 ′ \{𝑂} and 𝑌 ∈ 𝑇\{𝑂} , and in particular, 𝑋 ∗ 𝑂 ∗ 𝑌 for each
𝑋 ∈ 𝑆\{𝑂} and 𝑌 ∈ 𝑇\{𝑂}. But in this case we must have 𝑂 ∈ 𝐴𝐵, because otherwise we
have 𝑂 ∗ 𝐵 ∗ 𝑌0 for some point 𝑌0 ∈ 𝑇, which is a contradiction.
(Uniqueness)
The uniqueness of 𝑂 is trivial.
∎
Some consequences of Dedekind’s Axiom include the Segment-circle Continuity Theorem, the
Circle-circle Continuity Theorem and the Archimedean Principle. Now the Circle-circle
Continuity Theorem fills the loophole of Euclid’s proof on the existence of an equilateral triangle
(first lecture), and the Segment-circle Continuity Theorem fills the loophole of his proof on
segment relocation we saw in the first tutorial (Example 1.1).
𝐴 𝐵1
𝐶
∃𝑁 ∈ ℕ
𝐴
Segment-circle Continuity
Circle-circle Continuity
Page 5 of 12
𝐵1 𝐵2 𝐵3
⋯
𝐵𝑁 𝐶 𝐵𝑁+1
Archimedean Principle
MATH4221 Euclidean and Non-Euclidean Geometries (2012 Spring)
© Henry Cheng @ HKUST
Tutorial Note 6
Mar 14, 2012 (Week 7)
Let’s look at Euclid’s Proposition I.1 again!
Example 10.2 (Proposition I.1, Elements) Let 𝐴𝐵 be a segment. Show that there exists a
point 𝐶 such that Δ𝐴𝐵𝐶 is equilateral.
Proof:
Let 𝛾 ≔ {𝑃: 𝐴𝑃 ≅ 𝐴𝐵} be the circle with center 𝐴 and radius
𝐴𝐵, and 𝛿 ≔ {𝑃: 𝐵𝑃 ≅ 𝐵𝐴} be the circle with center 𝐵 and
radius 𝐵𝐴.
 Observe that 𝐵 ∈ 𝛾 and 𝐵 is a point in the interior of 𝛿.
So the intersection of 𝛾 and the interior of 𝛿 is non-empty.
 By Axioms (B2) and (C3), there exists a point 𝐷 such that
𝛾
𝐷
𝛿
𝐴
𝐵
𝐶
𝐷 ∗ 𝐴 ∗ 𝐵 and 𝐴𝐵 ≅ 𝐴𝐷, so 𝐷 ∈ 𝛾. But since 𝐵𝐷 > 𝐵𝐴, 𝐷 is in the exterior of 𝛿.
Thus the intersection of 𝛾 and the exterior of 𝛿 is also non-empty.
By the Circle-circle Continuity Theorem, 𝛾 ∩ 𝛿 is a doubleton. So there exists 𝐶 ∈ 𝛾 ∩ 𝛿.
Since 𝐶 ∈ 𝛾, we must have 𝐴𝐶 ≅ 𝐴𝐵. Since 𝐶 ∈ 𝛿, we must have 𝐵𝐶 ≅ 𝐵𝐴.
Now all three sides of Δ𝐴𝐵𝐶 are congruent, so Δ𝐴𝐵𝐶 is an equilateral triangle.
∎
It can be easily foreseen that apart from Segment-circle
Continuity Theorem and Circle-circle Continuity Theorem, we
should also have a Line-circle Continuity Theorem: Given a
circle 𝑐, if a line 𝑙 is incident to a point inside 𝑐, then 𝑙 and 𝑐
meet at exactly two points. You will need to prove this theorem
using Circle-circle Continuity Theorem in Assignment 3.
Line-circle Continuity
The subject we are studying so far is called “incidence geometry with betweenness, congruence
and continuity”. There are 6 undefined terms and 14 axioms in this subject so far. We can
further include the following parallelism axiom, called Hilbert’s Euclidean Parallel Postulate:
(Parallel Postulate)
For any line 𝑙 and any point 𝑃 not incident to 𝑙, there exists a unique
line 𝑚 incident to 𝑃 which is parallel to 𝑙.
If the above Parallel Postulate is also included, then the subject is commonly called the
Euclidean plane. If it is not included, then the subject is commonly called neutral geometry.
Page 6 of 12
MATH4221 Euclidean and Non-Euclidean Geometries (2012 Spring)
© Henry Cheng @ HKUST
Tutorial Note 6
Mar 14, 2012 (Week 7)
A Summary on Axioms in Incidence Geometry with Betweenness, Congruence and Continuity
3 Incidence Axioms:
(I1) Two distinct points determine a unique line.
(I2) Any line is incident to at least two points.
(I3) There exists three non-collinear points.
4 Betweenness Axioms:
(B1) If 𝐴 ∗ 𝐵 ∗ 𝐶, then 𝐶 ∗ 𝐵 ∗ 𝐴 and 𝐴, 𝐵, 𝐶 are collinear.
(B2) For any two distinct points 𝐴 and 𝐶, there exist 𝐵 and 𝐷 such that 𝐴 ∗ 𝐵 ∗ 𝐶 and
𝐴 ∗ 𝐶 ∗ 𝐷.
(B3) For any three distinct points on a line, one and only one is between the other two.
(B4) Plane Separation: For any two points not on a given line 𝑙, either they are on the same
side of 𝑙 or they are on opposite sides of 𝑙.
6 Congruence Axioms:
(C1) Segment congruence is an equivalence relation.
(C2) Angle congruence is an equivalence relation.
(C3) Segment Relocation: For any segment 𝐴𝐵, there exists 𝐵 ′ on a given ray ⃗⃗⃗⃗⃗⃗
𝐴′ 𝐶 such
that 𝐴𝐵 ≅ 𝐴′ 𝐵 ′.
(C4) Angle Relocation: For any angle ∠𝐴𝐵𝐶, there exists 𝐶 ′ on each side of a given line
⃡⃗⃗⃗⃗⃗⃗⃗
𝐴′ 𝐵 ′ such that ∠𝐴𝐵𝐶 ≅ ∠𝐴′ 𝐵 ′ 𝐶 ′ .
(C5) Segment Addition: If 𝐴 ∗ 𝐵 ∗ 𝐶, 𝐴′ ∗ 𝐵 ′ ∗ 𝐶 ′ , 𝐴𝐵 ≅ 𝐴′ 𝐵 ′, 𝐵𝐶 ≅ 𝐵 ′ 𝐶 ′ , then 𝐴𝐶 ≅ 𝐴′ 𝐶 ′ .
(C6) SAS: If Δ𝐴𝐵𝐶 and Δ𝐴′ 𝐵 ′ 𝐶 ′ exist, ∠𝐵𝐴𝐶 ≅ ∠𝐵 ′ 𝐴′ 𝐶 ′ , 𝐴𝐵 ≅ 𝐴′ 𝐵 ′ , and 𝐴𝐶 ≅ 𝐴′ 𝐶 ′ ,
then Δ𝐴𝐵𝐶 ≅ Δ𝐴′ 𝐵 ′ 𝐶 ′ .
1 Continuity Axiom:
(D) Dedekind’s Axiom: For any line 𝑙 with a Dedekind cut {𝑆, 𝑇}, there exists a unique point
𝑂 incident to 𝑙 such that 𝑋 ∗ 𝑂 ∗ 𝑌 for each 𝑋 ∈ 𝑆\{𝑂} and 𝑌 ∈ 𝑇\{𝑂}.
Some tips for your study:

Remember important definitions (e.g. equivalence relations, functions, segments, rays, angles),
axioms (those listed above, affine planes and projective planes) and theorems (e.g. projective
completion, isomorphisms, Pasch’s Theorem, Crossbar Theorem, etc.) from the lecture notes.
 Go over the examples and try out the exercise problems in the tutorial notes.
 Look at the feedback I’ve given for your work in the assignments.
And even more importantly...
 Ask for help whenever you get lost, either from Dr Ching or from me.
Wish you all the best of luck in your mid-term test! ^^
Page 7 of 12
MATH4221 Euclidean and Non-Euclidean Geometries (2012 Spring)
© Henry Cheng @ HKUST
Tutorial Note 6
Mar 14, 2012 (Week 7)
Exercise
1. (Greenberg 3.34) In the subject of incidence geometry with betweenness and congruence,
prove that
(a) Euclid’s Postulate 2 holds in the following form:
For every two segments 𝐴𝐵 and 𝐶𝐷, there exists a unique point on the line ⃡⃗⃗⃗⃗
𝐴𝐵 such that
𝐴 ∗ 𝐵 ∗ 𝐸 and 𝐶𝐷 ≅ 𝐵𝐸; and
(b) Euclid’s Postulate 3 holds in the following form:
The center of a circle is unique and the radius of a circle is unique up to congruence, i.e. if
the points 𝑂, 𝑂′ and radii 𝑂𝐴, 𝑂′ 𝐴′ respectively determine the same circle, then 𝑂 = 𝑂′
and 𝑂𝐴 ≅ 𝑂′ 𝐴′ .
2. (Greenberg 4.17) Let 𝛾 be a circle with center 𝑂 and let 𝐴𝐵 be a chord of 𝛾 (i.e. 𝐴 and
𝐵 are two distinct points on 𝛾). Let 𝑀 be the mid-point of 𝐴𝐵. Prove that
(a) If 𝑂 ≠ 𝑀, then ⃡⃗⃗⃗⃗⃗
𝑂𝑀 is perpendicular to ⃡⃗⃗⃗⃗
𝐴𝐵 .
(b) If a line ⃡⃗⃗⃗⃗⃗
𝑃𝑀 is perpendicular to ⃡⃗⃗⃗⃗
𝐴𝐵 , then 𝑂 is incident to ⃡⃗⃗⃗⃗⃗
𝑃𝑀.
Remark:
The uniqueness of the mid-point of a segment will be proved in Assignment 3. Now
just take this uniqueness for granted.
3. (Noronha 1.5.3) It can be shown that the angle bisector of any angle always exists and is
unique, and the perpendicular bisector of any segment always exists and is unique. Now given
a triangle Δ𝐴𝐵𝐶 , show that the line containing the angle bisector of ∠𝐵𝐴𝐶 is the
perpendicular bisector of the segment 𝐵𝐶 if and only if 𝐴𝐵 ≅ 𝐴𝐶.
Remark: An angle bisector of ∠𝐵𝐴𝐶 is a ray ⃗⃗⃗⃗⃗
𝐴𝐷 such that ∠𝐵𝐴𝐷 ≅ ∠𝐶𝐴𝐷 , while a
perpendicular bisector of a line segment 𝐵𝐶 is a line 𝑙 which is incident to the
mid-point of 𝐵𝐶 and is perpendicular to 𝐵𝐶.
4. Let 𝑙 be a line with a Dedekind cut {𝑆, 𝑇}. Then there exists a “cut point” 𝑂 on 𝑙 as
guaranteed by Dedekind’s Axiom. Show that if 𝐴 and 𝐵 are points incident to 𝑙 such that
𝑂 ∗ 𝐴 ∗ 𝐵, then either 𝐴, 𝐵 ∈ 𝑆 or 𝐴, 𝐵 ∈ 𝑇.
5. Let 𝐴𝐵 be a segment with a Dedekind cut {𝑆, 𝑇}. Then there exists a “cut point” 𝑂 on 𝐴𝐵
as guaranteed by Dedekind’s Axiom. Show that 𝑂 = 𝐴 if and only if either 𝑆 = {𝐴} or
𝑇 = {𝐴}.
6. (Noronha 1.7.5 adapted) Let 𝐴𝐵 and 𝐴′ 𝐵 ′ be two segments with 𝐴𝐵 < 𝐴′ 𝐵 ′ . Show that
there exists a right triangle whose hypotenuse is congruent to 𝐴′ 𝐵 ′ and one leg is congruent to
𝐴𝐵.
Hint:
The Line-circle Continuity Theorem may be useful.
Page 8 of 12