Download Spring 2007 Math 330A Notes Version 9.0

Page 1 of 106
Spring 2007 Math 330A Notes, Version 9
Reading Material for Foundations of Geometry I
(by Mark Barsamian)
1.
Relations..............................................................................................................................3
1.1.
Cartesian Products ...................................................................................................3
1.2.
Relations ..................................................................................................................5
1.3.
Exercises ................................................................................................................11
2.
Axiom Systems .................................................................................................................15
2.1.
Definition ...............................................................................................................15
2.2.
Primitive Relations and Primitive Terms ...............................................................16
2.3.
Interpretations and Models ....................................................................................19
2.4.
Properties of Axiom Systems.................................................................................22
2.5.
Exercises ................................................................................................................29
3.
Axiomatic Geometries .....................................................................................................31
3.1.
What is an analytic geometry? ...............................................................................31
3.2.
What is an axiomatic geometry? ............................................................................31
3.3.
Finite Geometries ...................................................................................................32
3.4.
More about terminology ........................................................................................34
3.5.
Fano’s and Young’s Finite Geometries .................................................................36
3.6.
Incidence Relations and Incidence Geometries .....................................................37
3.7.
Exercises ................................................................................................................42
4.
Building on the axiom list of Incidence Geometry ........................................................44
4.1.
The need for a larger list of axioms .......................................................................44
4.2.
Binary and Ternary Relations on a Set ..................................................................44
4.3.
Introducing Incidence and Betweenness Geometry...............................................45
4.4.
Line Segments and Rays ........................................................................................46
4.5.
Plane Separation.....................................................................................................49
4.6.
Line Separation ......................................................................................................50
4.7.
Angles and Triangles .............................................................................................51
4.8.
Exercises ................................................................................................................54
5.
Neutral Geometry I ..........................................................................................................55
5.1.
The need for a larger axiom system: Introducing Neutral Geometry ....................55
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5.2.
Triangle Congruence and its Role in the Neutral Geometry Axioms ....................57
5.3.
Two Theorems about Triangles .............................................................................62
5.4.
Line Segment Subtraction and the Ordering of Segments .....................................63
5.5.
Right Angles ..........................................................................................................65
5.6.
Angle Addition and Subtraction, and Ordering of Angles ....................................68
5.7.
Three More Theorems about Triangles..................................................................70
5.8.
Exercises ................................................................................................................72
6.
Neutral Geometry II ........................................................................................................73
6.1.
The Alternate Interior Angle Theorem and Some Corollaries...............................73
6.2.
The Exterior Angle Theorem and Some Theorems Whose Proofs Use It .............74
6.3.
Exercises ................................................................................................................77
7.
Measure of Line Segments and Angles ..........................................................................82
7.1.
Theorems Stating the Existence of Measurement Functions .................................82
7.2.
Two length functions for Neutral Geometry..........................................................83
7.3.
An example of a curvy-looking Hline ...................................................................85
7.4.
Theorems about segment lengths and angle measures ..........................................86
7.5.
Exercises ................................................................................................................91
8.
Euclidean Geometry ........................................................................................................93
8.1.
9.
Building Euclidean Geometry from Neutral Geometry .........................................93
For Reference: Axioms, Defintions, and Theorems of Neutral Geometry .................94
9.1.
The Axioms of Neutral Geometry .........................................................................94
9.2.
The Definitions of Neutral Geometry ....................................................................94
9.3.
The Theorems of Neutral Geometry ....................................................................101
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1. Relations
1.1. Cartesian Products
1.1.1. Definition and examples
Definition 1 Cartesian Product
• Symbol: A × B
• Spoken: The Cartesian Product of A and B.
• Usage: A and B are sets
• Meaning in words: A × B is the set consisting of all ordered pairs ( a , b ) , where a is an
element of A and b is an element of B.
• Meaning in symbols: A × B = {( a, b ) : a ∈ A and b ∈ B}
Example: Let A = { x, y} and B = {1, 2, 3}
a) Find A × B .
b) Find B × A
c) Answer the following true/false questions. If your answer is “false”, explain why.
i) 2x ∈ A × B true false
ii) x 2 ∈ A × B true false
iii) x × 2 ∈ A × B
true false
iv) ( x, 2 ) ∈ A × B
true false
v) ( 2, x ) ∈ A × B
true false
Notice that nothing in the definition of Cartesian Product requires that the two sets used in the
product be different. The following example illustrates this.
Example: Let A = { x, y} and B = {1, 2, 3}
a) Find B × B .
b) Answer the following true/false questions. If your answer is “false”, explain why.
i) 2x ∈ B × B true false
ii) x 2 ∈ B × B true false
true false
iii) 3 × 2 ∈ B × B
iv) ( 2, 2 ) ∈ B × B
true false
true false
v) (1, 3 ) = ( 3,1)
Our first two examples have involved finite sets. However, nothing in the definition of the
Cartesian Product requires that the sets be finite, and in fact, you have all already used the
Cartesian Product in a setting involving infinite sets.
Example: Let A = ℝ and let B = ℝ . Then A × B = ℝ × ℝ , which is commonly denoted as
ℝ2 . This is the set of ordered pairs of real numbers. Here is a proper definition.
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Definition 2 the Cartesian plane
• Symbol: ℝ2
• Spoken: “r two”, or “the x, y plane”, or “the Cartesian plane”.
• Meaning in symbols: ℝ × ℝ
• Meaning in words: The set of ordered pairs of real numbers.
Observations:
1) All real numbers are allowed, not just integers. So, pairs such as ( 5, 2.7 ) , (π , −10 ) , ( −2π , 0 ) ,
2)
3)
4)
5)
etc., are all elements of ℝ2 .
Parentheses and commas are used, because ℝ2 is a Cartesian product. So “ π ×10 ”, “ π ,10
”, etc., are not allowed.
Order is important: ( 5, 2.7 ) ≠ ( 2.7,5 ) .
The definition of the Cartesian product makes no provisions for “scalar multiplication”. In
the past, you may have seen computations such as 3 ( 2,5 ) = ( 6,15 ) or − ( 2, 5 ) = ( −2, −5 ) .
Computations such as these are perfectly valid, but they arise in contexts where one is
dealing with “vectors”. Vectors may be typeset in a way that makes them look just like the
elements of a Cartesian product, but be aware that a vector is something different. In a
Cartesian product, there is no scalar multiplication.
Similarly, the definition of the Cartesian product makes no provisions for “addition”. In the
past, you may have seen computations such as ( 2,5 ) + (1,8 ) = ( 3,13) . Again, computations
such as this are perfectly valid, but they arise in contexts where one is dealing with “vectors”.
In a Cartesian product, there is no addition.
1.1.2. Visualizing a Cartesian product
Cartesian products involving finite sets can be visualized using tables. There are a number of
conventions that can be followed. A common convention is that for a product such as A × B , the
elements of the set A correspond to the rows of the table; the elements of set B correspond to the
columns. Each cell of the table corresponds to a ( row, column ) pair. Elements of the Cartesian
product are denoted by putting some sort of mark, such as an “X”, in a cell.
Example: Let A = { x, y} and B = {1, 2, 3} . Then A × B is
visualized as the table shown at right. Notice that the upper left
corner gives the names of the sets used in constructing the
product. Nowhere in the table is it written that the x and the y are
from the set A, and that the 1,2,3 are from the set B. It doesn’t
have to be written, because it is a convention.
A× B
2
3
1
2
3
x
y
A× B
The element ( x, 3 ) ∈ A × B could be displayed as shown at right.
1
x
y
*
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B× A
x
y
x
y
1
On the other hand, B × A is visualized as the table
2
3
B× A
The element ( 3, x ) ∈ B × A is displayed as
1
2
3
X
Sometimes, Cartesian products involving infinite sets can be visualized as well. You have done
this for years, every time you drew a set of axes for the “x,y plane”. It is important to notice that
some of the conventions in this case differ from the conventions used above, when illustrating
finite Cartesian products with tables. Namely, in the case of the finite cartesian products, the
elements of the left set are listed vertically, along the left edge of the table, while elements of the
right set are listed horizontally, along the top of the table. In the x,y plane, however, elements of
the left set are displayed on the horizontal axis, while elements of the right set are displayed on
the vertical axis. Single elements in the “x,y plane” are displayed as little dots, or little crosses.
Remember that one of these little marks represents an ordered pair of real numbers.
Class Example.
y
(right coordinate)
2
(1.5, 2 )
( 2,1.5 )
1
1
1.2. Relations
1.2.1. Definitions and Examples
Definition 3 Relation
• Words: R is a relation from A to B.
• Usage: A and B are sets.
• Meaning in words: R is a subset of A × B .
• Meaning in symbols: R ⊂ A × B
2
x
(left coordinate)
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Definition 4 related to
• Symbol: xRy (Most often, other symbols besides R are used.)
• Spoken: x is related to y
• Usage: It assumed that R is a relation from A to B, where A and B are some sets.
• Meaning in words: the mathematical statement “ ( x, y ) is an element of the set R.”
•
Meaning in symbols: “ ( x, y ) ∈ R ”
Remark 1: Since it represents a mathematical statement, the symbol xRy can be true or false.
Remark 2: It is not assumed that x ∈ A and y ∈ B . That is, there is nothing “illegal” about
writing the symbol down in some case where x ∉ A or y ∉ B . This will be elaborated in the
examples.
Example: Let A = { x, y} and B = {1, 2, 3} . Above, we found that
A × B = {( x,1) , ( x, 2 ) , ( x,3) , ( y,1) , ( y, 2 ) , ( y,3)} .
a) Let R = {( x, 2 ) , ( y,1)} . Then R is a relation from A to B. Observe that ( x, 2 ) ∈ R . In other
words, we could write xR2, or “x is related to 2”, and it would be a true statement.
b) Let R = {( y,1) , ( y, 2 )} . Then R is a relation from A to B. Observe that the statement “y is
related to 2”, abbreviated yR2, is true. However, the statement “2 is related to y”,
abbreviated 2Ry, is false, because ( 2, y ) ∉ R . In fact, 2 ∉ A and y ∉ B , so there is no way
that ( 2, y ) could possibly be an element of R. Even so, there is nothing “illegal” about
the expression 2Ry. The symbol represents a false statement, but it is not illegal.
c) Let R = ∅ . Is R a relation from A to B? If not, say why not.
d) Let R = A × B = {( x,1) , ( x, 2 ) , ( x,3) , ( y,1) , ( y, 2 ) , ( y,3)} . Is R a relation from A to B? If
not, say why not.
e) Let R = {( x, 2 ) , (1, y )} . Is R a relation from A to B? If not, say why not.
f) Let R = {( x, 2 ) , ( x,3) , ( y,1) , ( y, 2 )} . Then R is a relation from A to B.
i) Find all a such that aR2.
ii) Find all a such that aR3.
iii) Find all b such that yRb.
1.2.2. Visualizing relations
Because a relation R from a set A to a set B is just a subset of the Cartesian product A × B , any
illustration for the A × B can just be “filled in” to produce a picture of relation R.
An alternate way to view a relation R from one finite set A to another finite set B is to use an
arrow diagram. Elements of set A are denoted by dots in some arrangement to the left, and
elements of set B are denoted by dots in some arrangement to the right. To indicate that aRb is
true, one draws an arrow from the dot for element a to the dot for element b.
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1
For example, the relation from Example (f) above can
be illustrated with the table and arrow diagram shown
at right.
1 2 3
x
x
* *
y * *
y
2
3
1.2.3. Relation on a set
Observe that in the definition of relation, we used the symbols A and B, but there is no
requirement that these two sets be different. That is, it could be that the set B is actually the same
as the set A. In that case, we say that the relation is a “relation on a set”. The following definition
make this more precise.
Definition 5 Relation on a Set
• Words: R is a relation on A.
• Usage: A is a set.
• Meaning: R is a relation from A to A.
• Equivalent meaning in words: R is a subset of A × A .
• Equivalent meaning in symbols: R ⊂ A × A
• Additional terminology: R is also called a binary relation on A.
Examples: In the following examples, Let A = {1, 2,3, 4, 5, 6} .
a) Let Ra = {(1,3 ) , ( 3,1) , ( 3, 5 ) , ( 5, 3) , ( 2, 4 ) , ( 4, 2 ) , ( 4, 6 ) , ( 6, 4 )} . Then Ra is a relation on the
set A.
b) Let Rb = {(1,3 ) , ( 3,5 ) , (1, 5 ) , ( 2, 4 ) , ( 4, 6 ) , ( 2, 6 )} . Then Rb is a relation on the set A.
(1,1) , ( 2, 2 ) , ( 3,3) , ( 4, 4 ) , ( 5,5) , ( 6, 6 ) , (1,3) , 
c) Let Rc = 
 Then Rc is a relation on the set A.
( 3,1) , ( 3,5) , ( 5,3) , ( 2, 4 ) , ( 4, 2 ) , ( 4, 6 ) , ( 6, 4 ) 
So far, our examples of relations on a set have all been given by explicit lists of elements of sets.
But relations on a set can also be described by a formula.
Examples: In the following examples, Let A = ℝ , the set of all real numbers.
d) Let x Rd y means x ≤ y . Then Rd is a relation on the set ℝ . (We don’t really need the R
symbol, do we?)
e) Let x Rey means xy = 0 .Then Re is a relation on the set ℝ .
f) Let x Rfy means y = 2 x + 1 . Then Rf is a relation on the set ℝ .
1.2.4. Visualizing a relation on a set
Since a relation on a set is just a particular type of relation, the same tools that can be used to
visualize regular relations (tables and graphs and arrow diagrams) can be used to visualize a
relation on a finite set. An special kind of arrow diagram for a relation on a set A is the directed
graph. In this kind of arrow diagram, elements of the set A are only shown once, rather than
twice. Arrows are drawn as you might expect. Let’s revisit our previous examples Ra, Rb, and Rc
from the previous section and make a table and a directed graph for each.
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1 2 3 4 5 6
1
*
Relation Ra
2
3 *
4
*
5
6
4
*
*
5
*
*
2
3
4
*
*
1
4
3
5
2
*
6
1 2 3 4 5 6
1 *
*
2
*
*
3 *
*
*
4
*
*
*
5
6
6
*
5
6
Relation Rc
2
*
1 2 3 4 5 6
1
*
*
Relation Rb
3
*
4
1
3
5
2
*
*
*
6
1
1.2.5. Properties that a relation on a set may or may not have
Throughout the definitions, it is assumed that R is a relation on a set A.
Definition 6 Reflexive Property
• Words: R is reflexive
• Meaning: ∀a ∈ A, aRa
• Meaning in words: Every element of set A is related to itself.
Definition 7 Symmetric Property
• Words: R is symmetric
• Meaning: ∀a, b ∈ A, IF aRb THEN bRa
Definition 8 Transitive Property
• Words: R is transitive
• Meaning: ∀a, b, c ∈ A, IF ( aRb AND bRc ) THEN aRc
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Definition 9 Equivalence Relation
• Words: R is an equivalence relation
• Meaning: R is Reflexive and Symmetric and Transitive
Examples:
a) Consider relation Ra described above.
i) Is 3 related to itself? That is, is 3Ra3 true? No, so Ra is not reflexive.
ii) Notice that Ra is symmetric.
iii) Notice that 2Ra4 and 4Ra6, but 2Ra6 is not true. Therefore, Ra is not transitive.
iv) Therefore, Ra is not an equivalence relation.
b) Consider relation Rb described above.
i) Notice that 3Rb3 is false, so Rb is not reflexive.
ii) Notice that 1Rb3 is true, but 3Rb1 is not true. Therefore, Rb is not symmetric.
iii) Notice that 1Rb3 and 3Rb5 are both true, and 1Rb5 is also true. And notice that 2Rb4
and 4Rb6 are both true, and 2Rb6 is also true. Therefore, Rb is transitive.
iv) Therefore, Rb is not an equivalence relation.
c) Consider relation Rc described above. Group work:
i) Is Rc reflexive?
ii) Is Rc symmetric?
iii) Is Rc transitive?
iv) Is Rc an equivalence relation?
Notice that the sets and relations used in examples a), b), and c) above were very basic, and yet it
was rather tedious to check for the reflexive, symmetric and transitive properties by looking only
at the sets that define each relation. Using a picture to visualize the relation helps to make the
checking of the reflexive and symmetric properties much easier. From studying the pictures for
the three relations above, we can easily make the following general observations:
When a relation is reflexive,
• All of the cells on the main diagonal of the table are filled in.
• Every dot in the directed graph has an arrow looping from the dot back to the dot.
This happens in the pictures for relation Rc.
When a relation is not reflexive,
• At least one of the cells on the main diagonal of the table is empty.
• At least one dot in the directed graph does not have an arrow looping around it.
This happens in the pictures for relation Ra and Rb.
When a relation is symmetric,
• The table is symmetric across the main diagonal.
• Every arrow in the diagram is a double arrow.
This happens in the pictures for relation Ra and Rc. (The little loop arrows are effectively double
arrows. Think about it.)
When a relation is not symmetric,
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• The table is not symmetric across the main diagonal.
• There is an arrow in the diagram that is not a double arrow.
This happens in the pictures for relation Rb .
When a relation is transitive,
• The table is not much help in looking for transitivity.
• Every “segmented path” in the arrows, there is a “direct path” that goes from the same
starting dot to the same ending dot..
This happens in the pictures for relation Rb . For instance, there is a segmented path that goes
1 → 3 → 5 . A direct path goes 1 → 5 .
When a relation is not transitive,
• The table is not much help in looking for failures of transitivity, either.
• There is a “segmented path” in the arrows, for which there is not a “direct path” that goes
from the same starting dot to the same ending dot..
This happens in the pictures for relation Ra and Rc. For instance, in the directed graph for Ra,
there is a segmented path that goes 1 → 3 → 5 , but there is not a direct path that goes 1 → 5 .
1.2.6. Defining Relations by giving a general description of the elements
Class examples: We will consider relations on the set of real numbers, ℝ . Remember that a
relation on the set of real numbers is just a fancy name for a subset of the cartesian plane, ℝ × ℝ .
For each of the following relations, draw a cartesian plane and sketch the points that are elements
of the relation. Then decide if the relation is reflexive, symmetric, transitive.
example
R0
R1
R2
R3
Relation
x R0 y means x – y = 2
x R1 y means x < y
x R2 y means x 2 + y 2 = 1
x R3 y means xy ≠ 0
R4
x R4 y means ( y − x )( y − 2 x ) = 0
R5
x R5 y means x ≤ y
R6
x R6 y means x − y ≤ 1
R7
x R7 y means x 2 = y 2
Reflexive? Symmetric? Transitive?
Big hint for R4. ( y − x )( y − 2 x ) = 0 is logically equivalent to ( y = x ) OR ( y = 2 x )
1.2.7. Remark on Relations as Predicates
Consider relation R3 from the previous section. The sentence “2 R3 5” is a true statement, and the
sentence “0 R3 5” is a false statement. The sentence “x R3 y” is neither true nor false because we
do not know the values of x and y. If we substitute in some actual values for x and y, the sentence
becomes a sentence that is either true or false, but not both. In other words, the sentence “x R3 y”
is a predicate. But the sentence “ ∀x, y ∈ ℝ, IF xRy THEN yR3 x ” is a true statement. and the
sentence “ ∀x ∈ ℝ , xR3 x ” is a false statement. So the properties of the relations are statements.
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1.3. Exercises
1) Let A = {4, 5} and B = {3, a, b}
a)
b)
c)
d)
Find A × B .
Find B × A .
Find A × A .
Answer the following true/false questions. If your answer is “false”, explain why.
i) b4 ∈ A × B
true false
ii) 4b ∈ A × B
true false
iii) 5 × a ∈ A × B
true false
iv) ( 5, a ) ∈ A × B
true false
v) ( a ,5 ) ∈ A × B
true false
vi) 5 ⋅ 3 ∈ A × B
true false
vii) 15 ∈ A × B
true false
viii) 4b ∈ A × A
true false
ix) 4 × 5 ∈ A × A
true false
true false
x) ( 5,5 ) ∈ A × A
xi) ( 4,5 ) ∈ A × A
true false
xii) ( 4, 5 ) = ( 5, 4 )
true false
2) Let C be the set of celebrities and M be the set of months. Answer the following true/false
questions. If your answer is “false”, explain why.
a) Michael Jordan × February ∈ M × C true false
b) September × Woody Allen ∈ M × C true false
c)
( December,Tiger Woods ) ∈ M × C
true
false
3) Let L = {l1 , l2 , l3 , l4 } be the set of lines in the figure below, and let P = { p1 , p2 , p3 , p4 , p5 , p6 }
be the set of dots. Answer the following true/false questions. If your answer is “false”,
explain why.
l1
p2
p4
p3
p5
p6
p1
l2
l3
a) l2 × l4 ∈ L × L
true
false
b)
true
false
true
false
c)
( p5 , l1 ) ∈ L × P
( p4 , l 2 ) ∈ P × L
l4
p7
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d)
e)
( p4 , l 4 ) ∈ P × L
( p5 , l1 ) ∈ P × L
true
false
true
false
4) Consider the cartesian product ℝ 2 = ℝ × ℝ . What elements of the cartesian product
ℝ 2 = ℝ × ℝ are denoted by the dots labelled a,b,c,d in the figure? (In your old jargon, you
would have been asked to give the coordinates of each of the four points.)
y
2
b
a
1
x
1
2
d
c
5) Draw a picture to illustrate the cartesian product ℝ 2 = ℝ × ℝ , along with the following
elements, a,b,c,d. (In your old jargon, you would have been asked to draw a set of axes, and
then put in the following four points.)
a) ( 4, −2 )
b)
c)
d)
( −3.5, 2 )
( −1, −3)
(π , 2 )
6) Let A = {4, 5} and B = {3, d , e}
a) Let R = {( 3, 4 ) , ( 4,3 )} . Is R a relation from A to B? If not, say why not.
b) Let R = {( 4,3) , ( 4, d ) , ( 5, d )} . Is R a relation from A to B? If not, say why not.
c) Let R = {( 4, e ) , ( 5,3) , ( 5, d ) , ( 5, e )} . Then R is a relation from A to B.
i) Is 4 related to d ?
ii) Is 5 related to d ?
iii) Is d related to 5?
iv) Is 3R5 true?
v) Find all a such that aRd.
vi) Find all a such that aRe.
vii) Find all b such that 5Rb.
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7) Let L = {l1 , l2 , l3 , l4 } , and let P = { p1 , p2 , p3 , p4 , p5 , p6 } be the set of dots. Let
R = {( p2 , l1 ) , ( p4 , l2 ) , ( p4 , l4 ) , ( p6 , l4 )} . Then R is a relation from P to L. See the picture in
exercise (3) for reference. Notice that the relation could be defined in words by saying that
the sentence pRl means “the dot p touches the line l ”. Answer the following questions.
a) Is p4 Rl4 ?
b) Is l4 related to p4 ?
c) Is p4 related to l4 ?
d) Is p4 related to l2 ?
e) Is p2 related to l3 ?
f) Is l2 Rl4 ?
g) Is l2 Rl1 ?
8) Using the same sets and same picture in exercise 3, and referring to relation R, we could
define a new relation S on the set L by saying that the statement laSlb means “there exists a
point p such that pRla and pRlb”. In other words, we would say that two lines are related by
relation S if there exists a point p that touches both of the lines.
a) Is l2Sl4?
b) Is l2Sl3?
c) Is l1Sl1?
d) Is l3Sl3?
9) Again using the same sets and same picture in exercise 3, and referring to relation R, we
could define a new relation T on the set P by saying that the statement paTpb means “there
exists a line l such that paRl and pbRl”. In other words, we would say that two points are
related by relation T if there exists a line l that both points touch.
a) Is p4Tp6?
b) Is p6Tp4?
c) Is p2Tp2?
d) Is p3Tp3?
10) Consider the following relations on the set of real numbers, ℝ . Remember that a relation on
the set of real numbers is just a fancy name for a subset of the cartesian plane, ℝ × ℝ .
Illustrate each relation by drawing it as a subset of the plane. Then decide if the relation is
reflexive, symmetric, transitive.
example
Ra
Relation
x Ra y means 2 x − y = 1
Rb
x Rb y means x = y
Rc
Rd
Re
x Rc y means x 2 + y 2 < 1
x Rd y means xy = 0
x Re y means xy > 0
Reflexive? Symmetric? Transitive?
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For the next two exercises, let A be the set of all lines in the Cartesian plane. (Here we are not
using the picture from exercise 3.)
11) Define “perpendicular lines” to mean two lines with the property that the products of their
slopes is -1 OR one line is vertical and one line is horizontal. Define a relation ⊥ on the set A
as follows. For l1 , l2 ∈ A , “ l1 ⊥ l2 ” means “ l1 is perpendicular to l2 ”. Is this relation
reflexive? Symmetric? Transitive? An Equivalence Relation? Explain your answers.
12) In this exercise, you will see an example of how the choice of wording of a definition can
affect the properties of a relation.
a) Define “parallel lines” to mean two lines that have the same slope or are both vertical.
Define a relation || on the set A as follows. For l1 , l2 ∈ A , “ l1 || l2 ” means “ l1 is parallel to
l2 ”. Is this relation reflexive? Symmetric? Transitive? An Equivalence Relation? Explain
your answers.
b) Define “parallel lines” to mean two lines that do not intersect. Define a relation || on the
set A as follows. For l1 , l2 ∈ A , “ l1 || l2 ” means “ l1 is parallel to l2 ”. Is this relation
reflexive? Symmetric? Transitive? An Equivalence Relation? Explain your answers.
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2. Axiom Systems
2.1. Definition
We will use the term axiom system to mean a finite list of statements that are assumed to be true.
The individual statements are the axioms. The word postulate is often used instead of axiom.
Example 1 of an axiom system
1. Elvis is dead.
2. Chocolate is the best flavor of ice cream.
3. 5 = 7.
Notice that the first statement is one that most people are used to thinking of as true. The second
sentence is clearly a statement, but one would not have much luck trying to find general
agreement as to whether it is true or false. But if we list it as an axiom, we are assuming it is true.
The third statement seems to be problematic. If we insist that the normal rules of arithmetic must
hold, then this statement could not possibly be true. There are two important issues here. The
first is that if we are going to insist that the normal rules of arithmetic must hold, then that
essentially means that our axiom system is actually larger than just the three statements listed:
the axiom system would also include the axioms for arithmetic. The second issue is that if we do
assume that the normal rules of arithmetic must hold, and yet we insist on putting this statement
on the list of axioms, then we have a “bad” axiom system in the sense that its statements
contradict each other. We will return to this when we discuss consistency of axiom systems.
So the idea is that regardless of whether or not we are used to thinking of some statement as true
or false, when we put the statement on a list of axioms we are simply assuming that the statement
is true.
With that in mind, we could create a slightly different axiom system by modifying our first
example.
Example 2 of an axiom system
1. Elvis is alive.
2. Chocolate is the best flavor of ice cream.
3. 5 = 7.
The statements of an axiom system are used in conjunction with the rules of inference to prove
theorems. Used this way, the axioms are actually part of the hypotheses of each theorem proved.
For example, suppose that we were using the axiom system from Example 2, and we were
somehow able to use the rules of inference to prove the following theorem from the axioms.
If Bob is Blue then Ann is Red.
Then what we really would have proven is the following statement:
If ((Elvis is alive) and (Chocolate is the best flavor of ice cream) and (5 = 7) and (Bob is
Blue)) then Ann is Red.
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2.2. Primitive Relations and Primitive Terms
As you can see from the examples in the previous section, axiom systems may be comprised of
statements that we are used to thinking of as true, or statements that we are used to thinking of as
false, or some mixture of the two. More interestingly, an axiom system can be made up of
statements whose truth we have no way of assessing. The easiest way to get such an axiom
system is to build statements using words whose meaning has not been defined. In this course,
we will be doing this in two ways.
2.2.1. Primitive Relations
The first way of building statements whose meaning is undefined is to use nouns whose meaning
is known in conjunction with transitive verbs whose meaning is not known. This can be
described nicely in the jargon of relations. For instance, consider the set ℤ of integers. Introduce
an undefined relation R on the set ℤ . That is, introduce a subset R ⊂ ℤ × ℤ , but don’t say what
that subset is. Then the symbol 5R7 means that ( 5, 7 ) ∈ R and would be spoken as “5 is related
to 7”. This is a sentence with the noun 5 as the subject, the noun 7 as the direct object, and the
words “is related to” as the transitive verb. We have no idea idea what this sentence might mean,
because the relation is undefined.
In the context of axiom systems, an undefined relation is sometimes called a primitive relation.
When one presents an axiom system that contains primitive relations—that is, undefined
transitive verbs—it is important to introduce those primitive relations before listing the axioms.
Here is an example of an axiom system consisting of sentences built using primitive relations in
the manner described above.
Axiom system #1
Primitive Relations:
• relation R on the set ℤ , spoken “x is related to y”.
Axioms:
1. 5 is related to 7
2. 5 is related to 8
3. For all real numbers x and y, if x is related to y, then y is related to x.
4. For all real numbers x, y, and z, if x is related to y and y is related to z, then x is
related to z.
With the symbols and terminology of relations that we learned in Chapter 1, we can easily
abbreviate the presentation of this axiom system.
Axiom system #1, abbreviated version
Primitive Relations:
• relation R on the set ℤ
Axioms:
1. 5R7
2. 5R8
3. Relation R is symmetric.
4. Relation R is transitive.
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Observe that each of the axioms is a statement whose truth we have no way of assessing, because
the relation R is undefined. But we can prove the following theorem.
Theorem for axiom system #1: 7 is related to 8.
Proof
1) 7 is related to 5 (by axioms 1 and 3)
2) 7 is related to 8 (by statement 1 and axioms 2 and 4)
End of proof
As mentioned in the previous section, the axioms could be stated explicitly as part of the
theorem. (Then we would not really need to state the axiom system separately.)
If ((R is a relation on ℤ ) and (5R7) and (5R8) and (R is symmetric) and (R is transitive),
then 7R8.
2.2.2. Primitive Terms
The second way of building statements whose meaning is undefined is to use not only undefined
transitive verbs, but also undefined nouns. A straightforward way to do this is to introduce sets A
and B whose elements are undefined. For instance, let A be the set of akes and B be the set of
bems, where ake and bem are undefined nouns. Introduce the following sentence: “the ake is
related to the bem”. Note that this is a sentence with the undefined noun ake as the subject, the
words “is related to” as the transitive verb, and the undefined noun bem as the direct object. Of
course we have no idea what this sentence might mean, because the nouns ake and bem are
undefined. But we now have the following building blocks that can be used to build sentences
• the undefined noun: ake
• the undefined noun: bem
• the undefined sentence: The ake is related to the bem.
Since we don’t know the meaning of the sentence “The ake is related to the bem”, we have
effectively introduced an undefined relation from set A to set B. We could call this undefined
relation R. Using the standard notation for relations, we could write R ⊂ A × B . The sentence
“The ake is related to the bem” would mean that ( ake, bem ) ∈ R and would be denoted by
symbol akeRbem.
In the context of axiom systems, an undefined noun is sometimes called an undefined term, or a
primitive term, or an undefined object, or a primitive object. In presentations of axiom system
that contains primitive terms, the primitive terms are customarily listed along with the primitive
relations, before the axioms. Here is an example of an axiom system consisting of sentences built
using primitive terms and primitive relations in the manner described above.
Axiom system #2
Primitive Terms:
• ake
• bem
Primitive Relations:
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• relation R from the set A of all akes to the set B of all bems.
Axioms:
1. There are four akes. These may be denoted ake1, ake2, ake3, and ake4.
2. For any set of two akes, denoted {akei , akek } where i ≠ k , there is exactly one
bem such that akei is related to the bem and akek is related to the bem.
3. For any bem, there is exactly one set of two akes, denoted {akei , akek } where
i ≠ k , such that akei is related to the bem and akek is related to the bem.
As with axiom system #1, each of these axioms in axiom system #2 is a statement whose truth
we have no way of assessing, because the words ake and bem are undefined and the relation R is
undefined. But we can prove the following theorem.
Theorem #1 for axiom system #2: There are exactly 6 bems.
Proof
1. Axioms #2 and #3 tell us that there is bijective correspondence between
the sets of two akes and the bems.
2. By axiom #1, there are four akes. Therefore, it is possible to build six
unique sets of two akes, denoted {akei , akek } where i ≠ k .
3. Therefore, there must be exactly 6 bems.
End of proof
As with the theorem that we prove in the previous section for Axiom System #1, we note that the
theorem just presented could be written with all of the primitive terms, primitive relations, and
axioms put into the hypothesis. The resulting statement would be quite long.
Theorem: If Blah Blah Blah then there are exactly 6 bems.
In the exercises, you will prove the following:
Theorem #2 for Axiom System #2: For any ake, there are exactly three bems that are
related to the ake.
Our examples of axiom systems with undefined terms and undefined relations seem rather
absurd, because their axioms are meaningless. What purpose could such abstract collections of
nonsense sentences possibly serve? Well, the idea is that we will use such abstract axiom
systems to represent actual situations that are not so abstract. Then for any abstract theorem that
we have been able to prove about the abstract axiom system, there will be a corresponding true
statement that can be made about the actual situation that the axiom system is supposed to
represent.
This begs the question: why study the axiom system at all, if the end goal is to be able to prove
statements that are about some actual situation? Why not just study the actual situation and prove
the statements in that context? The answer to that is twofold. First, a given abstract axiom system
can be recycled, used to represent many different actual situations. By simply proving theorems
once, in the context of the axiom system, the theorems don’t need to be reproved in each actual
context. Second, and more important, by proving theorems in the context of the abstract axiom
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system, we draw attention to the fact that the theorems are true by the simple fact of the axioms
and the rules of logic, and nothing else. This will be very important to keep in mind when
studying axiomatic geometry.
2.3. Interpretations and Models
As mentioned above, an axiom system with undefined terms and undefined relations is often
used to represent an actual situation. This idea of representation is made more precise in the
following definition. You’ll notice that the representing sort of gets turned around: we think of
the actual situation as a representation of the axiom system.
Definition 10 Interpretation of an axiom system
Suppose that an axiom system consists of the following four things
• an undefined object of one type, and a set A containing all of the objects of that type
• an undefined object of another type, and a set B containing all of the objects of that type
• an undefined relation R from set A to set B
• a list of axioms involving the primitive objects and the relation
An interpretation of the axiom systems is the following three things
• a designation of an actual set A’ that will play the role of set A
• a designation of an actual set B’ that will play the role of set B
• a designation of an actual relation R’ from A’ to B’ that will play the role of the relation R
As examples, for Axiom System #2 from the previous section we will investigate three different
interpretations invented by Alice, Bob, and Carol.
Recall Axiom System #2 included the following things
• an undefined term ake and a set A containing all the akes
• an undefined term bem and a set B containing all the bems
• a primitive relation R from set A to set B.
• a list of three axioms involving these undefined terms and the undefined relation
Alice’s interpretation of Axiom System #2.
• Let A’ be the set of dots in the picture at right. Let set A’ play
the role of set A.
• Let B’ be the set of segments in the picture at right. Let set B’
play the role of set B.
• Let relation R’ from A’ to B’ be defined by saying that the words
“the dot is related to the segment” mean “the dot touches the
segment”. Let relation R’ play the role of the relation R.
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Bob’s interpretation of Axiom System #2.
• Let A’ be the set of dots in the picture at right. Let set A’ play
the role of set A.
• Let B’ be the set of segments in the picture at right. Let set B’
play the role of set B.
• Let relation R’ from A’ to B’ be defined by saying that the words
“the dot is related to the segment” mean “the dot touches the
segment”. Let relation R’ play the role of the relation R.
Carol’s interpretation of Axiom System #2.
• Let A’ be the set whose elements are the letters v, w, x, and y. That is, A′ = {v, w, x, y} .
Let set A’ play the role of set A.
• Let B’ be the set whose elements are the sets {v, w} , {v, x} , {v, y} , and {w, x} , {w, y} ,
•
and { x, y} . That is, B′ = {{v, w} , {v, x} , {v, y} , {w, x} , {w, y} , { x, y}} . Let set B’ play the
role of set B.
Let relation R’ from A’ to B’ be defined by saying that the words “the letter is related to
the set” mean that “the letter is an element of the set”. Let relation R’ play the role of the
relation R.
Notice that Alice and Bob have slightly different interepretations of the axiom system. Is one
better than the other? It turns out that we will consider one to be much better than the other. The
criterion that we will use is to consider what happens when we translate the Axioms into
statements about dots and segments. Using a find & replace feature in a word processor, we can
simply replace every occurrence of ake with dot, every occurrence of bem with segment, and
every occurrence of is related to with touches. Here are the resulting three statements.
Statements:
1. There are four dots. These may be denoted dot1, dot2, dot3, and dot4.
2. For any set of two dots, denoted {doti , dotk } where i ≠ k , there is exactly one segment such
that doti touches the segment and dotk touches the segment.
3. For any segment, there is exactly one set of two dots, denoted {doti , dotk } where i ≠ k , such
that doti touches the segment and dotk touches the segment.
We see that in Bob’s interpretation, all three of these statements are true. In Alice’s interpretation
the first and third statements are true, but the second statement is false.
What about Carol’s interpretation? We should consider what happens when we translate the
Axioms into statements about letters and sets. We can simply replace every occurrence of ake
with letter, every occurrence of bem with set, and every occurrence of is related to with is an
element of. Here are the resulting three statements.
Statements:
1. There are four letters. These may be denoted letter1, letter2, letter3, and letter4.
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2. For any set of two letters, denoted {letteri , letterk } where i ≠ k , there is exactly one set such
that letteri is an element of the set and letterk is an element of the set.
3. For any set, there is exactly one set of two letters, denoted {letteri , letterk } where i ≠ k , such
that letteri is an element of the set and letterk is an element of the set.
We see that in Carol’s interpretation, the translations of the three axioms are three statements that
are all true.
Let’s formalize these ideas with two definitions.
Definition 11 successful interpretation
To say that an interpretation of an axiom system is successful means that when the undefined
terms and undefined relations in the axioms are replaced with the corresponding terms and
relations of the interpretation, the resulting statements are all true.
Definition 12 model of an axiom system
A model of an axiom system is an interpretation that is successful.
So we would say that Bob’s and Carol’s interpretations are successful: they are models. Alice’s
interpretation is unsuccessful: it is not a model.
Notice also that Bob’s and Carol’s models are essentially the same in the following sense: one
could describe a correspondence between the objects and relations of Bob’s model and the
objects and relations of Carol’s model in a way that all corresponding relationships are
preserved. Here is one such correspondence.
objects in Bob’s model
the lower left dot
the lower right dot
the upper right dot
the upper left dot
the segment on the bottom
↔
↔
↔
↔
↔
↔
objects in Carol’s model
the letter v
the letter w
the letter x
the letter y
the set {v, w}
the segment that goes from lower left to upper right ↔ the set {v, x}
the segment on the left side ↔ the set {v, y}
the segment on the right side ↔ the set {w, x}
the segment that goes from upper left to lower right ↔ the set {w, y}
the segment across the top ↔ the set { x, y}
relation in Bob’s model ↔ relation in Carol’s model
the dot touches the segment ↔ the letter is an element of the set
What did I mean above by the phrase “...in a way that all corresponding relationships are
preserved...”? Notice that the following statement is true in Bob’s model.
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The lower right dot touches the segment on the right side.
If we use the correspondence to translate the terms and relations from Bob’s model into terms
and relations from Carol’s model, that statement becomes the following statement.
The letter w is an element of the set {w, x} .
This statement is true in Carol’s model. In a similar way, any true statement about relationships
between dots and segments in Bob’s model will translate into a true statement about relationships
between letters and sets in Carol’s model.
The notion of two models being essentially the same, in the sense described above, is formalized
in the following definition.
Definition 13 isomorphic models of an axiom system
Two models of an axiom system are said to be isomorphic if it is possible to describe a
correspondence between the objects and relations of one model and the objects and relations of
the other model in a way that all corresponding relationships are preserved.
It should be noted that it will not always be the case that two models for a given axiom system
are isomorphic. We will return to this in the next section, when we discuss completeness.
2.4. Properties of Axiom Systems
In this section, we will discuss three important properties that an axiom system may or may not
have. They are consistency, completeness, and independence.
2.4.1. Consistency
We will use the following definition of consistency.
Definition 14 consistent axiom system
An axiom system is said to be consistent if it is possible for all of the axioms to be true. The
axiom system is said to be inconsistent if it is impossible for all of the axioms to be true.
This definition should bother you a little bit. Hopefully, the next few paragraphs will clarify it.
One proves that an axiom system is consistent by producing a model for the axiom system. For
Axiom System #2, we have two models—Bob’s and Carol’s—so the axiom system is definitely
consistent.
How would one prove that an axiom system is inconsistent? Consider the following rule of
inference from Math 306.
~ p→c
Contradiction Rule
∴p
In this rule, the symbol c stands for a contradiction—a statement that is always false. We used
this rule in to following way. To prove that statement p is true using the method of contradiction,
we started by assuming that statement p was false. We then showed that we could reach a
contradiction. Therefore, the statement p must be true.
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Suppose we replace the statement p with the statement ~q. Then the contradiction rule becomes
~ (~ q) → c
Contradiction Rule
∴( ~ q )
In other words,
q→c
∴~ q
This version of the rule states that if one can demonstrate that statement q leads to a
contradiction, then statement q must be false.
Contradiction Rule
There are other versions of this rule as well. Consider what happens if we use a statement q of
the form statement1 ∧ statement2 .
Contradiction Rule
( statement1 ∧ statement2 ) → c
∴~ ( statement1 ∧ statement2 )
If we apply DeMorgan’s law to the conclusion of this version of the rule, we obtain
( statement1 ∧ statement2 ) → c
Contradiction Rule
∴ ( ~ statement1 ) ∨ ( ~ statement2 )
This version of the rule states that if one can demonstrate that an assumption that statement1 and
statement2 are both true leads to a contradiction, then at least one of the statements must be false.
Finally, consider what happens if we use a whole list of statements.
( statement1 ∧ statement2 ∧ ⋯ ∧ statementk ) → c
Contradiction Rule
∴ ( ~ statement1 ) ∨ ( ~ statement2 ) ∨ ⋯ ∨ ( ~ statementk )
This version of the rule states that if one can demonstrate that an assumption that a whole list of
statements is true leads to a contradiction, then at least one of the statements must be false.
Now return to the notion of an inconsistent axiom system. Recall in an inconsistent axiom
system, it is impossible for all of the axioms to be true. In other words, at least one of the axioms
must be false. We see that the Contradiction Rule could be used to prove that an axiom system is
inconsistent. That is, if one can demonstrate that an assumption that a whole list of axioms is true
leads to a contradiction, then at least one of the axioms must be false.
We can create an example of an inconsistent axiom system by messing up Axiom system #2. We
mess it up by appending a fourth axiom in a certain way.
Axiom system #3, an example of an inconsistent axiom system
Primitive Terms:
• ake
• bem
Primitive Relations:
• relation R from the set A of all akes to the set B of all bems
Axioms:
1. There are four akes. These may be denoted ake1, ake2, ake3, and ake4.
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2. For any set of two akes, denoted {akei , akek } where i ≠ k , there is exactly one
bem such that akei touches the bem and akek touches the bem.
3. For any bem, there is exactly one set of two akes, denoted {akei , akek } where
i ≠ k , such that akei touches the bem and akek touches the bem.
4. There are exactly five bems.
Using Axiom System #3, we can prove the following theorem.
Theorem 1 for axiom system #3: There are exactly 6 bems.
The proof is the exact same proof that was used to prove the identical theorem for
axiom system #2. The proof only uses the first three axioms.
Now we can prove the following:
Theorem 2 for axiom system #3: There are exactly 5 bems and there are exactly 6 bems.
Proof: By Axiom #4 and Theorem 1
But this statement is a contradiction! So we have demonstrated that an assumption that the four
axioms from Axiom system 3 are true leads to a contradiction. Therefore, least one of the axioms
must be false. In other words, Axiom System #3 is inconsistent.
Note that it is tempting to say that it must be axiom #4 that is false, because there was nothing
wrong with the first three axioms before we threw in the fourth one. But in fact, there is not
really anything wrong with the fourth axiom in particular. For instance, if one discards axiom #3,
then the remaining list of axioms, consisting of axiom #1, axiom #2, and axiom #4 is perfectly
consistent. Here is such an axiom system, with the axioms re-numbered. You are asked to prove
that this axiom system is consistent in Exercise #6.
Axiom system #4
Primitive Terms:
• ake
• bem
Primitive Relations:
• relation R from the set A of all akes to the set B of all bems
Axioms:
1. There are four akes. These may be denoted ake1, ake2, ake3, and ake4.
2. For any set of two akes, denoted {akei , akek } where i ≠ k , there is exactly one
bem such that akei touches the bem and akek touches the bem.
3. There are exactly five bems.
So the problem is not with any one particular axiom. Rather, the problem is with the whole set of
four.
2.4.2. Independence
An axiom system that is not consistent could be thought of as one in which the axioms don’t
agree; an axiom system that is consistent could be thought of as one in which there is no
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disagreement. In this sort of informal language, we could say that the idea of independence of an
axiom system has to do with whether or not there is too much agreement. Better yet, we could
say that independence has to do with whether or not there is any redundancy in the list of
axioms. The following definitions will make this precise.
Definition 15 dependent and independent axioms
An axiom is said to be dependent if it is possible to prove that the axiom is true as a consequence
of the other axioms. An axiom is said to be independent if it is not possible to prove that it is true
as a consequence of the other axioms.
To prove that an axiom is dependent, one essentially removes the axiom from the list of axioms
and calls it a theorem, and demonstrates that one can prove the theorem with a proof that uses
only the other axioms.
For an example of a dependent axiom, consider the following list of axioms that was constructed
by appending an additional axiom to the list of axioms for Axiom System #2.
Axiom system #5
Primitive Terms:
• ake
• bem
Primitive Relations:
• relation R from the set of akes to the set of bems.
Axioms:
1. There are four akes. These may be denoted ake1, ake2, ake3, and ake4.
2. For any set of two akes, denoted {akei , akek } where i ≠ k , there is exactly one
bem such that akei touches the bem and akek touches the bem.
3. For any bem, there is exactly one set of two akes, denoted {akei , akek } where
i ≠ k , such that akei touches the bem and akek touches the bem.
4. There are exactly six bems.
We recognize the statement of axiom #4. It is the same statement as Theorem #1 for Axiom
System #2. In other words, it can be proven that axiom #4 is true as a consequence of the first
three axioms. So Axiom #4 is not independent; it is dependent.
To prove that a particular axiom is independent, one thinks of that axiom as just an extra
statement, instead of thinking of it as an axiom. For the remaining, smaller axiom system, one
must produce two models: one model in which all of the other axioms are true and the extra
statement is also true, and a second model in which all of the other axioms are true and the extra
statement is false.
For an example of an independent axiom, consider axiom #3 from Axiom System #2. It is an
independent axiom. To prove that it is independent, we treat it as an extra statement, instead of as
an axiom, and we consider models of the remaining, smaller axiom system.
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Axiom system #6: This is just Axiom System #2 with the third axiom treated as an extra
statement instead of as an axiom.
Primitive Terms:
• ake
• bem
Primitive Relations:
• relation R from the set of akes to the set of bems.
Axioms:
1. There are four akes. These may be denoted ake1, ake2, ake3, and ake4.
2. For any set of two akes, denoted {akei , akek } where i ≠ k , there is exactly one
bem such that akei touches the bem and akek touches the bem.
Extra statement that used to be an axiom
• For any bem, there is exactly one set of two akes, denoted {akei , akek } where i ≠ k , such
that akei touches the bem and akek touches the bem.
Now consider two models for Axiom System #6
Bob’s model of Axiom System #6.
• Let A’ be the set of dots in the picture at right. Let set A’ play
the role of set A.
• Let B’ be the set of segments in the picture at right. Let set B’
play the role of set B.
• Let relation R’ from A’ to B’ be defined by saying that the words
“the dot is related to the segment” mean “the dot touches the
segment”. Let relation R’ play the role of the relation R.
Dan’s model of Axiom System #6.
• Let A’ be the set of dots in the picture at right. Let set A’ play
the role of set A.
• Let B’ be the set of segments in the picture at right. Let set B’
play the role of set B.
• Let relation R’ from A’ to B’ be defined by saying that the words
“the dot is related to the segment” mean “the dot touches the
segment”. Let relation R’ play the role of the relation R.
Consider the translation of the extra statement into the language of the models:
• For any segment, there is exactly one set of two dots, denoted {doti , dotk } where i ≠ k ,
such that doti touches the segment and dotk touches the segment.
We see that in Bob’s model of Axiom System #6, the extra statement is true, while in Dan’s
model of Axiom System #6, the removed axiom is false. So
Based on these two examples, we would say that in Axiom System #2, the third axiom is
independent.
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The following definition is self-explanatory.
Definition 16 independent axiom system
An axiom system is said to be independent if all of its axioms are independent. An axiom system
is said to be not independent if one or more of its axioms are not independent.
To prove that an axiom system is independent, one must prove that each one of its axioms is
independent. That means that for each of the axioms, one must go through a process similar to
the one that we went through above for Axiom #3 from Axiom System #2. This can be a huge
task.
To prove that an axiom system is not independent, one need only prove that one of its axioms is
not independent.
2.4.3. Completeness
Recall that in Section 2.3, we found that Bob’s and Carol’s models of Axiom System #2 were
isomorphic models. It turns out that any two models for that axiom system are isomorphic. Such
a claim can be rather hard—or impossible—to prove, but it is a very important claim. It
essentially says that the axioms really “nail down” every aspect of the behavior of any model.
This is the idea of completeness.
Definition 17 complete axiom system
An axiom system is said to be complete if any two models of the axiom system are isomorphic.
An axiom system is said to be not complete if there exist two models that are not isomorphic.
It is natural to wonder why the word complete is used to describe this property. One might think
of it this way. On the other hand, if an axiom system is complete, then it is like a complete set of
specifications for a corresponding model. All models for the axiom system are essentially the
same: they are isomorphic. If an axiom system is not complete, then one does not have a
complete set of specifications for a corresponding model. The specifications are insufficient,
some details are not nailed down. As a result, there can be models that differ from each other:
models that are not isomorphic.
For an example, consider axiom system #6. In Section 2.4.2, we found two models that are not
isomorphic. (There are others as well.) In Bob’s model, there are six segments, But in Dan’s
model, there is only one segment. We can say that the statement “there are exactly six
segments”, or “there are exactly six bems” is an independent statement in Axiom System #6,
because the statement cannot be proven true on the basis of the axioms. If we wanted to, we
could construct a new axiom system #7 by appending an axiom to Axiom System #6 in the
following manner.
Axiom system #7: (This is just Axiom System #6 with an additional axiom added.)
Primitive Terms:
• ake
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• bem
Primitive Relations:
• relation R from the set A of all akes to the set B of all bems.
Axioms:
1. There are four akes. These may be denoted ake1, ake2, ake3, and ake4.
2. For any set of two akes, denoted {akei , akek } where i ≠ k , there is exactly one
bem such that akei touches the bem and akek touches the bem.
3. There are exactly six bems.
Of course, Bob’s successful interpretation for Axiom System #6 would be a successful
interpretation for Axiom System #7, as well. That is, Bob’s interpretation is a model for Axiom
System #6 and also for Axiom System #7. But Dan’s successful interpretation for Axiom
System #6 would not be a successful interpretation for Axiom System #7. That is, Dan’s
interpretation is a model for Axiom System #6, but not for Axiom System #7.
Keep in mind, that we could have appended a different axiom to Axiom System #6.
Axiom system #8: (This is just Axiom System #6 with a different additional axiom.)
Primitive Terms:
• ake
• bem
Primitive Relations:
• relation R from the set A of all akes to the set B of all bems.
Axioms:
1. There are four akes. These may be denoted ake1, ake2, ake3, and ake4.
2. For any set of two akes, denoted {akei , akek } where i ≠ k , there is exactly one
bem such that akei touches the bem and akek touches the bem.
3. There is exactly one bem.
We see that Dan’s interpretation is a model for Axiom Systems #6 and #8. On the other hand,
Bob’s interpretation is a model for Axiom System #6 but not for Axiom System #8.
Note that there are other axioms that could have been added to Axiom System #6. One obvious
one that we could have added is the one that we removed from Axiom System #2 to create
Axiom System #6 in the first place.
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2.5. Exercises
The first four problems are about Axiom System #1. This axiom system was introduced in
Section 2.2.1 and has an undefined relation.
1) Which of the following interpretations is successful? That is, which of these interpretations is
a model? Explain.
a. Interpret the words “x is related to y” to mean “ xy > 0 ”.
b. Interpret the words “x is related to y” to mean “ xy ≠ 0 ”.
c. Interpret the words “x is related to y” to mean “x and y are both even or are both odd”.
Hint: One of the three is unsuccessful. The other two are successful. That is, they are models.
2) Consider the two models of Axiom System #1 that you found in exercise (1). For each
model, determine whether the statement “1 is related to -1” is true or false.
3) Based on your answer to exercise (2) is the statement “1 is related to -1” an independent
statement? Explain.
4) Based on your answer to exercise (3), is Axiom System #1 complete? Explain.
5) In Section 2.2.2, we discussed Axiom System #2, which had undefined terms and relations.
Prove Theorem #2 for Axiom System #2.
6) In Section 2.4.1, we discussed Axiom System #4. Prove that this axiom system is consistent
by demonstrating a model. (Hint: Produce a successful interpretation involving a picture of
dots and segments.)
The next few exercises refer to the following new axiom system
Axiom system #9
Primitive Terms:
• cet
• dag
Primitive Relations:
• relation R from the set C of all cets to the set D of all dags.
Axioms:
1. There are exactly three cets. These may be denoted cet1, cet2, and cet3.
2. For any set of two cets, denoted {ceti , cetk } where i ≠ k , there is exactly one
dag such that ceti is related to the dag and cetk is related to the dag.
3. For any set of two dags, denoted {dagi , dag k } where i ≠ k , there is at least
one cet such that the cet is related to dagi and the cet is related to dagk.
4. For every dag, there is at least one cet that is not related to the dag.
7) Produce a model for Axiom System #9 that that uses dots and segments to correspond to cets
and dags, and that uses the words “the dot touches the segment” to correspond to the words
“the cet is related to the dag”. That is, draw a picture that works.
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8) Is Axiom System #9 consistent? Explain.
9) The goal is to prove that Axiom #1 is independent. In exercise #7, you produced a model
involving dots and segments that demonstrates that it is possible for all four axioms to be
true. Therefore, all you need to do is produce a model involving dots and segments that
demonstrates that it is possible for Axioms #2, #3, and #4 to be true and Axiom #1 to be
false.
10) The goal is to prove that Axiom #2 is independent. In exercise #7, you produced a model
involving dots and segments that demonstrates that it is possible for all four axioms to be
true. Therefore, all you need to do is produce a model involving dots and segments that
demonstrates that it is possible for Axioms #1, #3, and #4 to be true and Axiom #2 to be
false.
11) The goal is to prove that Axiom #3 is independent. In exercise #7, you produced a model
involving dots and segments that demonstrates that it is possible for all four axioms to be
true. Therefore, all you need to do is produce a model involving dots and segments that
demonstrates that it is possible for Axioms #1, #2, and #4 to be true and Axiom #3 to be
false.
12) The goal is to prove that Axiom #4 is independent. In exercise #7, you produced a model
involving dots and segments that demonstrates that it is possible for all four axioms to be
true. Therefore, all you need to do is produce a model involving dots and segments that
demonstrates that it is possible for Axioms #1, #2, and #3 to be true and Axiom #4 to be
false.
13) Is axiom system #9 independent? Explain.
14) Prove the following Theorem for Axiom System #9:
Theorem 1 for Axiom System #9: For any set of two dags, denoted {dagi , dag k }
where i ≠ k , there is exactly one cet such that the cet is related to dagi and the cet
is related to dagk.
Hint: Axiom #3 guarantees that there is at least one such cet. Suppose that there is more
than one such cet and show that you can reach a contradiction.
15) Prove the following Theorem for Axiom System #9:
Theorem #2 for Axiom System #9: There are exactly three dags.
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3. Axiomatic Geometries
For the remainder of the course, we will be studying axiomatic geometry. Before starting that
study, we should be sure and understand the difference between analytic geometry and axiomatic
geometry.
3.1. What is an analytic geometry?
Very roughly speaking, an analytic geometry consists of two things:
• a set of points that is represented in some way by real numbers
• a means of measuring the distance between two points
For example, in plane Euclidean analytic geometry, a point is represented by a pair ( x, y ) ∈ ℝ 2 .
That is, a point is a pair of real numbers. The distance between points P = ( x1 , y1 ) and
Q = ( x2 , y2 ) is obtained by the formula d ( P, Q ) =
( x2 − x1 ) + ( y2 − y1 )
2
2
. In three dimensions,
one adds a z-coordinate.
In analytic geometry, objects are described as sets of points that satisfy certain equations. A line
is the set of points ( x, y ) that satisfy an equation of the form ax + by = c; a circle is the set of
points that satisfy an equation of the form ( x − h ) + ( y − k ) = r 2 , etc. Every aspect of the
behavior of analytic geometric objects is completely dictated by rules about solutions of
equations. For example, any two lines either don’t intersect, or they intersect exactly once, or
they are the same line; there are no other possibilities. That this true is simply a fact about
simultaneous solutions of a pair of linear equations in two variables
ax + by = c

dx + ey = f
You will see that in axiomatic geometry, objects are defined in a very different way, and their
behavior is governed in a very different manner.
2
2
3.2. What is an axiomatic geometry?
Very roughly speaking, an axiomatic geometry is an axiom system with the following primitive
(undefined) things.
Primitive terms
• point
• line
Primitive relation
• the point lies on the line
Remark: It is not a very confident definition that begins with the words “…roughly speaking…”.
But in fact, one will not find general agreement about what constitutes an axiomatic geometry.
My description above contains some of the essentials. We will return to the notion of what
makes axiomatic points and lines behave the way we “normally” expect points and lines to
behave in Section 3.6, when we study incidence geometry.
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You’ll notice, of course, that the axiom system above is essentially the same sort of axiom
system that we discussed in Chapter 2. The only difference is that we stick to the particular
convention of using undefined terms called point and line, and the undefined relation spoken the
point is on the line. It is natural to wonder why we bothered with the meaningless terms ake,
bem, cet, and dag, when we could have used the more helpful terms point and line. The reason
for starting with the meaningless terms was to stress the idea that the terms are always
meaningless; they are not supposed to be helpful. When studying axiomatic geometry, it will be
very important to keep in mind that even though you may think that you know what a point and a
line are, you really don’t. The words are as meaningless as ake and bem. On the other hand,
when studying a model of an axiomatic geometry, we will know the meaning of the objects and
relations, but we will be careful to always give those objects and relations names other than point
and line. For instance, we used the names dot and segment in our models that involved drawings.
The word dot refers to an actual drawn spot on the page or chalkboard; it is our interpretation of
the word point, which is an undefined object.
Because the objects and relations in axiomatic geometry are undefined things, their behavior will
be undefined as well, unless we somehow dictate that behavior. That is the role of the axioms.
Every aspect of the behavior of axiomatic geometric objects must be dictated by the axioms. For
example, if we want lines to have the property that two lines either don’t intersect, or they
intersect exactly once, or they are the same line, then that will have to be specified in the axioms.
3.3. Finite Geometries
3.3.1. Three Point Geometry
A finite axiomatic geometry is one that has a finite number of points. Our first example has three
points.
Axiom System: Three-Point Geometry
Primitive Terms:
• point
• line
Primitive Relations:
• relation R from the set C of all points to the set D of all lines, spoken the point
lies on the line.
Axioms:
1. There are exactly three points.
2. For any set of two points, there is exactly one line that both points lie on.
3. For any set of two lines, there is at least one point that lies on both lines.
4. For every line, there is at least one point that does not lie on the line.
Notice that the Three Point Geometry is the same as Axiom System #9, presented in the
exercises of Section 2.5. Therefore, we can immediately state the following theorems, because
they are just translations of theorems that have already been proven.
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Three Point Geometry Theorem #1: For any two distinct lines, there is exactly one point that lies
on both lines. (This was proven in Exercise #14) of Chapter 2.)
Remark: You might be a little puzzled by this theorem. Why does axiom 3 say that there is at
least one point that lies on both lines, when in fact it turns out that there is exactly one point.
Why not just rephrase axiom 3 to say exactly one?!? The idea is that it is desirable to have an
axiom system that says as little as possible and yet conveys all the information necessary. A
better way of saying this is that it is desirable for an axiom system to be independent. Notice
that the conjunction of the following two statements
a) “there is at least one point that lies on both lines”, and
b) “there are not two points that lie on both lines”
is equivalent to the following statement
c) “there exists exactly one point that lies on both lines”
Three Point Geometry Theorem #1 essentially proved that the statement (b) is automatically
true as a consequence of the Three Point Geometry Axioms. In other words, statement (b) is
not independent. So, part of statement (c), the part that says that there is not more than one
point, is not independent. That’s why it is more desirable to use statement (a) as an axiom
than to use statement (c).
Three Point Geometry Theorem #2: There are exactly three lines. (This was proven in Exercise
#15) of Chapter 2.)
There are other theorems that can be proven as well. Here is a particularly simple theorem that is
in a similar spirit to Theorem #1.
Three Point Geometry Theorem #3: For any line, there is exactly one point that does not lie on
the line.
Proof
1. Suppose that L is a line.
2. There exists at least one point that does not lie on L. (by axiom 4)
3. We can call one such point P.
4. Assume that there is more than one point that does not lie on L. (assumption)
5. Then there is a second point, that we can call Q.
6. There is a line that both P and Q lie on. (by axiom 1)
7. We can call that line M.
8. There is a point that lies on both L and M (by axiom 3).
9. That point cannot be P or Q, because they do not lie on L.
10. So it must be some other point that lies on both L and M. We can call this point R.
11. Observe that points P, Q, and R all lie on line M. This contradicts axiom 4.
12. We have reached a contradiction. So our assumption was wrong. There cannot be more
than one point that lies on L. So there must be exactly one point that does not lie on the
line.
End of Proof
Three Point Geometry Theorem #4: For any line, there are exactly two points that lie on the line.
You will prove this theorem in the exercises.
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3.3.2. Four Point Geometry
Our next finite geometry has four points.
Axiom system: Four-Point Geometry
Primitive Terms:
• point
• line
Primitive Relations:
• relation R from the set A of all points to the set B of all lines, spoken the point
lies on the line.
Axioms:
1. There are four points.
2. For any set of two points, there is exactly one line that both points lie on.
3. Every line has exactly two points that lie on it.
Notice that the Four Point Geometry is the same as Axiom System #2, presented in Section 2.2.2.
Therefore, we can immediately state the following theorems, because they are just translations of
theorems that have already been proven.
Four Point Geometry Theorem #1: There are exactly six lines. (This was proven in Section
2.2.2.)
Four Point Geometry Theorem #2: For every point, there are exactly three lines that pass through
the point. (This was proven in Exercise #5) of Chapter 2.)
3.4. More about terminology
3.4.1. Is the primitive relation symmetric?
Is the primitive relation symmetric? That is, if point P lies on line L, will it also be true that line
L lies on point P? The answer must be no, and it has to do only with the conventions of the
terminology of relations. The undefined relation spoken “the point lies on the line” is a relation
from the set of points to the set of lines. Therefore, one does not say “the line lies on the point”.
So the relation cannot possibly be symmetric. Another way of explaining this is to say that the
symmetry property of relations is something that is defined only for relations on a set. The “lies
on” relation is not a relation on a set; it is a relation from one set to another. So it cannot be
symmetric and it cannot be turned around.
But restricting ourselves to always having to say “point P lies on line L” can lead to cumbersome
and bland-sounding writing. For that reason alone, it is worthwhile to introduce some additional
terminology.
Definition 18 passes through
• words: Line L passes through point P.
• meaning: Point P lies on line L.
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With this terminology, the axioms for the Four Point Geometry could be rephrased as follows
1.
There are four points.
2.
For any set of two points, there is exactly one line that passes through
throug both points.
3.
Every line passes through exactly two points.
It is worth remarking that most introductory axiomatic geometry books are a bit sloppy regarding
the order of words in expressions involving an undefined relation. One reason for this may be
that most of those books do not use the terminology of relations, and so they do not need to stick
to the conventions of that terminology. Here is a presentation of the Three Point Geometry from
one such textbook.
In your work, you will encounter a variety of wordings for the undefined relation. Here are five
more that we won’t use:
Definition 19 commonly used expressions that we will not use
• Line L contains point P.
• Line L is on point P.
• Point P is incident upon line L.
• Line L is incident upon point P.
3.4.2. Additional terminology
While we’re in the business of introducing new terminology, we may as well introduce some
more. The next two definitions are self
self-explanatory.
Definition 20 intersecting lines
• words: Line L intersects line M.
• meaning: There exists a point P that lies on both lines. (at least one point)
Definition 21 parallel lines
• words: Line L is parallel to line M.
• symbol: L M
•
meaning: Line L does not intersect line M.
It’s worth noting that we have essentially introduced a bunch of new relations in this
t section. The
words “Line
ine L passes through point P” indicate a relation from the set of all lines to the set of all
points. The words “Line L intersects line M” indicate a relation on the set of all lines. Similarly,
the words “Line L is parallel to lin
line M” indicate another relation on the set of all lines. These
relations are not primitive relations, because they have an actual meaning. Those meanings are
given by the above definitions, and those definitions refer to primitive terms and relations and to
previously defined words. We will refer to this kind of relation as a defined relation. It is
customary to not list the definitions of the defined relations when presenting the axiom system.
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This is understandable, because often there are many defined relations, and to list them all would
be very cumbersome. But it is unfortunate that the defined relations are not usually listed,
because it makes the defined relations seem less important than the other components of an
axiom system.
Here are two additional definitions, also straightforward.
Definition 22 collinear points
• words: The set of points { P1 , P2 ,⋯ , Pk } is collinear.
•
meaning: There exists a line L that passes through all the points.
Definition 23 concurrent lines
• words: The set of lines { L1 , L2 ,⋯ , Lk } is concurrent.
•
meaning: There exists a point P that lies on all the lines.
It’s worth noting that these two definitions are not relations. Rather, they are simply statements
that may or may not be true for a particular set of points or a particular set of lines. That is, they
are properties that a set of points or a set of lines may or may not have.
3.4.3. The Big Question
The following question is extremely important in axiomatic geometry.
Given a line L and a point P that does not lie on L, how many lines M exist that pass
through P and are parallel to L?
It is so important that we will refer to it as The Big Question. It will come up throughout the
remainder of the course.
3.5. Fano’s and Young’s Finite Geometries
Let’s return to finite geometries. A more complicated finite geometry is the following. (Note the
use of defined relations in the statements of the axioms.)
Axiom system: Fano’s Geometry
Primitive Terms:
• point
• line
Primitive Relations:
• relation R from the set A of all points to the set B of all lines, spoken the point
lies on the line.
Axioms:
1. There exists at least one line.
2. Every line passes through exactly three points.
3. Not all points of the geometry lie on the same line.
4. For any set of two points, there is exactly one line that passes through both.
5. Any two lines intersect.
In the exercises, you will prove the following things:
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•
•
•
•
Fano’s Geometry Theorem #1: There are exactly seven points and seven lines.
Fano’s Geometry Theorem #2: Fano’s Geometry is an Incidence Geometry (defined below)
Fano’s Geometry Theorem #3: In Fano’s Geometry, any two lines intersect exactly once.
Fano’s axioms are independent. (This one is hard!)
By changing just the fifth axiom in Fano’s Geometry, we obtain Young’s Geometry
Axiom system: Young’s Geometry
Primitive Terms:
• point
• line
Primitive Relations:
• relation R from the set A of all points to the set B of all lines, spoken the point
lies on the line.
Axioms:
1. There exists at least one line.
2. Every line passes through exactly three points.
3. Not all points of the geometry lie on the same line.
4. For any set of two points, there is exactly one line that passes through both.
5. For each line L, and for each point P that does not lie on L, there exists exactly
one line M that passes through P and is parallel to L.
In the exercises, you will prove the following things:
• Young’s Geometry Theorem #1: There are exactly nine points and twelve lines.
• Young’s Geometry Theorem #2: Young’s Geometry is an Incidence Geometry (defined
below)
• Young’s axioms are independent.
• Young’s Geometry Theorem #3: Two lines parallel to a third line are parallel to each other.
3.6. Incidence Relations and Incidence Geometries
You will notice that in each of the four finite geometries that we have encountered so far, the
axioms can be classified into two types. One type of axiom is just about the primitive terms.
Here are two examples.
• Four Geometry Axiom #1: There exist exactly four points.
• Fano’s Axiom #1: There exists at least one line.
A second type of axiom is about the behavior of the primitive relation. Here are two examples.
• Three Point Axiom #3. For any set of two lines, there is at least one point that lies on
both lines.
• Fano’s Axiom #3: Not all points of the geometry lie on the same line.
In some early books on axiomatic geometry, the primitive relation was spoken “the line is
incident upon the point”. Axioms such as the two above, that described the behavior of the
primitive relation, were called axioms of incidence. Even though most books no longer use the
words “…is incident upon…” for the primitive relation, it is still fairly common for any axioms
that describe the behavior of the primitive relation to be called axioms of incidence. This can be
confusing.
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To add to the confusion, it is common to refer to the following three incidence axioms as THE
incidence axioms.
Definition 24 THE incidence axioms
• Incidence Axiom 1: For every pair of distinct points, there exists exactly one line that
passes through both points.
• Incidence Axiom 2: Every line passes through at least two distinct points.
• Incidence Axiom 3: There exist three distinct non-collinear points. (at least three)
And it is common to use the term incidence geometry in the following way
Definition 25 incidence geometry
• words: an incidence geometry
• meaning: any axiomatic geometry in which THE incidence axioms are true statements.
The last sentence of this last definition needs some clarification. If a particular axiomatic
geometry has an axiom list that includes THE incidence axioms, then of course that geometry is
an incidence geometry. But quite often, an axiomatic geometry will not include THE incidence
axioms on its list of axioms, but the statements can be proven as a theorem and so they are still
true, and so the geometry is still an incidence geometry. As an example, let’s prove that the
Three Point Geometry is an incidence geometry.
Claim: the Three Point Geometry is an incidence geometry.
Proof:
1. Incidence axiom #1 is true in the three point geometry (by Three Point Geometry axiom
#2)
2. Incidence axiom #2 is true in the three point geometry (by Three Point Geometry theorem
#4)
3. Incidence axiom #3 is true in the three point geometry (by Three Point Geometry axiom
#1 and theorem #3)
End of Proof
What’s so special about THE incidence axioms? Very very roughly speaking, they guarantee that
the primitive points and lines in an abstract geometry will have some of the “normal” behavior
that we associate with points and lines in analytic geometry, or in drawings that we have made
all of our lives. (Indeed, I once saw axiomatic geometry defined as any axiom system with
primitive terms “point” and “line”, primitive relation “the point lies on the line”, in which THE
incidence axioms are true statements. That is not the definition of axiomatic geometry that we
are using, but it gives you an idea of the importance of the incidence axioms.)
For instance we are used to the fact that two lines can either be parallel, or intersect once, or be
the same line. That is, distinct lines that are not parallel only intersect once. It was mentioned in
Section 3.1 that in analytic geometry, lines behave this way as a consequence of behavior of
solutions of systems of linear equations. In Section 3.2, it was mentioned that in axiomatic
geometry, every aspect of the behavior of points and lines will need to be specified by the
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axioms. Well, THE incidence axioms do in fact guarantee that lines have the particular behavior
we are discussing. The following theorem articulates this fact.
Theorem 1 In Incidence Geometry, if L and M are distinct lines that are not parallel, then there
is exactly one point that both lines pass through.
Proof
Suppose that lines L and M are distinct, non-parallel lines. Because they are not parallel,
they must intersect. So there must be at least one point P that they both pass through.
Assume that there is also another point Q that they both pass through. Then lines L and M
both pass through points P and Q. This contradicts incidence axiom #1. We have reached
a contradiction. So our assumption was wrong: there can’t be a second point.
End of Proof.
Here are four more examples of Incidence Geometry theorems that guarantee that points and
lines have “normal” behavior.
Theorem 2 In Incidence Geometry, there exist three lines that are not concurrent
Theorem 3 In Incidence Geometry, given any line L, there exists a point not lying on L.
Theorem 4 In Incidence Geometry, given any point P, there exists a line that does not pass
through P.
Theorem 5 In Incidence Geometry, given any point P, there exist two lines that pass through P.
We proved above that the Three Point Geometry is an incidence geometry. In the exercises, you
will prove that the Four Point Geometry, Fano’s Geometry, and Young’s Geometry are also
incidence geometries
3.6.1. Infinite Incidence Geometries
All of our previous examples of geometries and of incidence geometries have been finite
geometries. But we will spend most of the quarter studying two axiomatic geometries that are
infinite geometries and that are incidence geometries.
3.6.2. A drawn model with straight-looking lines (Euclidean)
We’ll discuss the following interpretation in class.
Definition 26 the straight line interpretation of incidence geometry
• Points are interpreted as drawn dots.
• Lines are interpreted as drawn lines (drawn with a ruler).
• “the point is on the line” is interpreted as “the dot touches the drawn line”.
3.6.3. A drawn model with curvy-looking lines (Poincare disk)
Before introducing our second drawn model, we need to introduce some terms pertaining to the
unit circle and circles that intersect it in a certain way. The first of these is called the Hpoint.
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Definition 27 Hpoint
• word: Hpoint
• meaning: an ( x, y ) pair in the interior of the unit
circle
y
(0,1)
(1,0)
(-1,0)
x
The interior of the unit circle is the shaded region shown
in the figure at right. The boundary pairs, including the
four that are labeled, are NOT part of the interior, so
they are not Hpoints.
(0,-1)
We need to discuss something called “orthogonal circles”. To do that, we must first introduce the
idea of the angle of intersection of curvy objects. You are used to the idea of the angles created
by two intersecting lines: One simply considers the angle created by a pair of rays that emanate
from the intersection point and that lie on the two lines.
This same idea can be generalized to the intersection of curvy objects. Given two intersecting
objects, there are two straight lines that are tangent to the two objects at their point of
intersection.
We say that the two curvy objects are perpendicular (at the place where they intercept) if the
tangent lines are perpendicular.
Now consider two intersecting circles. The circles may intersect in just one point. But notice that
if the circles intersect at more than one point, then the angles created at the two intersection
points (created in the sense described above) will be congruent, in the sense that when the two
angles are placed on top of each other, they match. (To see this in the pictures below, draw a line
that connects the centers of two circles that are touching. Fold the paper along the line.) In
particular, if the two tangent lines at one intersection point are perpendicular, then the two
tangent lines at the other intersection point are also perpendicular.
Page 41 of 106
one intersection point
two intersection points
with non-perpendicular
tangent lines
two intersection points
with perpendicular
tangent lines
We will say that a circle is orthogonal to the unit circle if the tangent lines at the point of
intersection are perpendicular. Such circles will play a key role in our definition of Hlines.
Definition 28 Hlines
• word: Hline
• meaning: Either of the following particular types of sets of Hpoints
• A “straight-looking” Hline is the set of Hpoints that lie on a diameter
of the unit circle.
• A “curved-looking” Hline is the set of Hpoints that lie on a circle that
is orthogonal to the unit circle.
Note that because an Hline must be made up of Hpoints, and all
Hpoints lie in the interior of the unit circle, an Hline (of either flavor)
therefore does not include the two “ends”. We will refer to these as the
“missing ends”. To indicate that the ends are missing, we will put open
circles or arrowheads on the ends of our Hlines. (My typesetting
program won’t do open circles, so I’ll use arrowheads.) Shown at right
are some examples of Hlines.
Given two Hpoints A and B, there is exactly one Hline containing both points. This Hline will be
denoted by AB .
Two important observation about Hlines
• The only ones that are straight are the ones that go through the center of the circle.
• The ones that curve always curve away from the center of the circle.
We are now ready for our new interpretation of incidence geometry. It is called the Poincare disk
interpretation.
Definition 29 the Poincare disk interpretation of incidence geometry
• Points are interpreted as Hpoints.
• Lines are interpreted as Hlines.
• “the point is on the line” is interpreted as “the Hpoint is an element of the Hline”.
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In the exercises, you will explore the Poincare disk. There, you will hopefully convince yourself
that the Poincare disk interpretation is actually a model for incidence geometry. (Your
explorations won’t constitute a proof, but they are enough for our purposes.)
3.6.4. Spherical geometry is not an incidence geometry
We will also consider interpretations of incidence geometry involving the sphere. We will be
interested in determining which interpretations are successful. We will see that the “usual”
spherical geometry is not a successful interpretation of incidence geometry. Some reading
material may be added here, but not much. Most of our investigation of geometry on the sphere
will be in classroom discussions, using the Lenart Spheres.
3.7. Exercises
Recall that the Three Point Geometry was introduced in Section 3.3.1.
1)
2)
3)
4)
5)
Are the Three Point Geometry axioms independent? Explain.
Prove Three Point Geometry Theorem #4.
Prove that in the Three Point Geometry, a line cannot pass through three distinct points.
In the Three Point Geometry, do parallel lines exist? Explain.
In the Three Point Geometry, what is the answer to The Big Question? Explain.
Recall that the Four Point Geometry was introduced in Section 3.3.2.
6)
7)
8)
9)
Are the Four Point Geometry axioms independent? Explain.
Prove that the Four Point Geometry is an incidence geometry.
Prove that in the Four Point Geometry, parallel lines exist.
In the Four Point Geometry, what is the answer to The Big Question? Explain.
Recall that Fano’s and Young’s Geometries were introduced in Section 3.5.
10) Give a model for Fano’s geometry that interprets points as dots and interprets lines as curves.
11) Give a model for Fano’s geometry that interprets points as letters and interprets lines as sets.
12) Are Fano’s axioms consistent? Explain.
13) Prove Fano’s Geometry Theorem #1.
14) Prove Fano’s Geometry Theorem #2.
15) Prove Fano’s Geometry Theorem #3.
16) Prove that Fano’s axioms are independent. (All the steps are easy except for one killer.)
17) Are there parallel lines in Fano’s Geometry? Explain.
18) In Fano’s Geometry, what is the answer to The Big Question? Explain.
19) Give a model for Young’s geometry that interprets points as letters and interprets lines as
sets.
20) Are Young’s axioms consistent? Explain.
21) Prove Young’s Geometry Theorem #1.
22) Prove Young’s Geometry Theorem #2.
23) Prove Young’s Geometry Theorem #3. Hint: Proof by contradiction, involving axiom #5.
24) Prove that Young’s axioms are independent.
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25) Are there parallel lines in Young’s Geometry? Explain.
26) In Young’s Geometry, what is the answer to The Big Question? Explain.
Recall that incidence geometry was presented in Section 3.6.
27) Explain why each of the pictures below could not be an incidence geometry.
picture (a)
picture (b)
picture (c)
picture (d)
Recall that the Poincare disk interpretation of incidence geometry was presented in Section 3.6.3.
28) For each picture, draw Hline PQ . The center of the circle is marked with a small cross for
reference.
Q
Q
Q
P
P
P
Q
P
29) On the labeled circle below, (a) Draw Hline AB . Label this Hline L. (b) Draw Hline PQ .
Label this Hline M. (c) Draw Hline PR . Label this Hline N. Observe that Hpoint P is not on
Hline L, and that both Hlines M and N pass through P and do not intersect Hline L.
30) On the labeled circle below, draw Hlines AB , BC , and CA .
31) On the labeled circle below, draw Hlines AB , BC , and CA .
32) On the labeled circle below, draw Hlines AB , BC , CA , and CD .
P
Q
C
C
R
B
B
B
A
A
B
Exercise #29
A
Exercise #30
A
D
C
Exercise #31
Exercise #32
Page 44 of 106
4. Building on the axiom list of Incidence Geometry
4.1. The need for a larger list of axioms
We shall make it our goal to build an axiom system that prescribes the behavior of points and
lines in a way that accurately represents the “straight line” drawings that we have been making
all our lives. Remember, though, that in the language of axiom systems, we turn the idea of
representation around, and say that we want to the “straight line” interpretation to be a model of
our axiom system, and for that to be the only model of our axiom system. That is, we would like
our axiom system to be complete. (We’ve discussed the fact that the straight line interpretation is
fraught with problems: our paper can’t really go on forever, our pencils have fat tips, etc. But
that’s okay. We would like to have an axiom system that describes a sort of “idealized” straight
line world.)
In the previous chapter, we saw that the “straight line” interpretation was a model of Incidence
Geometry. But we studied a number of other models as well, and we saw that they varied widely
in their properties. As a result, we were able to conclude that the axioms of Incidence Geometry
are not complete. We need to consider what axioms to add to THE Incidence Axioms in order to
make a complete axiom system. We will do this in stages.
Notice that many of the models of Incidence Geometry were finite geometries. Our “straight
line” drawings clearly have an infinite number of points. In this chapter, we will start towards
our goal of building a complete axiom set by adding, in the first stage, a set of axioms that insure
that any model will have an infinite number of points. Much of the material is follows Marvin
Greenberg’s book, Euclidean and Non-Euclidean Geometry. The axioms that we will add are
called The Betweenness Axioms. Before we can introduce them, however, we need to quickly
learn about a new kind of relation.
4.2. Binary and Ternary Relations on a Set
In Section 1.2.3, you were introduced to the idea of a relation on a set. You learned that to say
that R is a relation on a set A means that R is a subset of the Cartesian product A × A . Such a
relation is more properly called a binary relation, because of the two “factors” of A in the
Cartesian product. (You’ll notice that I have updated Definition 5.) So for instance, the “less
than” symbol, <, represents a binary relation on the set of real numbers. When one uses the
symbol, it goes between two real numbers, such as 5 < 7. A binary relation of more interest to us
in our study of geometry is the “parallel” relation on the set of lines, studied in Section 1.3
Exercise 12). The idea of a relation on a set A generalizes to other numbers of “factors” of A, as
well. In particular, we will be interested in what are called ternary relations.
Definition 30 Ternary Relation on a Set
• Words: R is a ternary relation on A.
• Usage: A is a set.
• Meaning: R is a subset of A × A × A .
• Equivalent meaning in symbols: R ⊂ A × A × A
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For example, you are undoubtedly familiar with the following ternary relation on the set of real
numbers.
Definition 31 The Betweenness Relation on the set of real numbers
• Words: “x is between y and z.”
• Usage: x, y, and z are real numbers.
• Meaning: “x < y <z or z <y <x.”
• Remark: This is a ternary relation on the set of real numbers.
• Warning: This is NOT the same as betweenness for points, discussed in the next section.
When we introduce the batch of axioms that will be appended to the list of Incidence Axioms,
we will also introduce a new ternary relation. Unfortunately, it is an undefined relation.
4.3. Introducing Incidence and Betweenness Geometry
As mentioned earlier, our first “upgrade” of Incidence Geometry will be to add a batch of axioms
that will insure that any model will have an infinite number of points. In addition to the
undefined relation “the point lies on the line”, this axiom system has another undefined relation,
spoken “point B is between point A and point C”. In the jargon of relations, we would say that
this is a ternary relation on the set of points. Notice that the first batch of axioms is THE
Incidence Axioms, which dictate the behavior of the primitive relation “the point lies on the
line”. The second batch of axioms are called The Betweenness Axioms; they dictate the behavior
of the primitive relation “point B is between point A and point C”. This axiom system does not
have a name in the literature, but we will refer to it as the Incidence and Betweenness Geometry.
Axiom System: Incidence and Betweenness Geometry (abbreviated I&B Geometry)
Primitive Terms: point, line
Primitive Relations:
• relation from the set of points to the set of lines, spoken “the point lies on the line.”
• ternary relation on the set of points, spoken “point B is between point A and point C”,
denoted A*B*C.
Defined Terms, Relations, and Properties: “passes through”, “intersect”, “parallel”,
“concurrent”, “collinear”, “same side”, “opposite side”
Incidence Axioms
IA1: If P and Q are distinct points then there exists exactly one line that both lie on.
IA2: Every line passes through at least two distinct points.
IA3: There exist three distinct points that are non-collinear. (at least three)
Betweenness Axioms
BA1: If A*B*C, then A, B, and C are distinct collinear points, and C*B*A.
BA2: If B and D are distinct points, and L is the unique line that passes through both points,
then there exist points A, C, and E lying on L such that A*B*D and B*C*D and B*D*E.
BA3: If A, B, and C are three distinct points lying on the same line, then exactly one of the
points is between the other two.
BA4 (Plane Separation): If L is a line and A, B, and C are three points not lying on L then
(i) 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.
(ii) 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.
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Notice that in addition to the primitive terms and relations, the Incidence and Betweenness
axioms also use the defined relation “passes through” and the defined property “collinear”, both
of which were introduced in the previous chapter. But there are new defined terms as well, in
axiom BA4: “same side” and “opposite side”. You might wonder why I presented the axiom
system without first explaining what these new words mean. The reason is that the meaning of
the new words can only be understood after proving some theorems involving the earlier axioms.
In the next section, we will study those theorems and others of Incidence and Betweenness
Geometry, defining new terminology along the way.
It should be noted that any theorem of Incidence Geometry will also be a theorem of Incidence
and Betweenness Geometry, because the proofs will still work. So for instance, we know that in
Incidence and Betweenness Geometry, if two distinct lines intersect, they only intersect in one
point. Because of this, the theorem numbering that began in Section 3.6 will be continued as we
continue to build on the axioms of Incidence Geometry in this section and throughout the rest of
these notes.
Much of what we will discuss about Incidence & Betweenness Geometry will require the
definition of additional terminology. But there are some small theorems that we can discuss that
do not require any words that we don’t already know. Here is one, our first example of a theorem
for Incidence and Betweenness geometry. (Note that the theorem number continues the
numbering that began in Section 3.6.)
Theorem 6 In I&B Geometry, if A*B*C and A*C*D then A, B, C, D are distinct and collinear.
(You will prove this theorem in the exercises.)
But as I said, most of what we will discuss in this chapter will require new terms. In the next
section, we introduce line segments and rays, and prove some theorems about their basic
properties.
4.4. Line Segments and Rays
Our first three definitions make reference only to concepts from Incidence Geometry.
Definition 32 symbol
for a line
• symbol: AB
• meaning: the (unique) line that passes through points A and B
It is worth reminding ourselves of the extent to which points and lines are undefined in our
geometry. In many definitions of geometries, including some of the models that we study in our
course, a line will be defined as a particular kind of set of points. But that is not the case in our
geometry. Lines are not sets of points; lines are undefined things. It will often make sense to
refer to the set of points that lie on a line, and we should have a symbol for that. Here it is.
Definition 33 the set of points that lie on a line
• symbol: AB
{ }
•
meaning: the set of points that lie on line AB
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Again, in our geometry, a line is not the same thing as the set of points that lie on the line.
While we’re on the subject of things that are sets of points, we should introduce the definition of
the plane.
Definition 34 the plane
• meaning: the set of all points
It is worth noting that we are studying “plane geometry”. That is analogous to 2-dimensional
drawings. In plane geometry, points and lines are undefined, and the plane is defined to be the set
of all points. This is in contrast to what we could call “space geometry”, analogous to 3dimensional sculptures. In space geometry, points, lines, and planes are all undefined, and space
is defined to be the set of all points.
Our next three definitions refer to the Betweenness relation; they could not have been introduced
in the previous chapter. The terms they introduce are meant to be analogous to the identical
terms from the world of straight line drawings.
Definition 35 line segment, endpoints of a line segment
• symbol: AB
• spoken: “line segment A, B”, or “segment A, B”
• usage: A and B are points.
• meaning: the set AB = { A} ∪ {C : A * C * B} ∪ {B}
• additional terminology: points A and B are called endpoints of segment AB.
Definition 36 ray,
endpoint of a ray
• symbol: AB
• spoken: “ray A, B”
• usage: A and B are points.
• meaning: the set AB = { A} ∪ {C : A * C * B} ∪ { B} ∪ { D : A * B * D}
• additional terminology: Point A is called the endpoint of ray AB . We say that ray AB
emanates from point A.
Definition 37 opposite rays
• words: BA and BC are opposite rays
• meaning: A*B*C
Here are a few simple theorems that refer to some of these new terms.
Theorem 7 In I&B Geometry, every ray has an opposite ray.
Proof
1) Let QR be a ray.
2) There exists a points P such that P*Q*R (by axiom BA2)
3) Rays QP and QR are opposite rays. (by step 2 and the definition of opposite rays)
End of Proof
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Theorem 8 In I&B Geometry, for any two distinct points A and B, AB ∩ BA = AB .
Proof
We must prove that AB ⊂ AB ∩ BA and AB ∩ BA ⊂ AB . We start by proving AB ⊂ AB ∩ BA .
AB ∩ BA = {{ A} ∪ {C : A * C * B} ∪ { B} ∪ {D : A * B * D}}
∩ {{ E : E * A * B} ∪ { A} ∪ {C : A * C * B} ∪ { B}} definition of AB and AB
= { A} ∪ {C : A * C * B} ∪ {B}
definition of intersection
= AB
by definition of segment
End of Proof
Theorem 9 In I&B Geometry, for any two distinct points A and B, AB ∪ BA = AB .
{ }
Proof
Note that AB ∪ BA is just the union
{{ A} ∪ {C : A * C * B} ∪ {B} ∪ {D : A * B * D}} ∪ {{E : E * A * B} ∪ { A} ∪ {C : A * C * B} ∪ {B}} ,
which simplifies to { E : E * A * B} ∪ { A} ∪ {C : A * C * B} ∪ { B} ∪ { D : A * B * D} .
Step 1: prove that AB ∪ BA ⊂ AB . We will do a proof by method of division into cases.
Suppose that P ∈ AB ∪ BA . There are five cases.
• Case 1: P ∈ {E : E * A * B} . Axiom BA1 tells us that E,A, and B are collinear, so E lies
on line AB . Hence, P lies on AB . Therefore, P ∈ AB .by definition of AB .
• Case 2: P=A. Then P lies on line AB . Therefore, P ∈ AB .
{ }
•
•
{ }
{ }
{ }
{ }
{ }
{ }
{ }
{ }
Case 3: P ∈ {C : A * C * B} . Axiom BA1 tells us that A, C, and B are collinear, so C lies
on line AB . Hence, P lies on AB . Therefore, P ∈ AB .by definition of AB .
Case 4: P=B. Then P lies on line AB . Therefore, P ∈ AB .
Case 5: P ∈ { D : A * B * D} . Axiom BA1 tells us that A, B, and D are collinear, so D lies
on line AB . Hence, P lies on AB . Therefore, P ∈ AB .by definition of AB .
We see that in any case, P ∈ AB .
•
{ }
Step 2: prove that AB ⊂ AB ∪ BA . We will do a proof by method of division into cases.
Suppose that P ∈ AB . This just means that P lies on line AB . There are three cases.
• Case 1: P=A. Then P ∈ { A} . But this will mean that both P ∈ AB and P ∈ BA are true
(by definition of AB and BA ). So therefore, P ∈ AB ∪ BA .
• Case 2: P=B. Then P ∈ { B} . But this will mean that both P ∈ AB and P ∈ BA are true
(by definition of AB and BA ). So therefore, P ∈ AB ∪ BA .
{ }
{ }
Page 49 of 106
•
Case 3: A, B, and P are distinct points. Then Axiom BA3 tells usthat
one of the points
lies between the other two. If A lies between B and P, then P ∈ BA (by definition of BA
). If B lies between
A
and
P
,
then
P
∈
AB
(by
definition
of
AB ).
If P lies between A and
B, then P ∈ AB and P ∈ BA (by definition of P ∈ AB and P ∈ BA )
We see that in any case, P ∈ AB ∪ BA .
End of Proof
4.5. Plane Separation
Keep in mind that with the axioms, we are trying to prescribe behavior for our points and lines
(undefined things) that matches behavior that we have observed in our “straight line” drawings.
One thing that we observe in our drawings is that a drawn line splits the plane of the paper into
two halves. This is referred to as plane separation. The role of Axiom BA4 in I&B geometry is
to insure that our undefined points and lines behave the same way. Our next definitions introduce
new terms that are used in the Betweeness Axiom BA4. If we wanted to present the axioms in
logical, chronological order, we would put this definition in between BA3 and BA4.
Definition 38 same side
• words: “A and B are on the same side of L.”
• usage: A and B are points and L is a line that does not pass through either point.
• meaning: either ( A = B ) or ( A ≠ B and line segment AB does not intersect line L.)
Definition 39 opposite side
• words: “A and B are on the opposite side of L.”
• usage: A and B are points and L is a line that does not pass through either point.
• meaning: A ≠ B and line segment AB does intersect line L.
With these most recent definitions, we are equipped to understand the language of Axiom BA4.
(Go back and read that axiom now.)
In Greenberg’s book, the following theorem is called a corollary of Axiom BA4. That word can
mean a number of different things. In this case, it could be taken to mean that the theorem
follows from Axiom BA4 with a very small proof. You will do the proof in the exercises.
Theorem 10 In I&B Geometry, if A and B are on opposite sides of line L, and B and C are on
the same side of L, then A and C are on opposite sides of L.
Notice that Axiom BA4 does not seem to say anything about halves of the plane. The beauty of
the axiom, however, is that it doesn’t have to: the concept of half planes and their behavior can
be left to definitions and theorems, as follows.
Definition 40 half plane
• words: half-plane bounded by L, containing point A.
• meaning: the set whose elements are some point A not on L, along with all the other
points that are on the same side of L as point A
• symbol: HA
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Theorem 11 In I&B Geometry, every line L partitions the plane into three sets: (1) the set of
points that lie on L, (2) a half plane, and (3) a second half-plane. In other words, every
point of the plane either lies on L or is an element of one (not both) of the half planes.
Proof:
1) Let L be a line.
2) There is a point A not on L. (by Theorem 3)
3) There is a point O on L. (by IA2)
4) There is a point B such that A*O*B. (by BA2)
5) Point O is an element of segment AB. (by 4 and definition of segment)
6) segment AB intersects line L (by 3 and 5)
7) A and B are on opposite sides of L. (by 6 and definition of opposite side)
8) Suppose that C is any point in the plane. There are five possibilities.
i) C lies on L
ii) C = A.
iii) C ≠ A , but C and A are on the same side of L.
iv) C = B.
v) C ≠ B , but C and B are on the same side of L.
In every case, it is easy to show that exactly one of the following three statements is true:
• C lies on L
• C is in the same half plane as A, denoted HA.
• C is in the same half plane as B, denoted HB.
End of Proof
4.6. Line Separation
Just as we are familiar with the way a drawn line splits the plane in two, we are also familiar
with the way a dot on a drawn line splits the line in two. This is referred to as line separation.
The Incidence & Betweenness axioms can be used to prove that our undefined points and lines
have this same behavior. But by now, you are certainly getting the idea that it can be very tedious
to prove even simple behavior of points and lines using only the axioms. It turns out that the
proof of the line separation property is rather difficult. Its proof refers to two preliminary
theorems whose proofs are also rather hard. I will state the two preliminary theorems here, along
with the Line Separation theorem. You will prove only the middle one.
Theorem 12 In I&B Geometry, if A*B*C and A*C*D, then B*C*D and A*B*D.
(This theorem will not be proved in this course. We will take it as a given.)
Theorem 13 In I&B Geometry, if A*B*C and B*C*D, then A*B*D and A*C*D.
(In the exercises, you will prove this theorem, using the previous theorem as a given.)
Theorem 14 (Line Separation) In I&B Geometry, If A, B, C, and D are collinear points such
that A*B*C, and D ≠ B , then either D ∈ AB or D ∈ AC , but not both.
(This theorem will not be proved in this course. We will take it as a given.)
Another obvious behavior of drawn points and lines is that if a line goes into one side of a
triangle, it must come out through another side. We have not formally introduced triangles yet,
Page 51 of 106
but even so, this behavior of points and lines can be addressed, in what is called Pasch’s
Theorem. The proof of the theorem uses the plane separation property, Theorem 11.)
Theorem 15 (Pasch’s Theorem) In I&B Geometry, if A, B,and C are non-collinear points and
line L intersects segment AB at a point between A and B, then L also intersects either AC
or BC. Furthermore, if C does not lie on L, then L does not intersect both AC and BC.
Proof
1) Suppose that A, B,and C are non-collinear points and line L intersects segment AB at a point
between A and B.
2) There are two possibilities for point C: Either C lies on L, or it does not lie on L.
3) Suppose C lies on L. Then L intersects both AC andBC.
4) Suppose C does not lie on L.
5) A and B are on opposite sides of L (because line L intersects segment AB)
6) Now there are two more possibilities: C is either on the same side of L as point A, or
on the same side of L as point B. (by Theorem 11, the plane separation property)
7) Suppose C is on the same side of L as A. Then C is on the opposite side of L from
B. Therefore, L will not intersect AC but it will intersect BC.
8) Suppose C is on the same side of L as B. Then C is on the opposite side of L from
A. Therefore, L will not intersect BC but it will intersect AC.
End of Proof
Following are two more theorems that demonstrate that our undefined points and lines behave in
the same way that our drawn points and lines behave.
Theorem 16 In I&B Geometry, if A*B*C then AC = AB ∪ BC and B is the only point
common to segments AB and BC.
(You will prove this Theorem in the exercises.)
Theorem 17 In I&B Geometry, if A*B*C then B = BA ∩ BC and AB = AC .
(This theorem will not be proved in this course. We will take it as a given.)
Theorem 18 In I&B Geometry, if point A lies on line L, and point B does not lie on line L, then
every point of ray AB except point A is on the same side of L as B.
(You will prove this theorem in the exercises.)
4.7. Angles and Triangles
With the notion of rays and line segments, we are able to define angles and triangles. In this
section, we will explore some of their simpler behavior.
Definition 41 angle
• symbol: ∢ABC
• usage: A, B, and C are non-collinear points
• meaning: BA ∪ BC
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•
•
additional terminology: point B is called the vertex of ∢ABC , rays BA and BC are
called the sides.
observations: because BA ∪ BC = BC ∪ BA , the symbols ∢ABC and ∢CBA represent
the same angle.
Notice that an angle is a set of points, because it is defined as the union of two rays and rays are
sets of points. Furthermore, note that as sets of points, BA ∪ BC = BC ∪ BA , so the symbols
∢ABC and ∢CBA represent the same angle.
Definition 42 supplementary angles
• words: supplementary angles
• meaning: two angles that share a common side and whose other sides are opposite rays.
We won’t need the idea of supplementary angles in this chapter. I put the definition here to
emphasize that the notion of supplementary angles does not depend on any concept of angle
congruence or angle measure. This is worth restating explicitly: the definition of supplementary
angles does not define them as two angles whose measures add up to 180
! We don’t get to the
notion of angle congruence until the next chapter, and we won’t get to a notion of angle measure
until an even later chapter.
Definition 43 interior of an angle
• words: the interior of ∢ABC
• meaning: the set of points P such
that
(
P
is
on
the
same
side
of
line
BA as point C) and
(P is on the same side of line BC as point A).
Theorem 19 In I&B Geometry, given ∢BAC and point D lying on BC , point D is in the
interior of ∢BAC if and only if B*D*C.
(You will prove this theorem in the exercises.)
Theorem 20 In I&B Geometry, given angle ∢BAC ; point D in the interior of ∢BAC , and
point E such that C*A*E, the following three statements are all true:
(1) Every point on ray AD except A is in the interior of angle ∢BAC .
(2) No point on the ray opposite to AD is in the interior of angle ∢BAC .
(3) Point B is in the interior of angle ∢DAE .
(We won’t prove this theorem in this class, but will take it as a given.)
Definition 44 ray between two other rays
• words: ray BD is between BA and BC .
• meaning: Point D is in the interior of ∢ABC .
Theorem 21 (The Crossbar Theorem) In I&B Geometry, if ray AD is between rays AB and
AC then ray AD intersects segment BC.
(You will prove this theorem in the exercises.)
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Definition 45 triangle
• symbol: △ ABC
• spoken: triangle A, B, C
• usage: A, B, and C are non-collinear
• meaning: the set AB ∪ BC ∪ CA
• additional terminology:
o The points A, B, C are called the vertices of the triangle.
o The segments AB, BC, CA are called the sides of the triangle
o Side BC is said to be opposite vertex A. Similarly for the other sides.
o The angle ∢BAC is called angle A if there is no chance of this causing confusion.
Similarly for the other angles.
Definition 46 interior of a triangle
• words: the interior of △ ABC
• meaning: the set of points P such
that
(
P
is
on
the
same
side
of
line
BA as point
C) and
(P is on the same side of line BC as point A) and (P is on the same side of line AC as
point B)
Definition 47 exterior of a triangle
• meaning: the set of points Q that are neither an element of the triangle, itself, nor of the
interior of the triangle.
Theorem 22 In I&B Geometry, If ray r emanates from an exterior point of triangle △ ABC and
intersects side AB in a point between A and B, then ray r also intersects side AC or BC.
(You will prove this theorem in the exercises.)
Theorem 23 In I&B Geometry, if a ray emanates from an interior point of a triangle, then it
intersects one of the sides, and if it does not pass through a vertex, it intersects only one
side.
(You will prove this theorem in the exercises.)
Theorem 24 In I&B Geometry, a line cannot be contained in the interior of a triangle.
(You will prove this theorem in the exercises.)
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4.8. Exercises
[1] Prove Theorem 6.
[2] Prove Theorem 10.
[3] Prove Theorem 13. (Hint: Use Theorem 12, and just change the appropriate letters.)
[4] (Hard problem) The goal is to prove Theorem 16. Given A*B*C, do the following:
(a) Using Theorem 12, prove that AB ⊂ AC . Also prove that BC ⊂ AC . (Hint: change letters!)
Together, those statements prove that AB ∪ BC ⊂ AC .
(b) (Hard step) Prove that AC ⊂ AB ∪ BC . Hint: Suppose that P ∈ AC . In the case that P is
equal to one of the other points, A, B, C, then you’re done, because then P ∈ AB ∪ BC . In the
case that P is a new point, show (using the axioms) that there exists a line L that intersects
segment AC at point P. Use Axiom BA4 to show that either P ∈ AB or P ∈ BC (but not both).
(c) Prove that B is the only point common to segments AB and BC by using the method of
contradiction. More specifically, assume that there exists a point Q ≠ B that is common to
segments AB and BC. As in step (b), show that there exists a line M that intersects AC at point Q.
Continue, as you did in step (b), to show that either Q ∈ AB or Q ∈ BC (but not both). See if
you can spot the contradiction.
[5] Prove Theorem 18.
[6] Prove Theorem 19.
[7] Prove Theorem 21, the Crossbar Theorem. (Hint: Do a proof by contradiction. Assume that
the statement is false. Show that point B and C lie on the same side of AD . Then, using Theorem
20, derive a contradiction.)
[8] Prove Theorem 22
[9] Prove Theorem 23.
[10] Prove Theorem 24. Hint: Use the previous theorem.
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5. Neutral Geometry I
5.1. The need for a larger axiom system: Introducing Neutral Geometry
The Incidence and Betweenness Axioms discussed in previous chapters guarantee that our
undefined points and lines will have much of the behavior that we are accustomed to in our
drawings of points and straight lines. But there are many important aspects of the behavior of our
drawings that are not addressed at all in those axioms. One thing that we can observe about pairs
of drawings is whether or not one drawing can be slid on top of another drawing in a way that the
two drawings “fit”. If they do fit, the two drawings are called congruent. Using fancy jargon, we
could say that congruence is a binary relation on the set of drawings. We would like to have a
comparable notion of congruence for our primitive points and lines, and for triangles made from
them. For that, we will need to introduce congruence relations into our axiom system.
Recall that in the previous chapter, we introduced the notion of betweenness of points as an
undefined relation on the set of points, and we included a set of betweenness axioms that
specified how that undefined relation behaved. But later, we were able to introduce a notion of
betweenness of rays that was described by a definition. The behavior of that defined relation was
described in a few theorems. In this chapter, we will see the congruence of line segments and the
congruence of angles introduced as undefined relations that in our axiom system, and there will
be a number of congruence axioms that will specify how these undefined relations behave. But
we will be able to make an actual definition for congruence of triangles. One aspect of the
behavior of the congruence of triangles will be given by an axiom, but all other aspects will be
described by theorems.
In addition to the undefined congruence relations and the congruence axioms, we will also add
two axioms under the heading Axioms of Continuity. These two axioms clearly describe behavior
that we are familiar with in drawings. It is not clear why these statements need to be included as
axioms—that is, why they can’t simply be proven as theorems that are a consequence of the
other axioms—but it is a fact that they are independent of the other axioms. So, if we want these
two statements to be true in our abstract geometry, we do need to include them as axioms. In a
later chapter, we may revisit these axioms of continuity and some related theorems.
With the resulting collection of axioms—the incidence, betweenness, congruence, and
continuity—we have specified an enormous amount about the behavior of our abstract geometry.
What is most fascinating about this collection of axioms, however, is what they do not specify.
We will see that our “straight line” drawings a model of this geometry, but also the drawings of
Hpoints and Hlines in the Poincare disk are a model. It turns out that these are the only two kinds
of models of our axiom system that exist, but they are two non-isomorphic models, all the same.
In other words, this axiom system is not complete. Because of this, the axiom system is often
referred to as Neutral Geometry. It is presented on the following page.
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Axiom System: Neutral Geometry
Primitive Terms: point, line
Primitive Relations:
• relation from the set of points to the set of lines, spoken “the point lies on the line.”
• ternary relation on the set of points, spoken “point B is between point A and point C”,
denoted A*B*C.
• binary relation on the set of line segments, spoken “segment AB is congruent to segment
CD”, denoted AB ≅ CD .
• binary relation on the set of angles, spoken “angle A, B, C is congruent to angle D, E, F”,
denoted ∢ABC ≅ ∢DEF .
Defined Terms, Relations, and Properties: too many to list here
Incidence Axioms
IA1: If P and Q are distinct points then there exists exactly one line that both lie on.
IA2: Every line passes through at least two distinct points.
IA3: There exist three distinct points that are non-collinear. (at least three)
Betweenness Axioms
BA1: If A*B*C, then A, B, and C are distinct collinear points, and C*B*A.
BA2: If B and D are distinct points, and L is the unique line that passes through both points,
then there exist points A, C, and E lying on L such that A*B*D and B*C*D and B*D*E.
BA3: If A, B, and C are three distinct points lying on the same line, then exactly one of the
points is between the other two.
BA4 (Plane Separation): If L is a line and A, B, and C are three points not lying on L then
(i) 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.
(ii) 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.
Congruence Axioms
CA1: The congruence relation on the set of line segments is an equivalence relation
CA2: (segment construction axiom) For any segment AB and ray PQ , there exists a unique
point R on PQ such that PR ≅ AB .
CA3: (segment addition axiom) If A*B*C, A'*B'*C', AB ≅ A′B′ , and BC ≅ B′C ′ , then
AC ≅ A′C ′ .
CA4: The congruence relation on the set of angles is an equivalence relation
CA5: (angle construction axiom) Given an angle ∢BAC , distinct points A' and B', and a
choice of one of the two half-planes bounded by the line A′B′ , there exists a unique ray
A′C ′ such that C' lies in the chosen half plane and ∢BAC ≅ ∢B′A′C ′ .
CA6: (SAS axiom) If two sides and the included angle of one triangle are congruent to two
sides and the included angle of another triangle, then the triangles are congruent.
Axioms of Continuity
Archimedes Axiom: Given any segment CD and ray AB , there exists some positive integer n
and a set of points { P0 , P1 ,⋯ , Pn } on AB such that P0 = A ; for all k, Pk Pk +1 ≅ CD ; and either
Pn = B or ( A * Pn −1 * B and A * B * Pn ).
Circular Continuity Axiom: If a circle intersects both the interior and exterior of another
circle, then the two circles intersect in exactly two points.
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5.2. Triangle Congruence and its Role in the Neutral Geometry Axioms
As mentioned in the previous section, congruence of line segments and congruence of angles are
undefined relations in Neutral Geometry, but congruence of triangles is a defined relation. The
definition of triangle congruence—indeed the definition of any term or relation—is as important
in the axiom system as the primitive terms, primitive relations, and axioms. But the definitions
are traditionally left out of the presentation of the axiom system, because they would take up an
enormous amount of space. In this section, our goal is simply to present the definition of the
congruence relation on the set of triangles, and to understand the significance of the one Neutral
Geometry axiom that pertains to triangle congruence. We will focus on the terminology and will
not prove any theorems. Even so, the section is quite lengthy.
5.2.1. Correspondences between parts of triangles
The term correspondence is used in almost any discussion of triangle congruence, and in almost
any theorem about triangle congruence. So we start our presentation of the concept of triangle
congruence by defining what we mean by a correspondence.
Definition 48 “function”, “domain”, “codomain”, “image”, “machine diagram”;
“correspondence”
• Symbol: f : A → B
• Spoken: “ f is a function that maps A to B ”
• Usage: A and B are sets. Set A is called the domain and set B is called the codomain.
• Meaning: f is a machine that takes an element of set A as input and produces an element
of set B as output.
• More notation: If an element a ∈ A is used as the input to the function f , then the
symbol f (a ) is used to denote the corresponding output. The output f (a ) is called the
image of a under the map f .
• Machine Diagram:
a
input
•
f
f (a)
output
Domain:
Codomain:
the set A
the set B
Additional notation: If f is both one-to-one and onto (that is, if f is a bijection), then
the symbol f : A ↔ B will be used. In this case, f is called a correspondence between
the sets A and B.
Correspondences play a key role in the concept of triangle similarity and congruence, and they
will also play a key role in the concept of polygon similarity and congruence, so we should do a
few examples to get more familiar with them.
Examples
Page 58 of 106
1) Let f : ℝ → ℝ be the cubing function, f ( x ) = x3 . Then f is one-to-one and onto, so we
could say that f is a correspondence, and we would write f : ℝ ↔ ℝ .
2) Let S1 = { A, B, C , D, E} and S 2 = { L, M , N , O, P} . Define a function f : S1 → S 2 by this
picture:
S1
S2
A
L
f
B
M
C
N
D
O
E
P
Then we would say that f is a correspondence, and we would write f : S1 ↔ S 2 .
3) For the same example as above, we could display the correspondence more concisely:
A↔ N
B↔P
C↔L
D↔M
E ↔O
This takes up much less space, and is faster to write, than the picture. However, notice that
this way of displaying the correspondence still uses a lot of space.
4) There is an even more concise way to display the correspondence from the above example.
To understand the notation, though, we should first recall some conventions about brackets
and parentheses. When displaying sets (where order is not important), curly brackets are
used. For example, S1 = { A, B, C , D, E} = {C , A, E , D, B} . But when displaying an ordered
list, parentheses are used instead. So whereas the symbol { A, B} denotes the same set as the
symbol { B , A} , the symbol ( A, B ) denotes a different ordered pair from the symbol ( B, A ) .
With that notation in mind, we will use the symbol below to denote the function f described
in the previous examples.
( A, B, C , D, E ) ↔ ( N , P, L, M , O )
The parentheses indicate that the order of the elements is important, and the double arrow
symbol indicates that there is a correspondence between the lists. Notice that this way of
displaying the function is not as clear as the one in the previous example, but it takes up
much less space.
Definition 49 Correspondence between vertices of two triangles
• Words: “f is a correspondence between the vertices of triangles △ ABC and △ DEF .”
•
Meaning: f is a one-to-one, onto function with domain { A, B , C } and codomain { D , E , F } .
Examples of correspondences between the vertices of △ ABC and △ DEF .
1) ( A, B, C ) ↔ ( D, E , F )
2)
( A, B, C ) ↔ ( D, F , E )
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3)
4)
( B, A, C ) ↔ ( D, E , F )
( B , C , A ) ↔ ( D, F , E )
Notice that the third and fourth examples are actually the same. Each
could be illustrated by the figure shown at right.
A
B
C
f
D
E
F
If a correspondence between the vertices of two triangles has been given, then there is an
automatic correspondence between any other geometric items that are defined purely in terms of
those vertices. For example, suppose that we are given the following correspondence
( B, A, C ) ↔ ( D, E , F ) between the vertices of △ ABC and △ DEF . For clarity, we can display
the correspondence vertically. There is a correspondence between the sides of triangle △ ABC
and the sides of △ DEF , and a correspondence between the angles of triangle △ ABC and the
angles of △ DEF , since those items are defined only in terms of the vertices.
B↔D
the given correspondence between
A↔ E
vertices of △ ABC and vertices of △ DEF
C↔F
BA ↔ DE
AC ↔ EF
the automatic correspondence between
CB ↔ FD
∢BAC ↔ ∢DEF parts of △ ABC and parts of △ DEF
∢ACB ↔ ∢EFD
∢CBA ↔ ∢FDE
Based on the ideas of this discussion, we make the following definition.
Definition 50 Corresponding parts of two triangles
• Words: Corresponding parts of triangles △ ABC and △ DEF .
• Usage: A correspondence between the vertices of △ ABC and △ DEF has been given.
• Meaning: As discussed above, there is an automatic correspondence between the sides of
triangles △ ABC and the sides of △ DEF , and also between the angle of triangles △ ABC and
the angles of △ DEF . Suppose the correspondence between vertices were
( B, A, C ) ↔ ( D, E , F ) . Corresponding parts would be pairs such as the pair of sides,
AC ↔ EF , or the pair of angles, ∢ACB ↔ ∢EFD .
5.2.2. Triangle congruence
In Neutral Geometry, line segment congruence is an undefined binary relation on the set of line
segments, and angle congruence is an undefined binary relation on the set of angles. But triangle
congruence is a defined binary relation on the set of triangles. Here is the definition.
Definition 51 Triangle congruence
• Symbol: △ ABC ≅△ DEF
• Words: “ △ ABC is congruent to △ DEF .”
• Meaning: “There is a correspondence between the vertices of the two triangles such that
corresponding parts of the triangles are congruent.”
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•
•
Remark: Triangle congruence is a defined binary relation on the set of triangles.
Additional terminology: If a correspondence between vertices of two triangles has the
property that corresponding parts are congruent, then the correspondence is called a
congruence.
Remark: Many students remember the sentence “Corresponding parts of congruent triangles are
congruent” from their high school geometry course. The acronym is, of course, “CPCTC”. We
see now that “CPCTC” is really a summary of the definition of triangle congruence. That is, to
say that two triangles are congruent is the same as saying that corresponding parts of those two
triangles are congruent. This is worth restating: CPCTC is not an axiom and it is not a theorem; it
is merely shorthand for the definition of triangle congruence.
It is important to discuss notation at this point, because there is an abuse of notation common to
many geometry books. In most books, the symbol △ ABC ≅△ DEF is used to mean not only that
triangle △ ABC is congruent to △ DEF , but also that the correspondence between the vertices is
given by ( A, B, C ) ↔ ( D, E , F ) . That is, in most books, the order of the vertices in the
correspondence is assumed to be the same as the order of the vertices in the symbol
△ ABC ≅△ DEF . This can cause problems. To see why, notice
C
that given any three non-collinear points, A, B, and C, the
A
B
symbols △ ABC and △ ACB represent the same triangle. That is
because a triangle is defined to be the union of three line segments, and the three line segments
that make up △ ABC are exactly the same as the three line segments that make up △ ACB . Now
consider the correspondence ( A, B, C ) ↔ ( A, B, C ) of vertices of triangles △ ABC and △ ACB .
Below is a list of the resulting correspondence of parts.
A↔ A
the given correspondence between
B↔B
vertices of △ ABC and vertices of △ ACB .
C ↔C
AB ↔ AB
BC ↔ BC
the automatic correspondence between
CA ↔ CA
∢ABC ↔ ∢ABC parts of △ ABC and parts of △ ACB .
∢BCA ↔ ∢BCA
∢CAB ↔ ∢CAB
Clearly, each of the pairs of corresponding parts are congruent to each other. Therefore, we
would say that the correspondence ( A, B, C ) ↔ ( A, B, C ) is a congruence, and triangle △ ABC is
congruent to △ ACB .
Now consider the correspondence ( A, B, C ) ↔ ( A, C , B ) of vertices of triangles △ ABC and
△ ACB . Below is a list of the resulting correspondence of parts.
A↔ A
the given correspondence between
B↔C
vertices of △ ABC and vertices of △ ACB .
C↔B
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AB ↔ AC
BC ↔ CB
the automatic correspondence between
CA ↔ BA
∢ABC ↔ ∢ACB parts of △ ABC and parts of △ ACB .
∢BCA ↔ ∢CBA
∢CAB ↔ ∢BAC
Clearly, the pairs of corresponding parts are not congruent to each other. So the correspondence
( A, B, C ) ↔ ( A, C , B ) is not a congruence.
So we see that triangle △ ABC is congruent to △ ACB , but that the correspondence that should be
used is ( A, B, C ) ↔ ( A, B, C ) , not ( A, B, C ) ↔ ( A, C , B ) .
In most books, the symbol △ ABC ≅△ ACB is used to mean “triangle △ ABC is congruent to
△ ACB , and the correspondence used is ( A, B, C ) ↔ ( A, C , B ) .” This is bad, because although
the triangles △ ABC and △ ACB are congruent, the correspondence that should be used is
( A, B, C ) ↔ ( A, B, C ) , not ( A, B, C ) ↔ ( A, C , B ) . It would be useful to have a symbol that not
only indicates congruence but also tells which correspondence was used. But there is not such a
symbol, and so we have to make do with the symbols available. When the particular choice of
correspondence is an issue, we should write things like “ △ ABC ≅△ ACB using the
correspondence ( A, B, C ) ↔ ( A, B, C ) ”, instead of merely writing “ △ ABC ≅△ ACB ”.
5.2.3. Introducing CA6, the Side Angle Side Congruence Axiom
As mentioned earlier, we have a notion of congruence for drawn line segments and drawn
angles, having to do with sliding one drawn segment or angle on top of another and seeing if
they “fit”. The primitive relations of line segment congruence and angle congruence in the axiom
system of Neutral Geometry are meant to behave in the same way as our drawings. Congruence
axioms CA1 through CA5 prescribe much of that behavior.
But we can also draw triangles, of course, and we have a notion of congruence for drawn
triangles, again having to do with sliding one drawn triangle on top of another and seeing if they
“fit”. And in our drawings, we know that if enough parts of one drawing “fit” on top of the
corresponding parts of another drawing, then all of the other parts will fit, as well.
This can be said more precisely. To determine whether or not two drawn triangles fit on top of
each other, one would officially have to verify that every pair of corresponding line segments fit
on top of each other and also that every pair of corresponding angles fit on top of each other.
That is total of six fits that must be checked. But we know that with drawings, one does not
really need to check all six fits. Here are four examples of methods that can be used to verify that
two drawn triangles are congruent by checking only three fits instead of six:
• If two sides and the included angle of the first triangle fit on top of the corresponding
parts of the second triangle, then all the remaining corresponding parts always fit, as well.
• If two angles and the included side of the first triangle fit on top of the corresponding
parts of the second triangle, then all the remaining corresponding parts always fit, as well.
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•
•
If two angles and some non-included side of the first triangle fit on top of the
corresponding parts of the second triangle, then all the remaining corresponding parts
always fit, as well.
If all three sides the first triangle fit on top of the corresponding parts of the second
triangle, then all the remaining corresponding parts always fit, as well.
We would like our undefined relations of line segment congruence and angle congruence and our
defined relation of triangle congruence to have this same sort of behavior. But if we want them to
have that behavior, we must specify it in the Neutral Geometry axioms. One might think that it
would be necessary to include four axioms, to guarantee that the four kinds of behavior that we
observe in drawn triangles will also be observed in abstract triangles. But the amazing thing is
that we don’t need to include four axioms. We can include just one axiom, about just one kind of
behavior that we want abstract triangles to have, and then we can prove theorems that show that
triangles will also have the other three kinds of desired behavior.
Here is the axiom that will be included, along with the three theorems that will be proven later.
CA6 (SAS axiom): If there is a correspondence between parts of two triangles such that two
sides and the included angle of one triangle are congruent to the corresponding parts of the other
triangle, then all remaining corresponding parts are congruent as well, so the triangles are
congruent.
The ASA Congruence Theorem of Neutral Geometry. In Neutral Geometry, if there is a
correspondence between parts of two triangles such that two angles and the included side of one
triangle are congruent to the corresponding parts of the other triangle, then all the remaining
corresponding parts are congruent as well, so the triangles are congruent.
The AAS Congruence Theorem of Neutral Geometry. In Neutral Geometry, if there is a
correspondence between parts of two triangles such that two angles and a non-included side of
one triangle are congruent to the corresponding parts of the other triangle, then all the remaining
corresponding parts are congruent as well, so the triangles are congruent.
The SSS Congruence Theorem of Neutral Geometry. In Neutral Geometry, if there is a
correspondence between parts of two triangles such that the three sides of one triangle are
congruent to the corresponding parts of the other triangle, then all the remaining corresponding
parts are congruent as well, so the triangles are congruent.
Two of these three theorems, the ASA and SSS congruence theorems, will be proven in this
chapter. The third, the AAS congruence theorem, will be proven in the next chapter.
5.3. Two Theorems about Triangles
Before introducing any theorems of Neutral Geometry, it is important to note that we already
have a bunch of old theorems that will work as Neutral Geometry theorems. That is, any theorem
of Incidence Geometry or of Incidence & Betweenness Geometry will also be a theorem of
Neutral Geometry, because the proof will still work. For that reason, the theorem numbers in this
chapter will continue the numbering from the previous chapter.
Page 63 of 106
Theorem 25 The Triangle Construction Theorem
Given: Neutral Geometry, △ ABC , DE ≅ AB , and a point G not on DE .
Claim: There exists a unique point F in half plane HG such that △ ABC ≅△ DEF .
Proof
1. There is a unique ray DK such that K lies in half plane HG and ∢KDE ≅ ∢CAB . (by
CA5)
2. There is a unique point F on DK such that DF ≅ AC .(by CA2)
3. DK = DF . (by Theorem 17)
4. ∢FDE ≅ ∢CAB . (by 3, 1, and CA4)
5. △ ABC ≅△ DEF . (by given statement DE ≅ AB , step 4, step 2, and CA6)
End of proof
The next theorem is called the Isosceles Triangle Theorem. We should officially define the term
isosceles before stating the theorem. Notice that in the theorem, the isosceles triangle is part of
the given information, not part of the conclusion. The proof of the theorem uses a
correspondence of vertices in a clever way.
Definition 52 isosceles triangle
• Words: isosceles triangle
• Meaning: two sides of the triangle are congruent to each other (at least two)
Theorem 26 (The Isosceles Triangle Theorem)
Given: Neutral Geometry, △ ABC , AB ≅ AC (Given an isosceles triangle.)
Claim: ∢B ≅ ∢C (The two angles that are opposite the two congruent sides are congruent.)
Proof
1. Define a correspondence ( A, B, C ) ↔ ( A, C , B ) between the vertices of △ ABC and
△ ACB .
2. AB ≅ AC . (given)
3. AC ≅ AB . (by 2 and CA1)
4. ∢BAC ≅ ∢CAB . (by CA4)
5. △ ABC ≅△ ACB . (by steps 1 - 4 and CA6)
6. ∢B ≅ ∢C . (by 5 and definition of triangle congruence.)
End of proof
5.4. Line Segment Subtraction and the Ordering of Segments
The following theorem is a statement very similar to the segment addition axiom, CA3. The
proof is an indirect proof, using the method of contradiction. You will supply the justifications
for each step in an exercise. In the indicated step, make a drawing.
Theorem 27 (segment subtraction)
Given: Neutral Geometry; A*B*C; D*E*F; AB ≅ DE and AC ≅ DF .
Claim: BC ≅ EF .
Proof
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1. Assume that BC ≅/ EF . (justify)
2. There is a point G on EF such that BC ≅ EG . (justify) (drawing)
3. G ≠ F . (justify)
4. AC ≅ DG . (justify)
5. DG ≅ DF . (justify)
6. G = F . (justify)
7. G ≠ F and G = F . (justify)
8. BC ≅ EF (justify)
End of proof
The following theorem proves a claim that would be obvious for drawn line segments, but which
is a little tricky to prove for abstract segments.
Theorem 28 In Neutral Geometry, if AC ≅ DF and A*B*C, then there exists a unique point E
such that D*E*F and AB ≅ DE .
Proof
1. There is a unique point E on DF such that DE ≅ AB . (by CA2)
2. Assume that D*E*F is false.
3. Either E = F or D*F*E. (by definition of DF )
Case 1: E = F
4. If E = F, then observe that B and C are two distinct points on AC and yet
AC ≅ DF ≅ AB . (by given information and step 1) This contradicts CA2.
Case 2: D*F*E
5. If D*F*E then there is a point G on the ray opposite to ray CA such that FE ≅ CG . (by
CA2)
6. AG ≅ DE . (by CA3)
7. Observe that B and G are two distinct points on AC and yet AG ≅ DE ≅ AB . (by steps
1, 5, and 6.) This contradicts CA2.
Wrap-up
8. Either case leads to a contradiction. (steps 4 and 7)
9. D*E*F is true. (by steps 2 – 8 and the rule of contradiction.)
End of proof
Notice that numbers have played very little role in Neutral Geometry. In particular, we have not
introduced the notion of the length of a line segment, and we have not said that congruent
segments are ones that have the same length. Indeed, we have been very clear about the fact that
line segment congruence is an undefined relation. And we have shown that the axioms of
betweenness can be used to prove many things about points and lines that we might have thought
could only be proved by referring to length. In this section, we will see that it is possible to
compare two line segments without any notion of length. We will do this by introducing a new
relation on the set of line segments.
Definition 53 the order relation on the set of line segments
• Symbol: AB < CD
• Spoken: “Segment AB is less than segment CD.”
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•
•
Meaning: There exists a point E between C and D such that AB ≅ CE .
Remark: The order relation is a binary relation on the set of line segments.
The following theorem says that the order relation for line segments has behavior that is entirely
analogous to the behavior of line segment length, even the order relation does not use the notion
of length. You will prove this theorem in the exercises.
Theorem 29 Facts about the order relation on the set of line segments
Given: Neutral Geometry; line segments AB, CD, and EF.
Claim
(a) (trichotomy): Exactly one of the following is true: AB < CD, AB ≅ CD , or CD < AB.
(b): If AB < CD and CD ≅ EF , then AB < EF.
(c): (transitivity): If AB < CD and CD < EF, then AB < EF.
5.5. Right Angles
In the previous section, we saw that even though we have no notion of line segment length, we
are able to have an order relation on the set of line segments. In this section, we will do
something comparable with angles. Remember that in our axiom system, we do not have a
notion of measure of angles. Even so, we will see that it is possible to define right angles and to
define an order relation on the set of angles.
Recall that Supplementary Angles were defined in Definition 42, in Section 4.7. In the exercises,
you will justify the steps in the proof of the following theorem. In the indicated step, make a
drawing.
Theorem 30 In Neutral Geometry, supplements of congruent angles are congruent. That is, if
∢ABC and ∢CBD are supplementary, and ∢EFG and ∢GFH are supplementary, and
∢ABC ≅ ∢EFG , then ∢CBD ≅ ∢GFH .
Proof
1. We can assume that points E, G, and H have the property that AB ≅ EF , CB ≅ GF , and
DB ≅ HF . Otherwise, we can find three points on those rays that do have this property,
and rename those three points as E, G, and H. (justify) (draw a picture to illustrate)
2. △ ABC ≅△ EFG . (justify)
3. AC ≅ EG , and ∢A ≅ ∢E . (justify)
4. AD ≅ EH . (justify)
5. △CAD ≅△GEH . (justify)
6. CD ≅ GH and ∢D ≅ ∢H . (justify)
7. △CBD ≅△GFH . (justify)
8. ∢CBD ≅ ∢GFH (justify)
End of proof
Do any of you know why vertical angles are called vertical angles? Maybe we should have prize
for the best back story. Regardless of the reason for the name, the definition should be familiar,
as is the theorem that follows. The proof of the theorem is left to the exercises.
Definition 54 vertical angles
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•
•
words: vertical angles
Meaning: Two angles that share a vertex and whose sides are opposite rays. That is, two
angles that can be labeled ∢ABC and ∢DBE where A*B*D and C*B*E.
Theorem 31
In Neutral Geometry, vertical angles are congruent.
As mentioned earlier, it is possible to define right angles without any reference to angle measure.
This is worth re-stating more explicitly: Right angles are NOT defined to be angles whose
measure is 90
! We do not even have a notion of angle measure.
Remember that the term supplementary angle was introduced in Definition 42, in Section 4.7. It
is worth noting that every angle has a supplementary angle. You will prove this in the exercises.
The following definition of right angles refers to supplementary angles.
Definition 55 right angle
• words: right angle
• Meaning: An angle that is congruent to its supplementary angle
There are a bunch of obvious sounding statements about right angles that we will have to prove.
Some are hard to prove; some are easy. The following one is easy to prove. You will prove it in
the exercises.
Theorem 32
In Neutral Geometry, any angle congruent to a right angle is also a right angle.
Now that we have definition of right angles, we can define perpendicular lines. But before doing
that, we should describe how intersecting lines create angles. Suppose that L and M are distinct
lines that intersect at point P. By the Incidence & Betweenness axioms, there are points A and B
on line L such that A*P*B, and there are points C and D on line M such that C*P*D. Notice that
∢APC and ∢BPD are vertical angles, and that ∢APD and ∢BPC are vertical angles. So
when two distinct lines intersect, two pairs of vertical angles are created.
Definition 56 perpendicular lines
• symbol: L ⊥ M
• spoken: L and M are perpendicular
• Meaning: L and M are distinct, non-parallel lines that create a pair of vertical angles that are
right angles.
The following theorems are extremely important in Neutral Geometry. You will justify the proof
steps in the exercises. Some of the justifications are simple, amounting to a citation of a single
axiom, theorem or definition. But some of the justifications will be more complicated. In the
indicated steps, make drawings.
Theorem 33 In Neutral Geometry, for every line L and every point P not on L there exists a
line through P perpendicular to L.
Proof
1. There exist distinct points A and B on L. (justify)
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2. On the opposite side of L from point P, there exits a ray AX such that ∢XAB ≅ ∢PAB .
(justify)
3. There exists a point Q on AX such that AQ ≅ AP . (justify)
4. Segment PQ intersects line AB at a point that we can call R. (justify)
Case 1: R = A.
5. Suppose R=A. (draw picture for case 1)
6. ∢PAB is a right angle. (justify)
7. PQ ⊥ AB . (justify)
Case 2: R = B.
8. Suppose R=B. (draw picture for case 2)
9. △PAB ≅△QAB (justify)
10. ∢PBA ≅ ∢QBA (justify)
11. ∢PBA is a right angle. (justify)
12. PQ ⊥ L . (justify)
Case 3: A*R*B
13. Suppose A*R*B. (draw picture for case 3)
14. △PAR ≅△QAR . (justify)
15. ∢PRA ≅ ∢QRA . (justify)
16. ∢PRA is a right angle. (justify)
17. PQ ⊥ L . (justify)
Case 4: R*A*B.
18. Suppose R*A*B. (draw picture for case 4)
19. ∢PAR ≅ ∢QAR . (justify)
20. △PAR ≅△QAR . (justify)
21. ∢PRA ≅ ∢QRA . (justify)
22. ∢PRA is a right angle. (justify)
23. PQ ⊥ L . (justify)
Case 5: A*B*R.
24. Suppose A*B*R. (draw picture for case 5)
25. △PAR ≅△QAR . (justify)
26. ∢PRA ≅ ∢QRA . (justify)
27. ∢PRB is a right angle. (justify)
28. PQ ⊥ L . (justify)
Wrap up
29. In every case, we see that PQ ⊥ AB
End of proof
Theorem 34 In Neutral Geometry, for every line L and every point P on L there exists a unique
line through P perpendicular to L.
Proof
1. There exists a point Q not on L. (justify)
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2. There exists a line M that passes through Q and is perpendicular to L. (justify)
3. Lines L and M create a right angle. (justify) (draw a picture)
4. There exists a ray PR such that R lies on the same side of L as point Q and ray PR
forms a right angle with line L. (justify)
5. Line PR passes through P and is perpendicular to L. (justify)
6. Line PR is the only such line. (justify)
End of proof
5.6. Angle Addition and Subtraction, and Ordering of Angles
Recall that the notion of line segment addition is specified by an axiom (CA3). But the notions of
segment subtraction and ordering of segments were introduced in theorems. In this section, we
will see that notions of angle addition, angle subtraction and ordering of angles can all be
introduced in theorems; no axioms are necessary. In our first theorem, we will introduce angle
addition. The proof of this theorem is hard. In the exercises, you will be asked to supply pictures.
Theorem 35 (angle addition)
Given: Neutral Geometry; angle ∢ABC with point D in the interior; angle ∢EFG with
point H in the interior; ∢DBA ≅ ∢HFE ; ∢DBC ≅ ∢HFG
Claim: ∢ABC ≅ ∢EFG
Proof
1. By Theorem 21 (the crossbar theorem), we know that ray BD must intersect line
segment AC. Rename the intersection point as point D, so that we can say that A*D*C.
2. By CA2, we can assume that points E, G, and H are chosen so that BA ≅ FE , BC ≅ FG ,
and BD ≅ FH . (draw a picture)
The next eight steps are obnoxiously tricky. We must prove that E*H*G.
3. △ ABD ≅△ EFH and △ DBC ≅△HFG . (by CA6, the SAS axiom)
4. ∢ADB ≅ ∢EHF . (by 3 and the definition of triangle congruence)
5. ∢CDB ≅ ∢GHF . (by 3 and the definition of triangle congruence)
6. ∢ADB is supplementary to ∢CDB . (by 1 and the definition of supplementary angles.)
7. There is a point J such that E*H*J. (draw a picture)
8. ∢EHF is supplementary to ∢JHF . (by 7 and the definition of supplementary angles.)
9. ∢CDB ≅ ∢JHF . (by 4, 6, 8, and Theorem 30)
10. Rays HG and HJ must be the same ray. (by CA5, the angle construction axiom)
11. E*H*G. (by 7 and 10)
Show that another pair of triangles is congruent
12. AC ≅ EG . (by 3, the definition of triangle congruence, and CA3, the segment addition
axiom)
13. ∢BAC ≅ ∢FEG . (by 3, the definition of triangle congruence, 11, and Theorem 17)
14. △ ABC ≅△ EFG . (by 2, 13, 12, and CA6, the SAS axiom)
15. ∢ABC ≅ ∢EFG . (by 14 and the definition of triangle congruence.)
End of proof
Just as the principle of line segment subtraction followed from the principle of line segment
addition, the principle of angle subtraction follows from the principle of angle addition. You will
be asked to prove the following theorem in the exercises.
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Theorem 36 (Angle Subtraction).
Given: Neutral Geometry; angle ∢ABC with point D in the interior; angle ∢EFG with
point H in the interior; ∢DBA ≅ ∢HFE ; ∢ABC ≅ ∢EFG
Claim: ∢DBC ≅ ∢HFG
Notice that we have not introduced the notion of the measure of an angle and we have not said
that congruent angles are ones that have the same measure. Indeed, we have been very clear
about the fact that angle congruence is an undefined relation. In this section, we will see that it is
possible to compare two angles without any notion of angle measure. We will do this by
introducing a new relation on the set of angles.
(Missing Theorem: There should be a theorem here about rays that would be analogous to
Theorem 28. The theorem statement would be as follows:
Missing Theorem about Rays and Angles
Given: Neutral Geometry, ∢ABC ≅ ∢EFG , ray BD between rays BA and BC .
Claim: There is a unique ray FH between rays FE and FG such that ∢ABD ≅ ∢EFH .)
Definition 57 the order relation on the set of angles
• Symbol: ∢ABC < ∢DEF
• Spoken: “Angle ∢ABC is less than angle ∢DEF .”
• Meaning: There exists a ray EG between ED and EF such that ∢ABC ≅ ∢GEF .
• Remark: The order relation is a binary relation on the set of angles.
The following theorem says that the order relation for line segments has behavior that is entirely
analogous to the notion of angle measure, even the order relation does not use the notion of
measure. You will prove this theorem in the exercises.
Theorem 37 (facts about the order relation on the set of angles)
Given: Neutral Geometry; angles ∢P , ∢Q , ∢R .
Claim
(a) (trichotomy): Exactly one of the following is true: ∢P < ∢Q , ∢P ≅ ∢Q , or ∢Q < ∢P .
(b): If ∢P < ∢Q and ∢Q ≅ ∢R then ∢P < ∢R .
(c): (transitivity): If ∢P < ∢Q and ∢Q < ∢R then ∢P < ∢R .
Now that we have the order relation on the set of angles, we can prove one important remaining
theorem about right angles.
Theorem 38 (Euclid’s 4th postulate). All right angles are congruent to each other.
Proof
1. Suppose that ∢ABC and ∢DEF are right angles.
2. Exactly one of the following is true: ∢ABC < ∢DEF , ∢ABC ≅ ∢DEF , or
∢DEF < ∢ABC . (by Theorem 37a)
3. Assume that ∢ABC < ∢DEF .
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4. There exists a point G in the interior of ∢DEF such that ∢DEG ≅ ∢ABC . (by 3 and the
definition of the less than relation for angles)
5. There exists a point H such that A*B*H. (by BA2)
6. ∢ABC and ∢HBC are supplementary angles. (by definition of supplementary angles.)
7. ∢ABC ≅ ∢HBC . (because ∢ABC is a right angle)
8. There exists a point J such that D*E*J. (by BA2)
9. ∢DEG and ∢JEG are supplementary angles. (by definition of supplementary angles.)
10. ∢JEF ≅ ∢DEF . (because ∢DEF is a right angle)
11. ∢JEG ≅ ∢HBC . (by 4, 6, 9, and Theorem 30)
12. Point F is in the interior
of
∢
JEG
.
(Because
it
is
on
the
same
side
of
EJ as G and is on
the same side of of EG as J. Both of these statements require some thought.)
13. ∢JEF < ∢JEG . (by 12 and the definition of the less than relation for angles)
14. ∢JEF < ∢HBC . (by 13, 11, and Theorem 37)
15. ∢JEF < ∢ABC . (by 14, 7, and Theorem 37)
16. ∢JEF < ∢DEF . (by 15, 3, and Theorem 37)
17. Statement 16 contradicts statement 10. (Theorem 37 says that they cannot both be true)
18. We have reached a contradiction. Therefore, our assumption that ∢ABC < ∢DEF was
wrong. ∢ABC cannot be less than ∢DEF .
19. In a similar way, we could show that ∢DEF cannot be less than ∢ABC .
20. Therefore, ∢ABC ≅ ∢DEF . (by 2, 18, and 19)
End of proof
5.7. Three More Theorems about Triangles
In Section 5.2.3, it was mentioned that there were four important statements that we would like
to be true in Neutral Geometry: SAS congruence, ASA congruence, SSS congruence, and AAS
congruence. As mentioned in that section, the statement of SAS congruence is included in the list
of axioms, as CA6, while the other three statements will be proven as theorems. Here is the first
of those three theorems. You will justify the steps of this proof in the exercises.
Theorem 39 (ASA congruence). In Neutral Geometry, if there is a correspondence between
parts of two triangles such that two angles and the included side of one triangle are
congruent to the corresponding parts of the other triangle, then all the remaining
corresponding parts are congruent as well, so the triangles are congruent.
Proof
1. Suppose that △ ABC and △ DEF are triangles such that ∢A ≅ ∢D , ∢C ≅ ∢F , and
AC ≅ DF .
2. There is a unique point G on DE such that DG ≅ AB . (justify) (draw a picture)
3. △ ABC ≅△DGF . (justify)
4. ∢C ≅ ∢DFG . (justify)
5. Ray FE must be the same ray as ray FG . (justify)
6. Point E must be the same point as point G. (justify)
7. △ ABC ≅△ DEF . (justify)
End of proof
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The following theorem is the Converse of the Isosceles Triangle Theorem. It is worth noting that
it is the converse, not the original, that has the words “isosceles triangle” as part of the
conclusion. You will prove this theorem in the exercises. The proof is not hard.
Theorem 40 (Converse of the Isosceles Triangle Theorem)
Given: Neutral Geometry, △ ABC , ∢B ≅ ∢C (Given a triangle with two congruent angles.)
Claim: AB ≅ AC (The two sides that are opposite the two congruent angles are congruent. That
is, the triangle is isosceles.)
The final theorem of the Chapter is the SSS congruence theorem. Its proof is not easy, but it is a
good proof to end the chapter with because it uses many ideas from the chapter. You will prove
the theorem in the exercises.
Theorem 41 (SSS) In Neutral Geometry, if there is a correspondence between parts of two
triangles such that the three sides of one triangle are congruent to the corresponding parts
of the other triangle, then all the remaining corresponding parts are congruent as well, so
the triangles are congruent.
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5.8. Exercises
[1] Justify the steps in Theorem 27, the Segment Subtraction Theorem.
[2] The goal is to prove Theorem 29 about the order relation on the set of line segments.
(a) In order to prove Claim (a) of the theorem, suppose that AB and CD are not congruent.
Explain why there is a unique point G ≠ D on ray CD such that AB ≅ CG . Explain why there
are two cases to consider: C*G*D or C*D*G. In case C*G*D, show that AB < CD. In case
C*D*G, use Theorem 28 and some axioms to show that CD < AB. Draw pictures to illustrate.
(b) In order to prove Claim (b) of the theorem, use Theorem 28. Draw pictures to illustrate.
(c) In order to prove Claim (c) of the theorem, use some of the same ideas from parts (a) and (b),
and also use Theorem 12, from Section 4.5. Draw pictures to illustrate.
[3] Justify the steps in the proof of Theorem 30, which says that in Neutral Geometry,
supplements of congruent angles are congruent. Make a drawing where indicated.
[4] Prove Theorem 31, which says that in Neutral Geometry, vertical angles are congruent.
[5] The term supplementary angle was introduced in Definition 42, in Section 4.7. Prove that
every angle has a supplementary angle.
[6] Prove Theorem 32, which says that In Neutral Geometry, any angle that is congruent to a
right angle is also a right angle.
[7] Justify the steps in the proof of Theorem 33. Draw pictures where indicated.
[8] Justify the steps in the proof of Theorem 34. Draw pictures where indicated.
[9] Draw pictures to illustrate the proof of Theorem 35.
[10] Prove Theorem 36, the Angle Subtraction Theorem. (Hint: Use Theorem 35 and axiom
CA5, the Angle Construction Axiom. Try following the same structure as the proof of Theorem
27.) Draw pictures to illustrate your proof.
[11] Prove Theorem 37 about the order relation on the set of angles. (Hint: imitate exercise [2].)
[12] Justify the steps in the proof of Theorem 39, the ASA Congruence Theorem. Draw a picture
where indicated.
[13] Prove Theorem 40, the Converse of the Isosceles Triangle Theorem. Hint: use the ASA
congruence theorem. Draw a picture to illustrate your proof.
[14] Prove Theorem 41, the SSS Congruence Theorem. Hint: Suppose that we are given △ ABC
and △ DEF , with AB ≅ DE , BC ≅ EF , and CA ≅ FD . The goal is to prove that △ ABC ≅△ DEF
. Use Theorem 25, the Triangle Construction Theorem, to reduce to the case where A = D, C =
F, and the points B and E are on opposite sides of AC . Then consider the following three cases:
case 1: Points A and C are on opposite sides of BE . (Draw a picture to illustrate.)
case 2: Point A is between B and E. That is, B*A*E. (Draw a picture to illustrate.)
case 3: Points A and C are on the same sides of BE . (Draw a picture to illustrate.)
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6. Neutral Geometry II
In this chapter, we will continue our exploration of Neutral Geometry Theorems. The theorems
in this chapter are probably more familiar to you than the theorems in the last chapter.
6.1. The Alternate Interior Angle Theorem and Some Corollaries
The first theorem that we will study is the Alternate Interior Angle Theorem. In order to
understand the wording of the theorem, we need two definitions.
Definition 58 “transversal”
• Words: “Line T is transversal to lines L and M.”
• Meaning: “T intersects L and M in distinct points.”
Definition 59 “alternate interior angles”, “corresponding angles”
• Usage: Lines L, M, and transversal T are given.
• Labeled points: Let B be the intersection of lines T and L, and let E be the intersection of
lines T and M. (By definition of transversal, B and E are not the same point.) By the
betweenness axioms, there exist points A and C on line L such that A*B*C, points D and
F on line M such that D*E*F, and points G and H on line T such that G*B*E and B*E*H.
Without loss of generality, we may assume that points D and F are labeled such that it is
point D that is on the same side of line BE as point A.
H T
D
M
E
F
A
B
C
L
G
•
Meaning:
∢ABE
∢CBE
∢ABG
∢ABH
∢CBG
∢CBH
and ∢FEB is a pair of alternate interior angles.
and ∢DEB is a pair of alternate interior angles.
and ∢DEG is a pair of corresponding angles.
and ∢DEH is a pair of corresponding angles.
and ∢FEG is a pair of corresponding angles.
and ∢FEH is a pair of corresponding angles.
Theorem 42 The Alternate Interior Angle Theorem
Given: Neutral Geometry, lines L and M and a transversal T
Claim: If a pair of alternate interior angles is congruent, then lines L and M are parallel.
It should be pointed out that the name of the Alternate Interior Angle Theorem follows the usual
unspoken custom for theorem names. That is, the name of the theorem refers to something that is
in the hypotheses of the theorem, not the conclusion.
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The Alternate Interior Angle Theorem is a major theorem in two senses: its proof is a little
tricky, and it can be used to build fairly simple proofs of a bunch of other theorems. Those other
theorems could be considered corollaries of this theorem. You will be asked to justify the steps
of the proof of the Alternate Interior Angle Theorem in the exercises. The corollaries are listed
below; you will be asked to prove them in the exercises.
Theorem 43 The Corresponding Angle Theorem
Given: Neutral Geometry, lines L and M and a transversal T
Claim: If a pair of corresponding angles is congruent, then lines L and M are parallel.
Theorem 44
In Neutral Geometry, two distinct lines perpendicular to the same line are parallel.
Theorem 45 Uniqueness of the perpendicular from a point to a line
Given: Neutral Geometry, line L and a point P not on L
Claim: There is not more than one line that passes through P and is perpendicular to L.
(Remark: We know that there exists at least one perpendicular by Theorem 33.)
Theorem 46 Existence of parallel lines (The answer to THE BIG QUESTION)
Given: Neutral Geometry, line L and a point P not on L
Claim: There exists at least one line that passes through P and is parallel to L.
Although Theorem 46 is just a corollary of the Alternate Interior Angle Theorem, it is an
extremely important theorem. Notice that none of the Neutral Geometry axioms say anything
about THE BIG QUESTION. Even so, Theorem 46 proves that the answer is “at least one line”.
6.2. The Exterior Angle Theorem and Some Theorems Whose Proofs Use It
The next major theorem that we will study is the Exterior Angle Theorem. In order to understand
the wording of the theorem, we need one more definition.
Definition 60 “exterior angle”, “interior angle”, “remote interior angles”
• Words: An exterior angle of a triangle
• Meaning: An angle that is supplemental to one of the angles of the triangle
• Additional terminology: The angles of the triangle are also called “interior angles”. Given
an exterior angle, there will be an interior angle that is its supplement, and two other
interior angles that are not its supplement. Those other two interior angles are called
remote interior angles.
Theorem 47 The Exterior Angle Theorem
In Neutral Geometry, each of the remote interior angles is less than the exterior angle.
As with our previous major theorem, the Exterior Angle Theorem has a rather tricky proof, and it
allows us to prove a bunch of other theorems. In the remainder of this section, these additional
theorems will be introduced. These theorems have proofs that are rather tricky, so one would not
refer to these theorems as corollaries. You will be asked to justify the steps in proofs of the
Exterior Angle Theorem and some of the additional theorems in the exercises. You will only be
asked to create one proof.
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Theorem 48 The Angle Angle Side Congruence Theorem
In Neutral Geometry, if there is a correspondence between parts of two triangles such that two
angles and a non-included side of one triangle are congruent to the corresponding parts of the
other triangle, then all the remaining corresponding parts are congruent as well, so the triangles
are congruent.
Definition 61 “hypotenuse” and “legs” of a right triangle
• Meaning: The hypotenuse of a right triangle is the side that is opposite the right angle.
The other two sides are called legs.
Theorem 49 The Hypotenuse Leg Congruence Theorem
In Neutral Geometry, if the hypotenuse and one leg of one right triangle are congruent to the
hypotenuse and one leg of another right triangle, then all the remaining corresponding parts are
congruent as well, so the triangles are congruent.
Definition 62 “midpoint” of a line segment
• Words: C is a midpoint of segment AB
• Meaning: A*C*B and CA ≅ CB .
Theorem 50 Existence of Midpoint of a Line Segment
In Neutral Geometry, every segment has exactly one midpoint.
Definition 63 “bisector” of a line segment, “perpendicular bisector” of a line segment
• Words: Line L is a bisector of segment AB
• Meaning: L is distinct from line AB and passes through the midpoint of segment AB.
• Additional Terminology: If L is perpendicular to line AB and is also a bisector of
segment AB, then L is said to be a perpendicular bisector of segment AB.
Theorem 51
In Neutral Geometry, every line segment has exactly one perpendicular bisector.
Definition 64 “bisector” of an angle
• Words: a bisector of angle ∢ABC
• Meaning: a ray BD between rays BA and BC such that ∢DBA ≅ ∢DBC .
Theorem 52
In Neutral Geometry, every angle has exactly one bisector.
Theorem 53
In Neutral Geometry, in triangle △ ABC , if BC > AC then ∢A > ∢B .
Theorem 54
In Neutral Geometry, in triangle △ ABC , if ∢A > ∢B then BC > AC.
One can combine Theorem 26, Theorem 40, Theorem 53, and Theorem 54 into one theorem that
is easier to remember. We could call it the CACS and BABS Theorem. It is stated here without
proof. (It is really not a new theorem, but rather just a repackaging of previous theorems.)
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Theorem 55 The CACS and BABS Theorem
In triangles of Neutral Geometry, Congruent Angles are always opposite Congruent Sides, and
Bigger angles are always opposite Bigger Sides.
We end this section with the Hinge Theorem. The refers to our imagining two triangle legs of
fixed length forming an included angle that is not fixed but rather is a hinge that can be opened
varying amounts. The idea is that when the angle of the hinge is reduced, the ends of the legs
will be brought closer together. We are imagining drawings, where we have the notion of line
segment length and angle measure. In our axiomatic world of Neutral Geometry, the equivalent
is to consider two triangles that have two corresponding sides congruent and included angles not
congruent. In this world, remember that line segment congruence is undefined. The “less than”
relation for angles is defined, but its definition does not refer to angle measure. Rather, it refers
to earlier definitions that ultimately refer to betweenness and angle congruence, both undefined
concepts.
Theorem 56 The Hinge Theorem:
Given: Neutral Geometry, △ ABC , △ DEF , AB ≅ DE , AC ≅ DF
Claim: The following are equivalent
(1) ∢A < ∢D .
(2) BC < EF.
The proof of the Hinge theorem is the hardest proof of this section. You will justify the steps in
the exercises. (We will also do it in class.)
Page 77 of 106
6.3. Exercises
[1] Justify the steps in the following proof of Theorem 42. Make a drawing where indicated.
Proof:
1. Suppose that lines L and M are crossed by a transversal T; that T intersects L at point B
and intersects M at point E; that A and C are points on L such that A*B*C; that D and F
are points on M such that D*E*F; that points A and D are on the same side of line T; and
that ∢CBE ≅ DEB .
2. Assume that L and M intersect. (Justify) Let P be their point of intersection.
3. Point P lies on L and M. (justify)
4. Without loss of generality, we may assume that point P is on the same side of line T as
point C. (justify) (draw a picture)
Observe two pairs of supplementary angles and a consequence
5. The pair of angles ∢EBA and ∢EBC is supplementary. (justify)
6. The pair of angles ∢BED and ∢BEF is supplementary. (justify)
7. ∢EBA ≅ ∢BEF . (justify)
Build two triangles and show that they are congruent
8. There exists a point Q on ray ED such that EQ ≅ BP . (justify) (add to your picture)
9. Point Q lies on line M.
10. △QEB ≅△ PBE . (justify) (add to your picture)
11. ∢EBQ ≅ ∢BEP . (justify)
Find some more pairs of congruent angles and a consequence
12. ∢EBQ ≅ ∢BEF . (justify)
13. ∢EBQ ≅ ∢EBA . (justify)
14. Rays BQ and BA must be the same ray. (justify)
15. Point Q must lie on line L. (justify)
16. Point Q lies on lines L and M. (justify)
17. Lines L and M have two points in common. (justify)
Wrap-up
18. Step 17 is a contradiction (what does it contradict?) Therefore our assumption in step 2
was wrong. It cannot be that lines L and M intersect. Therefore, lines L and M must be
parallel.
End of Proof
[2] Prove Theorem 43. Hint: Use Theorem 42 in your proof.
[3] Prove Theorem 44. Hint: Use Theorem 42 or Theorem 43 in your proof.
[4] Prove Theorem 45. Hint: Use Theorem 44 in your proof.
[5] Prove Theorem 46. Hint: Show that there exists a line T that passes through P and is
perpendicular to L. (justify) Then show that there exists a line M that is perpendicular to line T at
point P. (justify) Then use Theorem 42, Theorem 43, or Theorem 44.
Page 78 of 106
[6] Justify the steps in the following proof of Theorem 47. Make a drawing where indicated.
Proof
1. Let △ ABC be given, and let D be a point such that A*B*D. Then ∢CBD is an exterior
angle, and angles ∢A and ∢C are its remote interior angles. (make a drawing) (Our
goal is to prove that ∢A < ∢CBD and ∢C < ∢CBD .)
Part 1: Prove that ∢C < ∢CBD
2. Either ∢C < ∢CBD or ∢C ≅ ∢CBD or ∢CBD < ∢C . (justify)
Case 1
3. Assume that ∢C ≅ ∢CBD .
4. Then lines AC and AD are parallel. (justify)
5. We have reached a contradiction. (what is the contradiction?) Therefore, our assumption
in step 2 was wrong. The statement ∢C ≅ ∢CBD cannot be true.
Case 2
6. Assume that ∢CBD < ∢C
7. There exists a ray CE between rays CA and CB such that ∢ECB ≅ ∢DBC . (justify)
8. Ray CE intersects segment AB at a point F such that A*F*B. (justify)
9. Line CE intersects line AD . (justify)
10. Line CE is parallel to line AD . (justify)
11. We have reached a contradiction. (what is the contradiction?) Therefore, our assumption
in step 6 was wrong. The statement ∢CBD < ∢C cannot be true.
Conclusion of Cases
12. ∢C < ∢CBD . (justification)
Part 2: Introduce a new point G in order to prove a new angle inequality.
13. There exists a point G such that C*B*G. (justify) (Add this new point to your drawing.)
Then ∢ABG is an exterior angle, and angles ∢C and ∢A are its remote interior angles.
14. A sequence of steps identical to steps 2 through 12, but with all the D’s replaced by G’s
and all the A’s & C’s interchanged, will result in the inequality ∢A < ∢ABG .
Part 3: Use the new angle inequality to prove the angle inequality that we really wanted to prove.
15. ∢ABG ≅ ∢CBD . (justify)
16. ∢A < ∢CBD . (justify)
End of Proof
[7] Justify the steps in the following proof of Theorem 48. Make a drawing where indicated.
Proof
1. Let △ ABC and △ DEF be triangles such that ∢A ≅ ∢D , ∢B ≅ ∢E , and AC ≅ DF .
2. Either AB ≅ DE or AB < DE or DE < AB. (justify)
Case 1
3. Assume that DE < AB.
4. There exists a point G between A and B such that AG ≅ DE . (justify) (draw a picture)
5. △ DEF ≅△ AGC . (justify)
6. ∢E ≅ ∢AGC . (justify)
Page 79 of 106
7. ∢B ≅ ∢AGC . (justify)
8. Observe that angle ∢AGC is an exterior angle for triangle △ BGC .
9. ∢B < ∢AGC . Justify
10. We have reached a contradiction. (what is the contradiction?) Therefore, our assumption
in step 3 was wrong. The inequality DE < AB cannot be true.
Case 2
11. Assume that AB < DE.
12. Steps similar to steps 3 through 10, involving a point H between D and E, would show
that another contradiction results. Therefore, the inequality AB < DE cannot be true.
Conclusion of Cases
13. AB ≅ DE . (justify)
14. △ ABC ≅△ DEF . (justify)
End of Proof
[8] Justify the steps in the following proof of Theorem 49. Draw a picture where indicated.
Proof
1. Let △ ABC and △ DEF be triangles such that angles ∢A and ∢D are right angles,
BC ≅ EF , and BA ≅ ED .
2. There exists a point G such that G*A*C and AG ≅ DF . (justify. This involves more than
one step) (draw a picture)
3. △ ABG ≅△ DEF . (Justify)
4. BG ≅ EF . (Justify)
5. BG ≅ BC . (Justify)
6. ∢G ≅ ∢C . (Justify)
7. △ ABG ≅△ ABC . (Justify)
8. △ DEF ≅△ ABC . (Justify)
End of Proof
[9] Justify the steps in the following proof of Theorem 50. Draw a picture where indicated.
Proof
1. Let segment AB be given.
2. There exists a point C that is not on line AB . (justify)
3. There exists a unique ray BX such that point X is on the opposite side of line AB from
point C and such that ∢XBA ≅ ∢CAB . (justify)
4. There exists a unique point D on ray BX such that BD ≅ AC . (Justify) (draw picture)
5. Line AB intersects segment CD. (Justify) We can call the intersection point E.
6. Point E is between A and B. (Justify. This will involve a few steps) (draw picture)
7. ∢AEC ≅ ∢BED . (Justify)
8. △ AEC ≅△ BED . (Justify)
9. AE ≅ BE . (Justify)
10. Point E is the midpoint of segment AB. (Justify)
End of Proof
[10] Prove Theorem 51.
Page 80 of 106
[11] Justify the steps in the following proof of Theorem 52. Draw a picture where indicated.
Proof
1. Let angle ∢ABC be given.
2. There exists a unique point D on ray BC such that BD ≅ BA . (Justify)
3. Segment AD has a midpoint, that we can call E. (Justify) (Draw a picture)
4. A*E*D. (Justify)
5. Point E is in the interior of angle ∢ABC . (Justify)
6. Ray BE is between rays BA and BC . (Justify)
7. △ EBA ≅△ EBD . (Justify)
8. ∢EBA ≅ ∢EBD . (Justify)
9. Ray BE is the bisector of angle ∢ABC . (Justify)
End of Proof
[12] Justify the steps in the following proof of Theorem 53. Draw a picture where indicated.
1. Suppose that triangle △ ABC is given, and that BC > AC. (draw a picture)
2. There exists a point D between B and C such that CD ≅ CA . (Justify) (add to your
picture)
3. ∢A > ∢DAC . (Justify) (add to your picture)
4. ∢DAC ≅ ∢ADC . (Justify)
5. ∢ADC > ∢B . (Justify)
6. ∢A > ∢B . (Justify. This will take two steps.)
End of proof
[13] Justify the steps in the following proof of Theorem 54.
Proof
1. Suppose that triangle △ ABC is given, and that ∢A > ∢B is true.
2. The statement ∢B > ∢A is not true. (Justify)
3. The statement AC > BC is not true. (Justify. This will involve the contrapositive of the
previous theorem.)
4. The statement ∢A ≅ ∢B is not true. (Justify)
5. The statement BC ≅ AC is not true. (Justify. This will involve the contrapositive of
another previous theorem.)
6. The statement BC > AC is true.(Justify)
End of Proof
[14] Justify the steps in the following proof of Theorem 56. Draw a picture where indicated.
Proof Part 1: Prove that (1) → ( 2 ) .
1. Suppose that ∢A < ∢D .
2. There exists a ray DG between rays DE and DF such that ∢EDG ≅ ∢A . (Justify)
3. There exists a point H on ray DG such that DH ≅ AC . (Justify)
4. △ DEH ≅△ ABC . (Justify)
5. EH ≅ BC . (Justify)
6. Ray DG intersects segment EF. (Justify) We may call the intersection point J.
7. Either H = J or D*J*H or D*H*J. (Justify)
Page 81 of 106
Case 1
8. Suppose that H = J. (Draw a picture)
9. EH < EF. (Justify)
10. BC < EF. (Justify)
Case 2
11. Suppose that D*J*H. (Draw a picture)
12. ∢EFH < ∢DFH . (Justify)
13. ∢DFH ≅ ∢DHF . (Justify)
14. ∢DHF < ∢EHF . (Justify)
15. ∢EFH < ∢EHF . (Justify)
16. EH < EF. (Justify)
17. BC < EF. (Justify)
Case 3
18. Suppose that D*H*J. (Draw a picture)
19. ∢HFE < ∢DHF . (Justify)
20. ∢DHF ≅ ∢DFH . (Justify)
21. ∢DFH < ∢JHF . (Justify)
22. ∢JHF < ∢EHF . (Justify)
23. ∢HFE < ∢EHF . (Justify)
24. EH < EF. (Justify)
25. BC < EF. (Justify)
Conclusion of Cases
26. In every case, we see that BC < EF.
Proof Part 2: Prove that ( 2 ) → (1) . It is easier to do an indirect proof by the method of
contraposition. That is, we will prove the contrapositive statement, ~ (1) →~ ( 2 ) .
27. Suppose that ∢A < ∢D is not true.
28. Either ∢A ≅ ∢D or ∢D < ∢A . (Justify)
Case 1
29. Suppose that ∢A ≅ ∢D .
30. △ ABC ≅△ DEF . (Justify)
31. BC ≅ EF . (Justify)
32. The statement BC < EF is not true. (Justify)
Case 2
33. Suppose that ∢D < ∢A .
34. Then a proof similar to the proof in part 1, but with the changes of variables A ↔ D ,
B ↔ E , and C ↔ F , would result in the statement EF < BC.
35. The statement BC < EF is not true. (Justify).
Conclusion of Cases
36. In either case, we see that the statement BC < EF is not true.
End of Proof
Page 82 of 106
7. Measure of Line Segments and Angles
7.1. Theorems Stating the Existence of Measurement Functions
So far, real numbers have played almost no role in our axiomatic geometry. We have used the
non-negative integers to count things, but have not used rational numbers and certainly not used
real numbers. This is in sharp contrast to the style of axiomatic geometry found in high school
textbooks. There, real numbers play a central role, even to the point of being mentioned
explicitly in the axiom system. The following two theorems show two ways that real numbers
can be introduced into our geometry: as line segment length and angle measure. We will not
prove these theorems in our class.
Theorem 57 Existence of a Length Function for Line Segments
Given: Neutral Geometry, line segment AB
Claim: There exists a unique function length ( ) : {line segments} → ℝ + with the following
properties.
1. The length function is onto. That is, for every positive real number x, there exists a line
segment CD such that length ( CD ) = x .
2. length ( AB ) = 1
3. length ( CD ) = length ( EF ) if and only if CD ≅ EF .
4. length ( CD ) < length ( EF ) if and only if CD < EF.
5. C*D*E if and only if C, D, E are collinear and length ( CE ) = length ( CD ) + length ( DE ) .
Theorem 58 Existence of a Measure Function for Angles
Given: Neutral Geometry
Claim: There exists a unique function degrees ( ) : {angles} → ( 0,180 ) with the following
properties.
1. The degrees function is onto. That is, for every 0 < x < 180 , there exists an angle ∢A
such that degrees ( ∢A ) = x .
2. degrees ( ∢A) = 90 if and only if ∢A is a right angle.
3. degrees ( ∢A ) = degrees ( ∢B ) if and only if ∢A ≅ ∢B .
4. degrees ( ∢A ) < degrees ( ∢A) if and only if ∢A < ∢B .
5. If ray AC is between rays AB and AD , then
degrees ( ∢BAD ) = degrees ( ∢BAC ) + degrees ( ∢CAD ) .
6. If angle ∢A is supplementary to angle ∢B , then degrees ( ∢A ) + degrees ( ∢B ) = 180
Notice that both the length function and the degrees function make use of a reference input. In
the case of the length function, a choice of a line segment AB whose length will be declared to
be 1 is central to the definition. In the case of the degrees function, it is declared that the measure
of a right angle will be 90.
Page 83 of 106
The length definition obviously allows an amount of choice. That is, if one uses a different
reference segement AB (more precisely, if one uses a different reference segment that is not
congruent to the first), then the resulting distance function will be different, as well.
There is choice in the definition of angle measure, as well, but not in the choice of the reference
input. The reference angle is always a right angle. But there is choice in the positive real number
that is assigned to be the measure of the right angle. In the definition of the degrees function, the
number 90 is used. Correspondingly, the degree measures of supplementary angles add up to
180, and the codomain of the degrees function is the set ( 0,180 ) . But we could have used some
number other than 90 for the measure of a right angle. When a number other than 90 is used, the
name of the angle measurement function is changed, as well. A common choice of measure for
right angles is the number π . When that number is used, the name of the angle measurement
2
function is changed to radians, the measures of supplementary angles add up to π , and the
codomain of the degrees function is the set ( 0, π ) . A less common choice of measure for right
angles is the number 100. When that number is used, the name of the angle measurement
function is changed to gradians, the measures of supplementary angles add up to 200, and the
codomain of the degrees function is the set ( 0, 200 ) . (You have probably noticed that most
calculators allow you to choose between degrees, radians, or gradians.)
7.2. Two length functions for Neutral Geometry
In this section, we will explore two length functions for Neutral Geometry. One will be familiar
to you: it comes from the standard distance formula that you have been using since high school.
But the other will be new.
Definition 65 the length function for line segments in straight-line drawings (Euclidean)
Meaning: the function lengthE : {Euclidean line segments} → ℝ + defined by the following
formula
lengthE ( AB ) =
( x1 − x2 ) + ( y1 − y2 ) ,
where AB is a Euclidean line segment with endpoints A = ( x1 , y1 )
2
2
and B = ( x2 , y2 ) . (This
implies that x1, y1, x2,and y2 are real numbers.)
Definition 66 the length function for line segments in the Hyperbolic plane
Meaning: the function lengthH : { Hyperbolic line segments} → ℝ + defined by the
following formula
 lengthE ( AR )


lengthE ( AS ) 
,
lengthH ( AB ) = ln 
 lengthE ( BR )



length
BS
(
)
E


where AB is a Hyperbolic line segment with endpoints that are the Hpoints A and B, and
R and S are the “missing endpoints” of the line AB .
Page 84 of 106
Two typical configurations of points A, B, R, and S are shown in the figures below.
R
R
S
A
A B
B
S
Notice that the Hyperbolic line segment AB can be a curvy thing that does does not look the
same as the Euclidean line segment AB. Also note that when computing the Hyperbolic length of
the Hyperbolic segment AB, the Euclidean lengths of four Euclidean line segments are used.
S
Hyperbolic line segment AB
is the curvy arc. It does not
look the same as Euclidean
line segment AB.
Euclidean line segments
AR, AS, BR, and BS are the
straight-looking things.
A
B
R
You might be a little disturbed by the fact that in one of the figures above, the ordering of points
on the line is R-A-B-S, while in the other, the ordering is S-A-B-R. Shouldn’t it matter what
names we give to the missing endpoints, at least when using the function dH? (Patrick asked
about this in class.) The answer is no, it doesn’t matter. To see why, consider the effect
interchanging the names of the missing endpoints R & S. That would mean that we should
interchange the symbols R & S in the formula for the distance between A & B. But rules of
logarithms can be used to show that if u, v, x, and y are positive numbers, then
u 
v 
v


ln
= ln  u  .
 x 
 y 
 y
 x
(You will prove this in the exercises.) Therefore, interchanging the symbols C & D in the
formula would have no effect on the outcome.
(
For example of the use of the distance function lengthH, let A = 0, − 1
)
( )
and let B = , 3 .
4
4
Both of these points qualify as Hpoints, and they lie on the Hline consisting of the portion of the
y-axis that lies inside the unit circle. The missing endpoints of this line are R = ( 0,1) and
S = ( 0, −1) . To compute the hyperbolic distance between A and B, we first compute the
Page 85 of 106
Euclidean distances needed in the formula for lengthH. The results are lengthE ( AR ) = 5 ,
4
3
7
1
, lengthE ( BR ) =
, and lengthE ( BS ) =
. Plugging these into the formula
lengthE ( AS ) =
4
4
4
for lengthH, we obtain
 lengthE ( AR )

 54 


lengthE ( AS )
 3 
 5⋅7 
 35 
 = ln  1 4  = ln 
lengthH ( AB ) = ln 
= ln   ≈ 2.46

 1⋅ 3 
 3 
 lengthE ( BR )

 47 


lengthE ( BS ) 
 4 

Notice that the hyperbolic length of line segment AB is not the same as the Euclidean length,
which is just lengthE ( AB ) = 1 .
7.3. An example of a curvy-looking Hline
When Hpoints A and B happen to lie on a diameter of the unit circle, the Hline AB is easy to
determine because it is straight: it will be described by the standard equation for a straight line
containing A and B. Because this straight line is a diameter of the unit circle, we know that it
must go through the origin. So the form of the equation will be y = mx for a non-vertical
diameter or x = 0 for the one diameter that is vertical. Determination of the missing endpoints R
and S is not terribly difficult: one would simply find the intersection of the straight non-vertical
line y = mx or the straight vertical line x = 0 with the unit circle. That is, one would simply solve
the pair of equations
 y = mx
 2
2
x + y = 1
in the case of a non-vertical diameter. In the case of a vertical diameter, of course, the missing
endpoints are the points ( 0,1) and ( 0, −1) .
In class on Tuesday, May 22, we discussed the fact that in general, it can be very difficult to find
the equation for a curvy-looking Hline. Given Hpoints A and B, one must find the equation for
the one circle that contains A and B and is orthogonal to the unit circle. This involves solving
some messy equations. Then, one must find the coordinates of the two points of intersection of
the orthogonal circle and the unit circle. This also involves solving messy equations. It is very
useful, therefore, to have at least one example of a curvy-looking Hline whose equation and
missing endpoints are not so hard to determine. The example that we discussed in class involved
an orthogonal circle that was simply the unit circle moved to the right. Recall that the unit circle
contains the following five famous points:

1 
3 1
3 1  1
 1 1  
−
,
,−
,
−
,
,
−
1,
0
,
−
, −  ,  −
(
)




  2 2


2 
2 2 
2
2


 2
A circle of radius 1 centered at the point
(
)
2, 0 will have five corresponding famous points
obtained by adding
2 to the x-coordinates of the points above.

1  
3
1
3
1
 1
−
+
2,
,
−
+
2,
,
,
−
+
2,
−
−
1
+
2,
0



,


 2
2 
2 
2
2   2


(
)
1 
 1
+ 2, −
−

2
2

Page 86 of 106
The coordinates of the first and last of these five famous points can be simplified:

1 
3
1
3
1  1
 1 1  
,
,−
,
−
+
2,
,
,
−
+
2,
−
−
1
+
2,
0



 , 

  2



2
2  2
2
 2 2 
 2
This new circle intersects the unit circle at the first and last of these five famous points.
Furthermore, the two circles are orthogonal. (Draw a picture of the two circles to convince
yourself of this fact.) So the set of points on the second circle that lie in the interior of the unit
circle qualifies as an Hline. That is, the set of points ( x, y ) that satisfy the following two
equations:
2
The point ( x, y ) lies on the circle of radius 1 centered at 2, 0 .
•
x − 2 + y2 = 1
(
(
•
)
x2 + y 2 < 1
)
(
)
The point ( x, y ) lies in the interior of the unit circle.
Three of the five famous points above qualify as Hpoints. They are are


3
1
3
1
A =  −
+ 2,  , B = −1 + 2, 0 , C =  −
+ 2, −  .
2
2
 2
 2
So those three points are collinear, all lying on the same Hline.
(
)
The remaining two points are the two points of intersection of the two circles. That means that
they are the missing endpoints of the Hline. The two points are
1 
 1 1 
 1
R=
,
,−
, S =
.
2
 2 2
 2
Three famous points on this Hline are


3
1
3
1
A =  −
+ 2,  , B = −1 + 2, 0 , C =  −
+ 2, −  .
2
2
 2
 2
In the exercises, you will compute the hyperbolic lengths of Hyperbolic segments AB and AC.
(
)
7.4. Theorems about segment lengths and angle measures
In this section, we will discuss five theorems of Neutral Geometry that use the concept of line
segment length or angle measure. Three of the theorems have fairly easy proofs, but two have
hard proofs.
The first of these theorems might seem a little strange to you. It says that in Neutral Geometry,
the sum of the measures of any two angles of a triangle is less than 180. This might seem
obvious: we all know that the sum of the measures of all three angles of a triangle is exactly 180,
so therefore the sum of any two of them must be less than 180, right? But we have not proven
that all triangle angle sums equal 180. What’s more, the statement that all triangle angle sums
equal 180 cannot be proven in Neutral Geometry, because it is not true! So, since we don’t yet
know anything about triangle angle sums, and since we will eventually know less than we
thought we knew, the following theorem is actually interesting. The proof of this theorem is not
hard; you will justify it in the exercises.
Page 87 of 106
Theorem 59 In Neutral Geometry, the sum of the measures of any two angles of a triangle is
less than 180.
Proof
1. Let two angles of a triangle be given. Without loss of generality, we can assume that the
triangle vertices are labeled such that the two given angles are ∢A and ∢B of △ ABC .
2. There exists a point D such that A*B*D. (Justify) (Draw a picture)
3. Angle ∢DBC is an exterior angle of △ ABC , and ∢A is one of its remote interior angles.
4. ∢A < ∢DBC . (Justify)
5. degrees ( ∢A) < degrees ( ∢DBC ) . (Justify)
6. degrees ( ∢A ) + degrees ( ∢B ) < degrees ( ∢DBC ) + degrees ( ∢B ) . (by 5 and algebra)
7. degrees ( ∢DBC ) + degrees ( ∢B ) = 180 . (Justify)
8. degrees ( ∢A ) + degrees ( ∢B ) < 180 . (by 6, 7, and algebra)
End of proof
The Triangle Inequality is well known, and also has a fairly simple proof that you will justify in
the exercises.
Theorem 60 (The Triangle Inequality) In Neutral Geometry, the length of any side of a triangle
is less than the sum of the lengths of the two other sides.
Proof
1. Let a side of a triangle be given. Without loss of generality, we can assume that the
triangle vertices are labeled such that the given side is side AC of △ ABC .
2. There exists a point D such that A*B*D and BC ≅ BD . (Justify) (Draw a picture)
3. length ( BC ) = length ( BD ) . (Justify)
4. length ( AB ) + length ( BC ) = length ( AB ) + length ( BD ) . (by 3 and algebra)
5. length ( AB ) + length ( BD ) = length ( AD ) . (Justify)
6.
7.
8.
9.
10.
∢ADC ≅ ∢DCB . (Justify)
∢DCB < ∢DCA . (Justify)
∢ADC < ∢DCA . (Justify)
AC < AD . (Justify)
length ( AC ) < length ( AD ) . (Justify)
11. length ( AC ) < length ( AB ) + length ( BC ) . (by 10, 5, 4, and algebra)
End of proof
In these notes, any fact that we prove is called a theorem.. In practice, there is a sort of a pecking
order for proven facts, and there are names that reflect the pecking order. Here is a rough
description:
• A Theorem is a very important fact.
• A Proposition is a less important fact.
• A Corollary is a fact that may or may not be important, but whose proof is really easy if
some prior Theorem or Proposition has been proven.
Page 88 of 106
•
A Lemma is a fact that is not particularly interesting by itself, but is needed in the proof
of a Theorem.
I have stuck to the use of only the term “theorem” in these notes just for simplicity. The next
theorem is an example of a fact that most books would call a “lemma”. It is hard to prove, and it
is introduced only because we want to use it in the proof of the Saccheri-Legendre Theorem.
Theorem 61 A Lemma about Triangles in Neutral Geometry: Given △ ABC , there exists a
triangle △PQR such that anglesum ( △ PQR ) = anglesum ( △ ABC ) and
degrees ( ∢P ) ≤
1
degrees ( ∢A ) .
2
Proof
1. Side BC has a midpoint, that we can call D. (Justify)
2. There exists a point E such that A*D*E and DA ≅ DE . (Justify) (Draw a picture)
3. ∢ADB ≅ ∢EDC . (Justify)
4. △ ADB ≅△EDC . (Justify)
5. ∢ABD ≅ ∢ECD and ∢BAD ≅ ∢CED . (Justify)
Show that triangle △ AEC has the same angle sum as triangle △ ABC .
6. angle sum ( △ AEC ) = degrees ( ∢CAE ) + degrees ( ∢AEC ) + degrees ( ∢ECA ) .
7. degrees ( ∢CAE ) = degrees ( ∢CAB ) − degrees ( ∢BAD ) . (Justify)
8. degrees ( ∢BAD ) = degrees ( ∢CED ) . (Justify)
9. degrees ( ∢CAE ) = degrees ( ∢CAB ) − degrees ( ∢CED ) . (by 7, 8)
10. degrees ( ∢ECA) = degrees ( ∢ECD ) + degrees ( ∢BCA ) . (Justify)
11. degrees ( ∢ECD ) = degrees ( ∢ABD ) . (Justify)
12. degrees ( ∢ECA ) = degrees ( ∢ABD ) + degrees ( ∢BCA ) . (by 10, 11)
13. angle sum ( △ AEC ) = degrees ( ∢CAB ) + degrees ( ∢ABD ) + degrees ( ∢BCA) (Substitued
9, 12 into 6 and cancelled)
14. angle sum ( △ AEC ) = angle sum ( △ ABC ) . (by 13)
Find a angle of triangle △ AEC that has measure less than or equal to half the measure of ∢A .
15. degrees ( ∢CAE ) + degrees ( ∢CED ) = degrees ( ∢CAB ) . (by 9)
16. So either degrees ( ∢CAE ) ≤
degrees ( ∢CED ) ≤
1
degrees ( ∢CAB ) or
2
1
degrees ( ∢CAB ) , or both.
2
Case 1
1
degrees ( ∢CAB ) is true, then let P=A, Q=E, and R=C. Then we
2
1
can say that anglesum ( △ PQR ) = anglesum ( △ ABC ) and degrees ( ∢P ) ≤ degrees ( ∢A ) .
2
17. If degrees ( ∢CAE ) ≤
Case 2
Page 89 of 106
18. If degrees ( ∢CAE ) ≤
1
degrees ( ∢CAB ) is not true, then it must be that
2
1
degrees ( ∢CAB ) is true. In this case, let P=E, Q=A, and R=C.
2
Then we can say that anglesum ( △ PQR ) = anglesum ( △ ABC ) and
degrees ( ∢CED ) ≤
degrees ( ∢P ) ≤
1
degrees ( ∢A ) .
2
Conclusion of cases
19. In either case, we have shown that there exists a triangle △PQR such that
1
anglesum ( △ PQR ) = anglesum ( △ ABC ) and degrees ( ∢P ) ≤ degrees ( ∢A ) .
2
End of proof
As mentioned earlier, you are all used to the fact that the angle sum of any triangle is exactly 180
degrees, but that fact cannot be proven in Neutral Geometry. All that can be proven is the
following very famous theorem.
Theorem 62 (The Saccheri-Legendre Theorem) In Neutral geometry, the angle sum of any
triangle is less than or equal to 180.
Proof (indirect proof by method of contradiction)
Step 1: Introduce triangle △ ABC and constants x and y.
Assume that triangle △ ABC has angle sum greater than 180. In symbols, we would write
degrees ( ∢A ) + degrees ( ∢B ) + degrees ( ∢C ) > 180 . This would mean that
degrees ( ∢A) + degrees ( ∢B ) + degrees ( ∢C ) = 180 + x , where x is some positive real
number. We know that degrees ( ∢B ) + degrees ( ∢C ) < 180 , by Theorem 59. Therefore,
degrees ( ∢A ) > x . Let y = degrees ( ∢A ) . We know that y is some number such that
y > x.
Step 2: Find a new triangle △ ABC .
By Theorem 61, we know that there exists a triangle △PQR such that
1
anglesum ( △ PQR ) = anglesum ( △ ABC ) and degrees ( ∢P ) ≤ degrees ( ∢A ) . That is,
2
1
anglesum ( △ PQR ) = 180 + x and degrees ( ∢P ) ≤ y . We can rename triangle △PQR as
2
the new triangle △ ABC . We now have a new triangle △ ABC such that
1
anglesum ( △ ABC ) = 180 + x and degrees ( ∢A ) ≤ y .
2
Step 3: Repeat the process.
We can again apply Theorem 61 to obtain yet another triangle △PQR such that
1
anglesum ( △ PQR ) = anglesum ( △ ABC ) and degrees ( ∢P ) ≤ degrees ( ∢A ) . That is,
2
Page 90 of 106
11  1
anglesum ( △ PQR ) = 180 + x and degrees ( ∢P ) ≤  y  = y . We can rename triangle
22  4
△PQR as the new △ ABC . We now have a new triangle △ ABC such that
1
anglesum ( △ ABC ) = 180 + x and degrees ( ∢A ) ≤ y .
4
Step 4: Observe the pattern in the repetitive process.
After one application of Theorem 61, we had a new triangle △ ABC such that
1
anglesum ( △ ABC ) = 180 + x and degrees ( ∢A ) ≤ y . After two applications of Theorem
2
61, we had a new triangle △ ABC such that anglesum ( △ ABC ) = 180 + x and
1
y . After k applications of Theorem 61, we would have a new triangle
4
1
△ ABC such that anglesum ( △ ABC ) = 180 + x and degrees ( ∢A ) ≤ k y .
2
Step 5: Show how the process would end.
Eventually, when k is high enough (after enough applications of Theorem 61), we will
1
reach a point where k y < x . At that point, we will have obtained a triangle △ ABC such
2
1
that anglesum ( △ ABC ) = 180 + x and degrees ( ∢A ) ≤ k y < x .
2
Step 6: Observe the contradiction and state the conclusion.
In this new triangle △ ABC , we know that degrees ( ∢B ) + degrees ( ∢C ) < 180 , by
degrees ( ∢A ) ≤
Theorem 59. Therefore, degrees ( ∢A) + degrees ( ∢B ) + degrees ( ∢C ) < x + 180 . But this
contradicts the fact that for the new triangle △ ABC , anglesum ( △ ABC ) = 180 + x .
Therefore, our assumption way back in Step 1 was wrong. In that step we assumed that
the original triangle △ ABC had angle sum greater than 180. This cannot be true.
Therefore, the angle sum of that original triangle △ ABC must have been less than or
equal to 180.
End of proof
In class, we will draw examples of triangles whose angle sums are less than 180.
The following theorem is not very hard to prove. You will prove it in the exercises.
Theorem 63 In Neutral Geometry, the sum of the degree measures of any two angles of a
triangle is less than or equal to the degree measure of their remote exterior angle.
Page 91 of 106
7.5. Exercises
u 
v 
v


= ln  u  . Prove this.
[1] In the reading, you were told that ln
 x 
 y 
 y
 x
[2] Let l be the Hline consisting of the portion of the x-axis that lies inside the unit circle. The
“missing endpoints” of this line are R = ( −1, 0 ) and S = (1, 0 ) . (They are not Hpoints.) Let A, B,
(
)
(
(
)
)
C, and D be the Hpoints A = ( 0, 0 ) , B = 1 , 0 , C = 1 , 0 , and D = 3 , 0 . Find
4
2
4
lengthH ( AB ) and lengthH ( CD ) . (Be sure to use R & S as the missing endpoints!)
[3] In Section 7.3, an example of a curvy-looking Hline was presented. It has missing endpoints
R and S, and famous points A, B, and C.
(a) Draw the unit circle, and then add to your drawing the curvy-looking Hline, with missing
endpoints R and S and famous points A, B, and C clearly labeled with their coordinates.
(b) Compute lengthH ( AB ) and lengthH ( AC ) .
[4] Let A = ( 0, 0 ) , and let B = ( x, 0 ) where x is some unspecified real number such that 0 < x < 1
. Compute lengthH ( AB ) .
[5] Given some unspecified positive number L, find a number x such that 0 < x < 1 and such that
the hyperbolic length of the line segment with endpoints A = ( 0, 0 ) and B = ( x, 0 ) is
lengthH ( AB ) = L . (Hint: Think of the result of the previous problem as a function that computes
length as a function of x. Call the resulting length L. Then you have an equation involving x and
L, and that equation is solved for L. The current problem is simply asking you to solve that
equation for x. The fact that you know 0 < x < 1 enables you deal with the absolute value sign.)
This exercise proves that the lengthH function is onto.
[6] Justify the steps in the proof of Theorem 59.
[7] Justify the steps in the proof of Theorem 60.
[8] Justify the steps in the proof of Theorem 61.
[9] Prove Theorem 63.
(
)
(
)
(
)
(
)
[10] Define points A = ( 0, 0 ) , B = 1 , 0 , C = 3 , 0 , D = 0, 1 , and E = 0, 3 . These
4
4
4
4
points lie inside the unit circle, so they may be considered as Euclidan Points or as Hpoints.
(a) Draw a large (5 to 6 inch diameter) unit circle, and plot points A, B, C, D, and E.
(b) Draw Hlines AC , AE , BD , and CE .
(c) Draw Euclidean segments BD and CE. In your drawing you should now be able to see
Euclidean triangles △ ABD and △ ACE , as well as Hyperbolic triangles △ ABD and △ ACE .
Page 92 of 106
(d) Using a protractor, estimate the angle sum of Hyperbolic triangle △ ABD .
(e) Using a protractor, estimate the angle sum of Hyperbolic triangle △ ACE .
The next few problems are about proof structure.
[11] What is the statement that is proven by a proof with the following structure?
Proof
1) In Neutral Geometry, suppose that triangle △ ABC has Property X.
2)
3)
4)
5) △ ABC has Property P.
End of Proof
[12] What is the statement that is proven by a proof with the following structure?
Proof (Indirect proof by method of contraposition.)
1) In Neutral Geometry, suppose that triangle △ ABC does not Property X.
2)
3)
4) △ ABC does not have Property P.
End of Proof
[13] What is the statement that is proven by a proof with the following structure?
Proof
Part 1
1)In Neutral Geometry, suppose that triangle △ ABC has Property Y.
2)
3)
4)
5) △ ABC has Property M.
Part 2
6) In Neutral Geometry, suppose that triangle △ ABC does not Property Y.
7)
8)
9) △ ABC does not have Property M.
End of Proof
Page 93 of 106
8. Euclidean Geometry
Intro will go here
8.1. Building Euclidean Geometry from Neutral Geometry
Theorem 64 Thirteen Statements that are Equivalent in Neutral Geometry
Given: The axioms for Neutral Geometry
Claim: The following statements are equivalent:
1) Euclid’s Fifth Postulate: “That, if a straight line falling on two straight lines makes the
interior angles on the same side less than two right angles, the straight lines, if produced
indefinitely, meet on that side on which are the angles less than the two right angles.”
2) Our reworded version of SMSG Postulate #16, which we will call EPP: The Euclidean
Parallel Postulate: For any line L and for any point P not on L, there is at most one line
M such that P LIES ON M and M is parallel to L.
3) Playfair’s Postulate: “For every line L and every point P not on L, there is a unique line
M that contains P and is parallel to L.”
4) “If a line intersects one of two parallel lines, then it intersects the other.”
5) “If a line is perpendicular to one of two parallel lines, then it is perpendicular to the
other.”
6) “Given any △PQR and any line segment AB , there exists a triangle △ ABC having AB
as one of its sides such that △ ABC is similar to △PQR but not congruent to △PQR .”
7) “If two parallel lines are intersected by a transversal, then the measures of pairs of
interior angles on the same side of the transversal add up to 180.”
8) “If two parallel lines are cut by a transversal, then any pair of alternate interior angles
created is congruent.” (This statement is the converse of Theorem 42, The Alternate
Interior Angle Theorem.)
9) “If two parallel lines are intersected by a transversal, then any pair of corresponding
angles formed is congruent.” (This statement is the converse of the statement of Theorem
43, The Corresponding Angle Theorem,)
10) “Every triangle has angle sum exactly 180.”
11) “The opposite sides of a parallelogram are congruent.”
12) “The opposite angles of a parallelogram are congruent.”
13) “The diagonals of a parallelogram bisect each other.”
Remark: Because these statements are equivalent, if one of them is true, then all thirteen of them
are true, and if one of them is false, then all thirteen of them are false. That means that if we
make any one of these thirteen statements an axiom (that is, we declare it to be true), and we add
it to the set of axioms describing Neutral Geometry, then regardless of which statement we have
chosen to make an axiom, the resulting larger axiom set will describe the same thing. The thing
that they describe is what we call “Euclidean Geometry”.
Page 94 of 106
9. For Reference: Axioms, Defintions, and Theorems of Neutral Geometry
9.1. The Axioms of Neutral Geometry
Incidence Axioms
IA1: If P and Q are distinct points then there exists exactly one line that both lie on.
IA2: Every line passes through at least two distinct points.
IA3: There exist three distinct points that are non-collinear. (at least three)
Betweenness Axioms
BA1: If A*B*C, then A, B, and C are distinct collinear points, and C*B*A.
BA2: If B and D are distinct points, and L is the unique line that passes through both points,
then there exist points A, C, and E lying on L such that A*B*D and B*C*D and B*D*E.
BA3: If A, B, and C are three distinct points lying on the same line, then exactly one of the
points is between the other two.
BA4 (Plane Separation): If L is a line and A, B, and C are three points not lying on L then
(i) 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.
(ii) 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.
Congruence Axioms
CA1: The congruence relation on the set of line segments is an equivalence relation
CA2: (segment construction axiom) For any segment AB and ray PQ , there exists a unique
point R on PQ such that PR ≅ AB .
CA3: (segment addition axiom) If A*B*C, A'*B'*C', AB ≅ A′B′ , and BC ≅ B′C ′ , then
AC ≅ A′C ′ .
CA4: The congruence relation on the set of angles is an equivalence relation
CA5: (angle construction axiom) Given an angle ∢BAC , distinct points A' and B', and a
choice of one of the two half-planes bounded by the line A′B′ , there exists a unique ray
A′C ′ such that C' lies in the chosen half plane and ∢BAC ≅ ∢B′A′C ′ .
CA6: (SAS axiom) If two sides and the included angle of one triangle are congruent to two
sides and the included angle of another triangle, then the triangles are congruent.
Axioms of Continuity
Archimedes Axiom: Given any segment CD and ray AB , there exists some positive integer n
and a set of points { P0 , P1 ,⋯ , Pn } on AB such that P0 = A ; for all k, Pk Pk +1 ≅ CD ; and either
Pn = B or ( A * Pn −1 * B and A * B * Pn ).
Circular Continuity Axiom: If a circle intersects both the interior and exterior of another
circle, then the two circles intersect in exactly two points.
9.2. The Definitions of Neutral Geometry
Definition 18 passes through
• words: Line L passes through point P.
• meaning: Point P lies on line L.
Page 95 of 106
Definition 19 commonly used expressions that we will not use
• Line L contains point P.
• Line L is on point P.
• Point P is incident upon line L.
• Line L is incident upon point P.
Definition 20 intersecting lines
• words: Line L intersects line M.
• meaning: There exists a point P that lies on both lines. (at least one point)
Definition 21 parallel lines
• words: Line L is parallel to line M.
• symbol: L M
•
meaning: Line L does not intersect line M.
Definition 22 collinear points
• words: The set of points { P1 , P2 ,⋯ , Pk } is collinear.
•
meaning: There exists a line L that passes through all the points.
Definition 23 concurrent lines
• words: The set of lines { L1 , L2 ,⋯ , Lk } is concurrent.
•
meaning: There exists a point P that lies on all the lines.
Definition 27 Hpoint
• word: Hpoint
• meaning: an ( x, y ) pair in the interior of the unit circle
Definition 28 Hline
• word: Hline
• meaning: Either of the following particular types of sets of Hpoints
• A “straight-looking” Hline is the set of Hpoints that lie on a diameter of the unit circle.
• A “curved-looking” Hline is the set of Hpoints that lie on a circle that is orthogonal to
the unit circle.
Definition 31 The betweenness relation on the set of real numbers
• Words: “x is between y and z.”
• Usage: x, y, and z are real numbers.
• Meaning: “x < y <z or z <y <x.”
• Remark: This is a ternary relation on the set of real numbers.
• Warning: This is NOT the same as betweenness for points, discussed in the next section.
Definition 32 symbol for a line
• symbol: AB
• meaning: the (unique) line that passes through points A and B
Page 96 of 106
Definition 33 the set of points that lie on a line
• symbol: AB
{ }
•
meaning: the set of points that lie on line AB
Definition 34 the plane
• meaning: the set of all points
Definition 35 line segment, endpoints of a line segment
• symbol: AB
• spoken: “line segment A, B”, or “segment A, B”
• usage: A and B are points.
• meaning: the set AB = { A} ∪ {C : A * C * B} ∪ {B}
•
additional terminology: points A and B are called endpoints of segment AB.
Definition 36 ray, endpoints of a ray
• symbol: AB
• spoken: “ray A, B”
• usage: A and B are points.
• meaning: the set AB = { A} ∪ {C : A * C * B} ∪ { B} ∪ { D : A * B * D}
• additional terminology: Point A is called the endpoint of ray AB . We say that ray AB
emanates from point A.
Definition 37 opposite rays
• words: BA and BC are opposite rays
• meaning: A*B*C
Definition 38 same side
• words: “A and B are on the same side of L.”
• usage: A and B are points and L is a line that does not pass through either point.
• meaning: either ( A = B ) or ( A ≠ B and line segment AB does not intersect line L.)
Definition 39 opposite side
• words: “A and B are on the opposite side of L.”
• usage: A and B are points and L is a line that does not pass through either point.
• meaning: A ≠ B and line segment AB does intersect line L.
Definition 40 half plane
• words: half-plane bounded by L, containing point A.
• meaning: the set whose elements are some point A not on L, along with all the other
points that are on the same side of L as point A
• symbol: HA
Page 97 of 106
Definition 41 angle
• symbol: ∢ABC
• usage: A, B, and C are non-collinear points
• meaning: BA ∪ BC
•
•
additional terminology: point B is called the vertex of ∢ABC , rays BA and BC are
called the sides.
observations: because BA ∪ BC = BC ∪ BA , the symbols ∢ABC and ∢CBA represent
the same angle.
Definition 42 supplementary angles
• words: supplementary angles
• meaning: two angles that share a common side and whose other sides are opposite rays.
Definition 43 interior of an angle
• words: the interior of ∢ABC
• meaning: the set of points P such
that
(
P
is
on
the
same
side
of
line
BA as point C) and
(P is on the same side of line BC as point A).
Definition 44 ray between two other rays
• words: ray BD is between BA and BC .
• meaning: Point D is in the interior of ∢ABC .
Definition 45 triangle
• symbol: △ ABC
• spoken: triangle A, B, C
• usage: A, B, and C are non-collinear
• meaning: the set AB ∪ BC ∪ CA
• additional terminology:
o The points A, B, C are called the vertices of the triangle.
o The segments AB, BC, CA are called the sides of the triangle
o Side BC is said to be opposite vertex A. Similarly for the other sides.
o The angle ∢BAC is called angle A if there is no chance of this causing confusion.
Similarly for the other angles.
Definition 46 interior of a triangle
• words: the interior of △ ABC
• meaning: the set of points P such
that
(
P
is
on
the
same
side
of
line
BA as point
C) and
(P is on the same side of line BC as point A) and (P is on the same side of line AC as
point B)
Definition 47 exterior of a triangle
Page 98 of 106
•
meaning: the set of points Q that are neither an element of the triangle, itself, nor of the
interior of the triangle.
Definition 49 Correspondence between the vertices of two triangles
• Words: “f is a correspondence between the vertices of triangles △ ABC and △ DEF .”
•
Meaning: f is a one-to-one, onto function with domain { A, B , C } and codomain { D , E , F } .
Definition 50 Corresponding parts of two triangles
• Words: Corresponding parts of triangles △ ABC and △ DEF .
• Usage: A correspondence between the vertices of △ ABC and △ DEF has been given.
• Meaning: As discussed above, there is an automatic correspondence between the sides of
triangles △ ABC and the sides of △ DEF , and also between the angle of triangles △ ABC and
the angles of △ DEF . Suppose the correspondence between vertices were
( B, A, C ) ↔ ( D, E , F ) . Corresponding parts would be pairs such as the pair of sides,
AC ↔ EF , or the pair of angles, ∢ACB ↔ ∢EFD .
Definition 51 Triangle Congruence
• Symbol: △ ABC ≅△ DEF
• Words: “ △ ABC is congruent to △ DEF .”
• Meaning: “There is a correspondence between the vertices of the two triangles such that
corresponding parts of the triangles are congruent.”
Definition 52 isosceles triangle
• Words: isosceles triangle
• Meaning: two sides of the triangle are congruent to each other (at least two)
Definition 53 the order relation on the set of line segments
• Symbol: AB < CD
• Spoken: “Segment AB is less than segment CD.”
• Meaning: There exists a point E between C and D such that AB ≅ CE .
• Remark: The order relation is a binary relation on the set of line segments.
Definition 54 vertical angles
• words: vertical angles
• Meaning: Two angles that share a vertex and whose sides are opposite rays. That is, two
angles that can be labeled ∢ABC and ∢DBE where A*B*D and C*B*E.
Definition 55 right angle
• words: right angle
• Meaning: An angle that is congruent to its supplementary angle
Definition 56 perpendicular lines
• symbol: L ⊥ M
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•
•
spoken: L and M are perpendicular
Meaning: L and M are distinct, non-parallel lines that create a pair of vertical angles that are
right angles.
Definition 57 the order relation on the set of angles
• Symbol: ∢ABC < ∢DEF
• Spoken: “Angle ∢ABC is less than angle ∢DEF .”
• Meaning: There exists a ray EG between ED and EF such that ∢ABC ≅ ∢GEF .
• Remark: The order relation is a binary relation on the set of angles.
Definition 58 transversal
• Words: “Line T is transversal to lines L and M.”
• Meaning: “T intersects L and M in distinct points.”
Definition 59 “alternate interior angles”; “corresponding angles”
• Usage: Lines L, M, and transversal T are given.
• Labeled points: Let B be the intersection of lines T and L, and let E be the intersection of
lines T and M. (By definition of transversal, B and E are not the same point.) By the
betweenness axioms, there exist points A and C on line L such that A*B*C, points D and
F on line M such that D*E*F, and points G and H on line T such that G*B*E and B*E*H.
Without loss of generality, we may assume that points D and F are labeled such that it is
point D that is on the same side of line BE as point A.
H T
M
D
E
A
L
B
F
C
G
•
Meaning:
∢ABE
∢CBE
∢ABG
∢ABH
∢CBG
∢CBH
and ∢FEB is a pair of alternate interior angles.
and ∢DEB is a pair of alternate interior angles.
and ∢DEG is a pair of corresponding angles.
and ∢DEH is a pair of corresponding angles.
and ∢FEG is a pair of corresponding angles.
and ∢FEH is a pair of corresponding angles.
Definition 60 “exterior angle”, “interior angle”, “remote interior angles”
• Words: An exterior angle of a triangle
• Meaning: An angle that is supplemental to one of the angles of the triangle
• Additional terminology: The angles of the triangle are also called “interior angles”. Given
an exterior angle, there will be an interior angle that is its supplement, and two other
interior angles that are not its supplement. Those other two interior angles are called
remote interior angles.
Page 100 of 106
Definition 61 “hypotenuse” and “legs” of a right triangle
• Meaning: The hypotenuse of a right triangle is the side that is opposite the right angle.
The other two sides are called legs.
Definition 62 “midpoint” of a line segment
• Words: C is a midpoint of segment AB
• Meaning: A*C*B and CA ≅ CB .
Definition 63 “bisector” of a line segment, “perpendicular bisector” of a line segment
• Words: Line L is a bisector of segment AB
• Meaning: L is distinct from line AB and passes through the midpoint of segment AB.
• Additional Terminology: If L is perpendicular to line AB and is also a bisector of
segment AB, then L is said to be a perpendicular bisector of segment AB.
Definition 64 “bisector” of an angle
• Words: a bisector of angle ∢ABC
• Meaning: a ray BD between rays BA and BC such that ∢DBA ≅ ∢DBC .
Definition 65 the length function for line segments in straight-line drawings (Euclidean)
Meaning: the function lengthE : {Euclidean line segments} → ℝ + defined by the following
formula
lengthE ( AB ) =
( x1 − x2 ) + ( y1 − y2 ) ,
where AB is a Euclidean line segment with endpoints A = ( x1 , y1 )
2
2
and B = ( x2 , y2 ) . (This
implies that x1, y1, x2,and y2 are real numbers.)
Definition 67 Definition 66 the length function for line segments in the Hyperbolic plane
Meaning: the function lengthH : { Hyperbolic line segments} → ℝ + defined by the
following formula
 lengthE ( AR )


lengthE ( AS ) 
,
lengthH ( AB ) = ln 
 lengthE ( BR )



length
BS
(
)
E


where AB is a Hyperbolic line segment with endpoints that are the Hpoints A and B, and
R and S are the “missing endpoints” of the line AB .
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9.3. The Theorems of Neutral Geometry
Theorem 1: In Incidence Geometry, if L and M are distinct lines that are not parallel, then there
is exactly one point that both lines pass through.
Theorem 2: In Incidence Geometry, there exist three lines that are not concurrent.
Theorem 3: In Incidence Geometry, given any line L, there exists a point not lying on L.
Theorem 4: In Incidence Geometry, given any point P, there exists a line that does not pass
through P.
Theorem 5: In Incidence Geometry, given any point P, there exist two lines that pass through P.
Theorem 6: In I&B Geometry, if A*B*C and A*C*D then A, B, C, D are distinct and collinear.
Theorem 7: In I&B Geometry, every ray has an opposite ray.
Theorem 8: In I&B Geometry, for any two distinct points A and B, AB ∩ BA = AB .
Theorem 9: In I&B Geometry, for any two distinct points A and B, AB ∪ BA = AB .
{ }
Theorem 10: In I&B Geometry, if A and B are on opposite sides of line L, and B and C are on the
same side of L, then A and C are on opposite sides of L.
Theorem 11: In I&B Geometry, every line L partitions the plane into three sets: (1) the set of
points that lie on L, (2) a half plane, and (3) a second half-plane. In other words, every point of
the plane either lies on L or is an element of one (not both) of the half planes.
Theorem 12: In I&B Geometry, if A*B*C and A*C*D, then B*C*D and A*B*D.
Theorem 13: In I&B Geometry, if A*B*C and B*C*D, then A*B*D and A*C*D.
Theorem 14: (Line Separation) In I&B Geometry, If A, B, C, and D are collinear points such that
A*B*C, and D ≠ B , then either D ∈ AB or D ∈ AC , but not both.
Theorem 15: (Pasch’s Theorem) In I&B Geometry, if A, B,and C are non-collinear points and
line L intersects segment AB at a point between A and B, then L also intersects either AC or BC.
Furthermore, if C does not lie on L, then L does not intersect both AC and BC.
Theorem 16: In I&B Geometry, if A*B*C then AC = AB ∪ BC and B is the only point common
to segments AB and BC.
Theorem 17: In I&B Geometry, if A*B*C then B = BA ∩ BC and AB = AC .
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Theorem 18: In I&B Geometry, if point A lies on line L, and point B does not lie on line L, then
every point of ray AB except point A is on the same side of L as B.
Theorem 19: In I&B Geometry, given ∢BAC and point D lying on BC , point D is in the
interior of ∢BAC if and only if B*D*C.
Theorem 20: In I&B Geometry, given angle ∢BAC ; point D in the interior of ∢BAC , and point
E such that C*A*E, the following three statements are all true:
(1) Every point on ray AD except A is in the interior of angle ∢BAC .
(2) No point on the ray opposite to AD is in the interior of angle ∢BAC .
(3) Point B is in the interior of angle ∢DAE .
Theorem 21: (The Crossbar Theorem) In I&B Geometry, if ray AD is between rays AB and
AC then ray AD intersects segment BC.
Theorem 22: In I&B Geometry, If ray r emanates from an exterior point of triangle △ ABC and
intersects side AB in a point between A and B, then ray r also intersects side AC or BC.
Theorem 23: In I&B Geometry, if a ray emanates from an interior point of a triangle, then it
intersects one of the sides, and if it does not pass through a vertex, it intersects only one side.
Theorem 24: In I&B Geometry, a line cannot be contained in the interior of a triangle.
Theorem 25: The Triangle Construction Theorem
Given: Neutral Geometry, △ ABC , DE ≅ AB , and a point G not on DE .
Claim: There exists a unique point F in half plane HG such that △ ABC ≅△ DEF .
Theorem 26: The Isosceles Triangle Theorem
Given: Neutral Geometry, △ ABC , AB ≅ AC (Given an isosceles triangle.)
Claim: ∢B ≅ ∢C (The two angles that are opposite the two congruent sides are congruent.)
Theorem 27: (segment subtraction)
Given: Neutral Geometry; A*B*C; D*E*F; AB ≅ DE and AC ≅ DF .
Claim: BC ≅ EF .
Theorem 28: In Neutral Geometry, if AC ≅ DF and A*B*C, then there exists a unique point E
such that D*E*F and AB ≅ DE .
Theorem 29: Facts about the order relation on the set of line segments
Given: Neutral Geometry; line segments AB, CD, and EF.
Claim
(a) (trichotomy): Exactly one of the following is true: AB < CD, AB ≅ CD , or CD < AB.
(b): If AB < CD and CD ≅ EF , then AB < EF.
(c): (transitivity): If AB < CD and CD < EF, then AB < EF.
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Theorem 30: In Neutral Geometry, supplements of congruent angles are congruent. That is, if
∢ABC and ∢CBD are supplementary, and ∢EFG and ∢GFH are supplementary, and
∢ABC ≅ ∢EFG , then ∢CBD ≅ ∢GFH .
Theorem 31: In Neutral Geometry, vertical angles are congruent.
Theorem 32: In Neutral Geometry, any angle congruent to a right angle is also a right angle.
Theorem 33: In Neutral Geometry, for every line L and every point P not on L there exists a line
through P perpendicular to L.
Theorem 34: In Neutral Geometry, for every line L and every point P on L there exists a unique
line through P perpendicular to L.
Theorem 35: (angle addition)
Given: Neutral Geometry; angle ∢ABC with point D in the interior; angle ∢EFG with
point H in the interior; ∢DBA ≅ ∢HFE ; ∢DBC ≅ ∢HFG
Claim: ∢ABC ≅ ∢EFG
Theorem 36: (angle subtraction)
Given: Neutral Geometry; angle ∢ABC with point D in the interior; angle ∢EFG with
point H in the interior; ∢DBA ≅ ∢HFE ; ∢ABC ≅ ∢EFG
Claim: ∢DBC ≅ ∢HFG
(Missing Theorem: There should be a theorem here about rays that would be analogous to
Theorem 28. The theorem statement would be as follows:
Missing Theorem about Rays and Angles:
Given: Neutral Geometry, ∢ABC ≅ ∢EFG , ray BD between rays BA and BC .
Claim: There is a unique ray FH between rays FE and FG such that ∢ABD ≅ ∢EFH .)
Theorem 37: (facts about the order relation on the set of angles)
Given: Neutral Geometry; angles ∢P , ∢Q , ∢R .
Claim
(a) (trichotomy): Exactly one of the following is true: ∢P < ∢Q , ∢P ≅ ∢Q , or ∢Q < ∢P .
(b): If ∢P < ∢Q and ∢Q ≅ ∢R then ∢P < ∢R .
(c): (transitivity): If ∢P < ∢Q and ∢Q < ∢R then ∢P < ∢R .
Theorem 38: (Euclid’s 4th postulate). All right angles are congruent to each other.
Theorem 39: (ASA congruence). In Neutral Geometry, if there is a correspondence between
parts of two triangles such that two angles and the included side of one triangle are congruent to
the corresponding parts of the other triangle, then all the remaining corresponding parts are
congruent as well, so the triangles are congruent.
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Theorem 40: (Converse of the Isosceles Triangle Theorem)
Given: Neutral Geometry, △ ABC , ∢B ≅ ∢C (Given a triangle with two congruent angles.)
Claim: AB ≅ AC (The two sides that are opposite the two congruent angles are congruent. That
is, the triangle is isosceles.)
Theorem 41: (SSS congruence) In Neutral Geometry, if there is a correspondence between parts
of two triangles such that the three sides of one triangle are congruent to the corresponding parts
of the other triangle, then all the remaining corresponding parts are congruent as well, so the
triangles are congruent.
Theorem 42: The Alternate Interior Angle Theorem
Given: Neutral Geometry, lines L and M and a transversal T
Claim: If a pair of alternate interior angles is congruent, then lines L and M are parallel.
Theorem 43: The Corresponding Angle Theorem
Given: Neutral Geometry, lines L and M and a transversal T
Claim: If a pair of corresponding angles is congruent, then lines L and M are parallel.
Theorem 44: In Neutral Geometry, two distinct lines perpendicular to the same line are parallel.
Theorem 45: Uniqueness of the perpendicular from a point to a line
Given: Neutral Geometry, line L and a point P not on L
Claim: There is only one line that passes through P and is perpendicular to L.
Theorem 46: Existence of parallel lines (The answer to THE BIG QUESTION)
Given: Neutral Geometry, line L and a point P not on L
Claim: There exists at least one line that passes through P and is parallel to L.
Theorem 47: The Exterior Angle Theorem
In Neutral Geometry, each of the remote interior angles is less than the exterior angle.
Theorem 48: The Angle Angle Side Congruence Theorem
In Neutral Geometry, if there is a correspondence between parts of two triangles such that two
angles and a non-included side of one triangle are congruent to the corresponding parts of the
other triangle, then all the remaining corresponding parts are congruent as well, so the triangles
are congruent.
Theorem 49: The Hypotenuse Leg Congruence Theorem
In Neutral Geometry, if the hypotenuse and one leg of one right triangle are congruent to the
hypotenuse and one leg of another right triangle, then all the remaining corresponding parts are
congruent as well, so the triangles are congruent.
Theorem 50: Existence of Midpoint of a Line Segment
In Neutral Geometry, every segment has exactly one midpoint.
Theorem 51: In Neutral Geometry, every line segment has exactly one perpendicular bisector.
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Theorem 52: In Neutral Geometry, every angle has exactly one bisector.
Theorem 53: In Neutral Geometry, in triangle △ ABC , if BC > AC then ∢A > ∢B .
Theorem 54: In Neutral Geometry, in triangle △ ABC , if ∢A > ∢B then BC > AC.
Theorem 55: The CACS and BABS Theorem
In triangles of Neutral Geometry, Congruent Angles are always opposite Congruent Sides, and
Bigger angles are always opposite Bigger Sides.
Theorem 56: The Hinge Theorem
Given: Neutral Geometry, △ ABC , △ DEF , AB ≅ DE , AC ≅ DF
Claim: The following are equivalent
(1) ∢A < ∢D .
(2) BC < EF.
Theorem 57 Existence of a Length Function for Line Segments.
Given: Neutral Geometry, line segment AB
Claim: There exists a unique function length ( ) : {line segments} → ℝ + with the following
properties.
1. The length function is onto. That is, for every positive real number x, there exists a line
segment CD such that length ( CD ) = x .
2. length ( AB ) = 1
3. length ( CD ) = length ( EF ) if and only if CD ≅ EF .
4. length ( CD ) < length ( EF ) if and only if CD < EF.
5. C*D*E if and only if C, D, E are collinear and length ( CE ) = length ( CD ) + length ( DE ) .
Theorem 58 Existence of a Measure Function for Angles
Given: Neutral Geometry
Claim: There exists a unique function degrees ( ) : {angles} → ( 0,180 ) with the following
properties.
1. The degrees function is onto. That is, for every 0 < x < 180 , there exists an angle ∢A
such that degrees ( ∢A ) = x .
2. degrees ( ∢A) = 90 if and only if ∢A is a right angle.
3. degrees ( ∢A ) = degrees ( ∢B ) if and only if ∢A ≅ ∢B .
4. degrees ( ∢A ) < degrees ( ∢A) if and only if ∢A < ∢B .
5. If ray AC is between rays AB and AD , then
degrees ( ∢BAD ) = degrees ( ∢BAC ) + degrees ( ∢CAD ) .
6. If angle ∢A is supplementary to angle ∢B , then degrees ( ∢A ) + degrees ( ∢B ) = 180
Page 106 of 106
Theorem 59 In Neutral Geometry, the sum of the measures of any two angles of a triangle is less
than 180.
Theorem 60 (The Triangle Inequality) In Neutral Geometry, the length of any side of a triangle is
less than the sum of the lengths of the two other sides.
Theorem 61 (A Lemma about Triangles in Neutral Geometry) Given △ ABC , there exists a
triangle △PQR such that anglesum ( △ PQR ) = anglesum ( △ ABC ) and
degrees ( ∢P ) ≤
1
degrees ( ∢A ) .
2
Theorem 62 (The Saccheri-Legendre Theorem) In Neutral geometry, the angle sum of any
triangle is less than or equal to 180.
Theorem 63 In Neutral Geometry, the sum of the degree measures of any two angles of a triangle
is less than or equal to the degree measure of their remote exterior angle.