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Spring 2008 Math 330A Notes, Version 10.1
Reading Material for Foundations of Geometry I
(by Mark Barsamian)
1.
Relations..............................................................................................................................5
1.1.
Cartesian Products ...................................................................................................5
1.2.
Relations ..................................................................................................................8
1.3.
Exercises ................................................................................................................16
2.
Axiom Systems .................................................................................................................19
2.1.
Introduction ............................................................................................................19
2.2.
Definition of Axiom System ..................................................................................21
2.3.
Primitive Relations and Primitive Terms ...............................................................22
2.4.
Interpretations and Models ....................................................................................25
2.5.
Properties of Axiom Systems.................................................................................28
2.6.
Exercises ................................................................................................................36
3.
Axiomatic Geometries .....................................................................................................39
3.1.
Introduction ............................................................................................................39
3.2.
Two Simple Finite Geometries ..............................................................................40
3.3.
Terminology...........................................................................................................42
3.4.
Fano’s and Young’s Finite Geometries .................................................................44
3.5.
Incidence Geometry ...............................................................................................45
3.6.
Models of Incidence Geometry..............................................................................46
3.7.
Exercises ................................................................................................................49
4.
Models of Incidence Geometry Whose Sets of Points are Infinite...............................51
4.1.
A Model of Incidence Geometry Involving Straight-Line Drawings ....................51
4.2.
A Model of Incidence Geometry Involving the Usual Analytic Geometry ...........52
4.3.
A Model of Incidence Geometry Involving the Poincare Disk .............................52
4.4.
Other Interpretations of Incidence Geometry ........................................................55
4.5.
Exercises ................................................................................................................56
5.
Incidence and Betweenness Geometry ...........................................................................61
5.1.
Binary and Ternary Relations on a Set ..................................................................61
5.2.
Introducing Incidence and Betweenness Geometry ...............................................62
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5.3.
Line Segments and Rays ........................................................................................64
5.4.
Plane Separation.....................................................................................................65
5.5.
Line Separation ......................................................................................................67
5.6.
Angles and Triangles .............................................................................................68
5.7.
Exercises ................................................................................................................71
6.
Neutral Geometry I ..........................................................................................................73
6.1.
The need for a larger axiom system: Introducing Neutral Geometry ....................73
6.2.
Line Segment Subtraction and the Ordering of Segments .....................................75
6.3.
Triangle Congruence and its Role in the Neutral Geometry Axioms ....................76
6.4.
Right Angles ..........................................................................................................82
6.5.
Angle Addition and Subtraction, and Ordering of Angles ....................................83
6.6.
The Alternate Interior Angle Theorem and Some Corollaries...............................85
6.7.
Exercises ................................................................................................................87
7.
Neutral Geometry II: Triangles ......................................................................................91
7.1.
The Isosceles Triangle Theorem and Its Converse ................................................91
7.2.
The Exterior Angle Theorem .................................................................................92
7.3.
Comparison Theorems ...........................................................................................94
7.4.
Midpoints and bisectors .........................................................................................95
7.5.
Exercises ................................................................................................................96
8.
Measure of Line Segments and Angles ..........................................................................99
8.1.
Theorems Stating the Existence of Measurement Functions .................................99
8.2.
Two length functions for Neutral Geometry........................................................100
8.3.
An example of a curvy-looking Hline .................................................................102
8.4.
Theorems about segment lengths and angle measures ........................................104
8.5.
Exercises for Chapter 8 ........................................................................................106
9.
Building Euclidean Geometry from Neutral Geometry .............................................111
9.1.
Thirteen Statements that are Logically Equivalent in Neutral Geometry ............112
9.2.
What’s up with the answer to THE BIG QUESTION?!? ....................................113
9.3.
Proving Theorems about Equivalent Statements .................................................114
9.4.
Finally, Euclidean Geometry ...............................................................................116
9.5.
Exercises ..............................................................................................................117
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10.
Euclidean Geometry I ....................................................................................................119
10.1.
Some results from the previous chapter presented as theorems ..........................119
10.2.
Circles, chords, and diameters and some Neutral Geometry Theorems ..............121
10.3.
Some Euclidean Geometry Theorems about Circles and Triangles ....................122
10.4.
Could Theorem 84 be a theorem of Neutral Geometry? ......................................123
10.5.
Exercises ..............................................................................................................127
11.
For Reference: Axioms, Definitions, and Theorems of Euclidean Geometry ..........135
11.1.
The Axioms of Euclidean Geometry ...................................................................135
11.2.
The Definitions of Euclidean Geometry ..............................................................136
11.3.
The Theorems of Euclidean Geometry ................................................................143
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1. Relations
Consider the following three mathematical expressions:
 y  x 2 (an equation that is also a function)
 x 2  y 2  1 (an equation that is not a function)
 x < y (an inequality)
These are three very different kinds of expressions. But there is a mathematical framework that
encompasses all three and even encompasses expressions like ―Line L1 is perpendicular to line
L2‖ that don’t have anything to do with real numbers or real variables. All of these expressions
are examples of what are called Relations. Relations play an important role in axiomatic
geometry. They are on the list of topics for Math 306, but are not covered there in enough detail
for our purposes. So we will start Math 330A by studying them.
1.1. Cartesian Products
Central to the definition of relations is the terminology and notation of the Cartesian Product of
two sets. This is a generalization of the so-called Cartesian plane—the set of order pairs (x,y) of
real numbers.
1.1.1. Definition and examples
The definition of Cartesian Product uses ordered pairs, but the pairs can be pairs of elements
from any two sets, not just sets of real numbers.
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) x2  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.
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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) x2  B  B true false
iii) 3  2  B  B
true false
iv)  2, 2   B  B
true false
v)
1,3   3,1
true
false
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  and let B  . Then A  B   , which is commonly denoted as
2
. This is the set of ordered pairs of real numbers. Here is a proper definition.
Definition 2 the Cartesian plane
 Symbol: 2
 Spoken: ―r two‖, or ―the x, y plane‖, or ―the Cartesian plane‖.
 Meaning in symbols: 
 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  ,  , 10  ,  2 , 0  ,
etc., are all elements of 2 .
2) Parentheses and commas are used, because 2 is a Cartesian product. So ―  10 ‖, ―  ,10
‖, etc., are not allowed.
3) Order is important:  5, 2.7    2.7,5  .
4) 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.
5) 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
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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
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
*
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.
Example:
y
(right coordinate)
2
1.5, 2 
 2,1.5
1
1
2
3
x
(left coordinate)
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1.2. Relations
Now that we are familiar with the Cartesian Product of sets, we are ready for the definition of
Relations.
1.2.1. Definitions and Examples
At the start of this chapter, it was mentioned that the concept of relations could include
mathematical exparessions such as y  x 2 and x 2  y 2  1 and x < y. Notice, however, that the
initial definition of relation makes no mention at all of either an x or a y. And note that in the
second definition, when an x and a y do appear, they do not stand for real numbers.
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
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.
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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 of the set 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.
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





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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  , the set of all real numbers.
d) Let x Rd y means x  y . The symbol Rd would mean ―the set of all pairs (x,y) such that
x  y .‖ That is, Rd   x, y    : x  y . Then Rd is a relation on the set . because
it is a subset of  . It could be called the ―less than or equal to‖ relation.
e) Let x Rey means xy = 0. The symbol Re would mean ―the set of all pairs (x,y) such that
xy  0 .‖ That is, Re   x, y    : xy  0 . Then Re is a relation on the set .
f) Let x Rf y means y  2 x  1. The symbol Rf would mean ―the set of all pairs (x,y) such that
y  2 x  1.‖ That is, R f 
 x, y  
 : y  2x  1 . Then Rf is a relation on the set
.
1.2.4. Visualizing a relation on a set
A relation on a set is just a particular type of relation, so the same tools that used to visualize
regular relations (tables and arrow diagrams) can be used to visualize a relation on a set.
In the case of a relation on a finite set A, there is an additional visualization tool available, called
the directed graph. A directed graph is a special kind of arrow diagram, in which 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.
Relation
Table
1 2 3 4 5 6
1
5
*
*
5
*
*
1 2 3 4 5 6
2
Relation Rb
3
4
2
*
6
1
3
*
3 *
4
4
*
2
Relation Ra
Directed Graph
*
*
*
6
1
4
3
*
*
5
2
*
5
6
6
1
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1 2 3 4 5 6
1 *
2
Relation Rc
*
*
3 *
4
5
6
3
4
*
*
*
5
*
*
*
2
*
*
*
*
6
1
To visualize relations on the set of real numbers—an infinite set—we start with a picture of the
underlying Cartesian plane and then simply darken all the (x,y) pairs in the plane that are
elements of the relation. The resulting picture is called the graph of the relation. For examples,
we can visualize the relations Rd, Re, and Rf above.
Relation
Graph
y
Rd   x, y    : x  y
y=x
x
y
Re   x, y    : xy  0
x
y
y=2x+1
R f   x, y    : y  2x  1
x
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1.2.5. Properties that a relation on a set may or may not have
There are four common properties that a relation on a set may or may not have. The definitions
follow. Notice that each definition is presented as a logical statement. 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
Definition 9 Equivalence Relation
 Words: R is an equivalence relation
 Meaning: R is Reflexive and Symmetric and Transitive
Each of the four properties above is a logical statement. Each may be true or false. If the
statement of one of the properties is true for a certain relation, then we say that the given relation
has that property. If the statement of one of the properties is false for a certain relation, then we
say that the given relation does not have that property.
Class Drill for Section Error! Reference source not found.: If the statement of one of the
properties is false, then the negation of that statement will be true. Therefore, it is important to
understand how to find the negations of each of the four statements above. Write the negations of
each of the four statements here:
Words Meaning
R is reflexive: a  A, aRa
R is not reflexive:
R is symmetric:
a, b  A, IF aRb THEN bRa
R is not symmetric:
R is transitive:
a, b, c  A, IF  aRb AND bRc  THEN aRc
R is not transitive:
R is an equivalence relation: R is Reflexive and Symmetric and Transitive
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R is not an equivalence relation:
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. Class Drill:
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,
 The table is not symmetric across the main diagonal.
 There is an arrow in the diagram that is not a double arrow.
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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. Relations on the Set of Real Numbers
In the drill on the following page, you will explore relations on the set of real numbers, .
Remember that a ―relation on the set of real numbers‖ is just a fancy name for a ―subset of the
cartesian plane,  ‖. Note that just as the tables and directed graphs for a relation on a finite
set could be used to check properties of the relation, the graph for a relation on the set of real
numbers can be used to check properties of the relation:






When a relation is reflexive, the entire line y = x is colored in.
When a relation is not reflexive, at least one point on the line y = x is not colored in.
When a relation is symmetric, the graph is symmetric across the line y = x.
When a relation is not symmetric, the graph is not symmetric across the line y = x.
The graph is not much help in looking for transitivity.
The graph is not much help in looking for failures of transitivity, either.
1.2.7. Remark on Relations as Predicates
Consider relation R5 from the previous section. The sentence ―2 R5 7‖ means ― 2  7 ‖. This is a
true statement. The sentence ―7 R5 3‖ means ― 7  3 ‖.The sentence ―x R5 y‖ means ― x  y ‖.
This sentence 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 R5 y‖ is a predicate.
But the sentence ― x  , xR5 x ‖ means ― x  , x  x ‖. This is a true statement. And the
sentence ― x, y  , IF xR5 y THEN yR5 x ‖ means ― x, y  , IF x  y THEN y  x ‖. This is a
false statement. So the properties of the relations are statements.
Page 15 of 150
R0
x R0 y means x – y = 2
R1
x R1 y means x < y
R2
x R2 y means x  y  1
R3
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
2
Transitive?
Relation
Symmetric?
Reflexive?
Class Drill for Section 1.2.6: 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. If a relation has one of the properties, then write ―yes‖ in the box. If a
relation does not have one of the properties, write ―no‖ in the box and give a counterexample that
shows that the relation does not have the property. That is, if you say that a relation is not
reflexive, then you need to give an example of an x that shows that the relation is not reflexive. If
you say that a relation is not symmetric, then you need to give an example of an x and a y that
show that the relation is not symmetric. If you say that a relation is not transitive, then you need
to give an example of an x, y, and z that shows that the relation is not transitive.
no
let
 x, y   8,6 
no, let
 x, y, z   1,0,1
2
no
let x  1
Big hint for R4.  y  x  y  2 x   0 is logically equivalent to  y  x  OR  y  2 x 
Page 16 of 150
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
vi) 5  3  A B
vii) 15  A B
viii) 4b  A  A
ix) 4  5  A  A
x)  5,5  A  A
true
true
false
true
true
true
false
false
xi)  4,5  A  A
true
false
xii)  4,5   5, 4 
true
false
true
false
false
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) Consider the cartesian product 2   . What
elements of the cartesian product 2   are
denoted by the dots labelled a,b,c,d in the figure at
right? (In your old jargon, you would have been
asked to give the coordinates of each of the four
points.)
4) Draw a picture to illustrate the cartesian product
2
  , 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)  3.5, 2
(c)  1, 3
(d)  , 2 
y
2
b
a
1
x
1
2
d
c
Page 17 of 150
5) 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.
6) Consider the following relations on the set of real numbers, . Remember that a relation on
the set of real numbers is just a fancy name for a subset of the cartesian plane,  .
Illustrate each relation by drawing it as a subset of the plane. Then decide if the relation is
reflexive, symmetric, transitive.
example
Relation
Reflexive? Symmetric? Transitive?
x Ra y means 2 x  y  1
Ra
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
For the next two exercises, let A be the set of all lines in the Cartesian plane.
7) 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.
8) In this exercise, you will explore three different definitions of ―parallel lines‖. Each
definition can be recast as a relation. You will see that even though all three relations are
spoken as ― l1 is parallel to l2 ‖, they are very different relations, with different properties.
a) Define ―parallel lines‖ to mean two lines that have the same slope. 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 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.
c) 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.
Page 18 of 150
For the next four exercises, refer to
the figure at right.
l1
Let L  l1 , l2 , l3 , l4  be the set of
lines in the figure.
Let P   p1 , p2 , p3 , p4 , p5 , p6  be
the set of dots in the figure.
p2
p4
p3
p5
p6
p1
l2
l3
l4
p7
9) Answer the following true/false questions. If your answer is ―false‖, explain why.
a) l2  l4  L  L
true false
b)
 p5 , l1   L  P
 p4 , l2   P  L true
d)  p4 , l4   P  L true
c)
true
false
false
false
true false
 p5 , l1   P  L
10) Let R   p2 , l1  ,  p4 , l2  ,  p4 , l4  ,  p6 , l4  . Then R is a relation from P to L. Notice that the
e)
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 ?
11) Using relation R from the previous exercise, 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?
12) Using relation R again, we could define another 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 is a line l that both points touch.
a) Is p4Tp6?
b) Is p6Tp4?
c) Is p2Tp2?
d) Is p3Tp3?
Page 19 of 150
2. Axiom Systems
2.1. Introduction
2.1.1. Looking Back at Math 306
In Math 306, you studied conditional statements—statements of the form
If Statement A is true then Statement B is true.
For example, you proved this Theorem:
If n is odd, then n2 is odd.
You saw that the proof of such a conditional statement has the following form:
Proof:
1. Statement A is true. (given)
2. Some statement (with some justification provided)
3. Some statement (with some justification provided)
4. Some statement (with some justification provided)
5. Statement B is true. (with some justification provided)
End of Proof
For example, the proof of the above theorem about odd integers is
Proof:
1.
2.
3.
4.
Suppose that n is odd.
There exists some integer k such that n = 2k + 1. (by 1 and the definition of odd)
n2 = (2k + 1)2 = (2k + 1)(2k + 1) = 4k2 + 4k + 1 = 2(2k2 + 2k) + 1 (arithmetic)
(2k2 + 2k) is an integer (because the set of integers is closed under addition and
multiplication)
5. n2 is odd. (by step 4 and the definition of odd)
End of Proof
In this proof, the only assumption is the given Statement A. The proof does not prove that
Statement A is true; it only proves that if statement A is true, then statement B is also true.
But in most of your proofs in Math 306, certainly in all of the proofs involving basic number
theory, there were actually unspoken assumptions in addition to the explicitly given Statement A.
The unspoken assumptions have to do with the basic properties of integerers. These basic
properties of integers are not statements that can be proven. Rather, they are statements that are
assumed to be true. Statements such as
The set of integers is closed under addition and multiplication
Page 20 of 150
or
There is a special number called 1 with the following properties
 1 is not equal to 0
 for all integers n, n*1 = n
It may seem silly that we are even discussing the above statements. They are so obviously true,
and we’ve known that they are true since grade school math. We never think about the
statements, even though we do integer arithmetic all the time.
Well, the statements can not in fact be proven true (or false). We can’t imagine what we would
do if they were false, though: all of the math we’ve lived with all of our lives would be out the
window.
So we just assume that all the basic properties of the integers are true and we go from there. But
to be totally honest about all the given information in the Theorem about odd numbers, we would
need to write it something like this:
Theorem: If all of the axioms of the integers are true [they would have to be listed explicitly
here] and n is odd, then n2 is also odd.
And the proof would need to look something like this
Proof:
1. Suppose that all of the axioms of the integers are true [they would have to be listed
explicitly here] and n is odd
2. There exists some integer k such that n = 2k + 1. (by 1 and the definition of odd)
3. n2 = k2 + 4k + 1 (arithmetic)
4. k2 + 4k is an integer (because the set of integers is closed under addition and
multiplication)
5. n2 is odd. (by 4 and the definition of odd)
End of Proof
But in Math 306, the emphasis is on learning how to build a proof. That is, you would have been
concerned with steps 2 through 5 of the above proof. Since we all know how to work with the
integers without having to refer to their underlying axioms, the axioms were not mentioned.
2.1.2. Looking Forward
Having studied in Math 306 the building of proofs of basic number theory theorems, a setting
where the underlying axioms could go unmentioned, we will study in Math 330A the building of
proofs of theorems about geometry. In geometry, the underlying axioms are not obvious and
cannot go unmentioned. In fact, Math 330A will primarily be about the axioms, themselves. We
will start by studying very simple axioms systems, in order to learn how one builds proofs based
on an explicit axiom system and to learn some of the basic terminology of axiom systems. Then,
we will turn our attention to developing an axiom system to describe Euclidean geometry.
Page 21 of 150
2.2. Definition of Axiom System
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.
Theorem: If Bob is Blue then Ann is Red.
Then what we really would have proven is the following statement:
Theorem: If ((Elvis is alive) and (Chocolate is the best flavor of ice cream) and (5 = 7) and
(Bob is Blue)) then Ann is Red.
Page 22 of 150
2.3. 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.
2.3.1. 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
of integers. Introduce
an undefined relation R on the set . That is, introduce a subset R   , 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: Axiom System #1
Primitive Relations: relation on the set , 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: Axiom System #1, abbreviated version
Primitive Relations: relation R on the set , spoken ―x is related to y‖
Axioms: 1) 5R7
2) 5R8
3) Relation R is symmetric
4) Relation R is transitive
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.
Page 23 of 150
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
statement of the theorem. (Then we would not really need to state the axiom system separately.)
Theorem: If ((R is a relation on ) and (5R7) and (5R8) and (R is symmetric) and (R is
transitive), then 7R8.
2.3.2. 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: Axiom System #2
Primitive Terms:  ake
 bem
Primitive Relations: relation from the set A of all akes to the set B of all bems spoken ―the ake
is related to the bem‖
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.
Page 24 of 150
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. 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 .
2. Axioms #2 and #3 tell us that there is bijective correspondence between
the sets of two akes and the bems.
3. Therefore, there must be exactly 6 bems.
End of proof
As with the theorem that we proved 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 theorem 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 every ake, there are exactly three bems that the
ake is related to.
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
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.
Page 25 of 150
2.4. 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.
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.
Page 26 of 150



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.
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.
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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.
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.
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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.
2.5. 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.
2.5.1. 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 not possible for all of the axioms to be true.
We will be interested in determining if a given axiom system is consistent or inconsistent. It is
worthwhile to think now about how one would prove that an axiom system is consistent, or how
one would prove that an axiom system is inconsistent.
Suppose that one suspects that an axiom system is consistent and wants to prove that it is
consistent. 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.
Suppose that one suspects that an axiom system is inconsistent and wants to prove that it is
inconsistent. One would be trying to prove that something is not possible. It is not obvious how
that would be done. The key is found in 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.
Suppose we replace the statement p with the statement ~q. Then the contradiction rule becomes
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Contradiction Rule
~ ~ q  c
~ q
In other words,
qc
Contradiction Rule
~ 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.
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
Contradiction Rule
 statement1  statement2   c
 ~ 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.
Contradiction Rule
 statement1  statement2   statementk   c
 ~ 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.
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Axiom System: Axiom System #3, an example of an inconsistent axiom system
Primitive Terms:  ake
 bem
Primitive Relations: relation from the set A of all akes to the set B of all bems spoken ―the ake
is related to the bem‖
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.
4) There are exactly five bems.
Using Axiom System #3, we can prove the following two theorems.
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.
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 for example, if one discards axiom
#3, then it turns out that 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: Axiom System #4, an example of an inconsistent axiom system
Primitive Terms:  ake
 bem
Primitive Relations: relation from the set A of all akes to the set B of all bems spoken ―the ake
is related to the bem‖
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) There are exactly five bems.
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So the problem is not with any one particular axiom. Rather, the problem is with the whole set of
four.
2.5.2. 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
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 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.
We will be interested in determining if a given axiom is dependent or independent. It is
worthwhile to think now about how one would prove that an axiom is dependent, or how one
would prove that an axiom is independent.
Suppose that one suspects that a given axiom is dependent and wants to prove that it is
dependent. To do that, one proves that the statement of the axiom must be true, with a proof that
uses only the other axioms. That is, one stops assuming that the given axiom is a true statement
and downgrades it to just an ordinary statement that might be true or false. If it is possible to
prove the statement is true, using a proof that uses only the other axioms, then the given axiom is
dependent.
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: Axiom System #5
Primitive Terms:  ake
 bem
Primitive Relations: relation from the set A of all akes to the set B of all bems spoken ―the ake
is related to the bem‖
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 first three axioms. They are the axioms from Axiom System #2. And we also
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.
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Suppose that one suspects that a given axiom is independent and wants to prove that it is
independent. To do that, one stops assuming that the given axiom is a true statement and
downgrades it to just an ordinary statement that might be true or false. 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.
Axiom System: Axiom System #6: This is just Axiom System #2 minus the third axiom
Primitive Terms:  ake
 bem
Primitive Relations: relation from the set A of all akes to the set B of all bems spoken ―the ake
is related to the bem‖
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 For any bem, there is exactly one set of two akes, denoted akei , akek  ,
that used to be an
where i  k , such that akei touches the bem and akek touches the bem.
axiom:
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:
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
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
can say that in Axiom System #2, the third axiom is independent.
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.
2.5.3. Completeness
Recall that in Section 2.4, 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. 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 2.5.2, 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.
Question for class: Why does the fact that Bob’s model has six segments and Dan’s model has
only one segment necessarily mean that the two models are not isomorphic?
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The discussion of Bob’s and Dan’s models for axiom system #6 can be generalized to the extent
that it is possible to formulate an alternate wording of the definition of a complete axiom system.
Remember that axiom system #6 has two axioms. Consider a feature that distinguished Bob’s
mode from Dan’s model. One obvious feature is the number of segments. As observed above, in
Bob’s model, there are six segments, but in Dan’s model, there is only one segment. Now
consider the following statement:
There are exactly six segments.
This statement is an independent statement in axiom system #6. That is, there is a model for
axiom system #6 in which the statement is true (Bob’s model) and there is a model for axiom
system #6 in which the statement is false (Dan’s model). The fact that an independent statement
can be made regarding the number of line segments indicates that axiom system #6 does not
sufficiently specify the number of line segments. That is, axiom system #6 is incomplete.
More generally, if it is possible to make a statement regarding the primitive objects in an axiom
system that is an independent statement, then the axiom system is not complete (and vice-versa).
Thus, an alternate way of wording the definition of a complete axiom system is as follows:
Alternate definition of a complete axiom system:
 words: The axiom system is complete
 meaning: all models of the axiom system are isomorphic
 alternate wording: There are no independent statements regarding the primitive objects
Alternate definition of an incomplete axiom system:
 words: The axiom system is incomplete
 meaning: there exist two models of the axiom system that are not isomorphic
 alternate wording: There is an independent statement regarding the primitive objects
We have discussed the fact 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: Axiom System #7: This is Axiom System #6 with an added axiom
Primitive Terms:  ake
 bem
Primitive Relations: relation from the set A of all akes to the set B of all bems spoken ―the ake
is related to the bem‖
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.
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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: Axiom System #8: This is Axiom System #6 with a different axiom added
Primitive Terms:  ake
 bem
Primitive Relations: relation from the set A of all akes to the set B of all bems spoken ―the ake
is related to the bem‖
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.
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2.6. Exercises
The first four problems are about Axiom System #1. This axiom system was introduced in
Section 2.3.1 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 2.3.2, we discussed Axiom System #2, which had undefined terms and relations.
Prove Theorem #2 for Axiom System #2.
6) In Section 2.5.1, 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: Axiom System #9
Primitive Terms:  cet (pronounced “ket”)
 dag
Primitive Relations: relation from the set of all cets to the set of all dags spoken ―the cet is
related to the dag‖
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 , dagk  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.
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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.
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 , dagk 
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
3.1. Introduction
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.
3.1.1. 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  
2
.
That is, a point is an ordered 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
3.1.2. 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: relation from the set of all points to the set of all lines, spoken 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.
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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 an 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 will be 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.
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 3.5, when we study incidence geometry.
3.2. Two Simple Finite Geometries
3.2.1. Three Point Geometry
A finite axiomatic geometry is one that has a finite number of points. Our first example has three.
Axiom System: The Three Point Geometry
Primitive Terms:  point
 line
Primitive Relation: relation from the set of all points to the set 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 almost the same as Axiom System #9, which was
presented in the Exercises of Section 2.6. The differences are minor, just choices of names. In
Axiom System #9, the primitive terms are ake and bem; in the Three-Point Geometry, the
primitive terms are point and line. In Axiom System #9, the primitive relation is spoken the ake
is related to the bem; In the Three-Point Geometry, the primitive relation is spoken the point lies
on the line.
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We will study the following theorems of Three-Point Geometry. Notice that Theorems #1 and #2
for Three Point Geometry are essentially the same as Theorems #1 and #2 for Axiom System #9.
Three Point Geometry Theorem #1: For any two distinct lines, there is exactly one point
that lies on both lines.
You will prove this Theorem in the exercises.
Three Point Geometry Theorem #2: There are exactly three lines.
You will prove this Theorem in the exercises.
Three Point Geometry Theorem #3: For any line, there is exactly one point that does not
lie on the line.
You will study a proof of this theorem in a class drill.
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.
3.2.2. Four Point Geometry
Our next finite axiomatic geometry has four points.
Axiom System: Four Point Geometry
Primitive Terms:  point
 line
Primitive Relations: relation from the set of all points to the set of all lines, spoken the point
lies on the line
Axioms: 1) There are exactly 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 2.3.2.
The differences are minor, just the same differences in choices of names that we encountered
before when we compared the Three Point Geometry to Axiom System #9. Following are two
theorems of Four Point Geometry. Notice that they are almost the same as Theorems #1 and #2
for Axiom System #9.
Four Point Geometry Theorem #1: There are exactly six lines.
You will prove this Theorem in the exercises.
Four Point Geometry Theorem #2: For every point, there are exactly three lines that pass
through the point.
You will prove this Theorem in the exercises.
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3.3. Terminology
3.3.1. 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 a different set. 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. But we won’t introduce very much. We will introduce just one additional term that
will enable us to change the word order when talking about points that lie on lines.
Definition 18 passes through
 words: Line L passes through point P.
 meaning: Point P lies on line L.
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 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
copied verbatim from one such textbook.
Axioms for Three-Point Geometry, copied from a Textbook
1) There exist exactly three distinct points in the geometry.
2) Each two distinct points are on exactly one line.
3) Not all the points of the geometry are on the same line.
4) Each two distinct lines are on at least one point.
In your work, you will encounter a variety of wordings for the undefined relation. Here are four
that we won’t use.
Commonly used expressions that we will not use
1) Line L contains point P.
2) Line L is on point P.
3) Point P is incident upon line L.
4) Line L is incident upon point P.
3.3.2. New Definitions
While we’re in the business of introducing new terminology, we may as well introduce some
more. The next two definitions are self-explanatory.
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Definition 19 intersecting lines
 words: Line L intersects line M.
 meaning: There exists a point P that lies on both lines. (at least one point)
Definition 20 parallel lines
 words: Line L is parallel to line M.
 symbol: L M

meaning: Line L does not intersect line M.
Remember that you have encountered other definitions of parallel. In Exercise 8) of Section 1.3,
you considered three definitions. The third definition presented in that exercise is the one that is
presented here as the definition of parallel lines in axiomatic geometry.
Discussion Question: Of the three definitions of parallel that were presented in Exercise 8) of
Section 1.3, why do you suppose that the third one is the one that is being used here? Would
there be any problem with using either of the other two definitions presented in that exercise?
It’s worth noting that we have essentially introduced three new relations in this section. The
words ―Line 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 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. This is understandable, because often there are usually many defined relations, and to list
them all would be very cumbersome. But it is unfortunate that the defined relations are not listed,
because it makes the defined relations seem less important than the other components of an
axiom system, and because the reader often does not remembe the definitions and must go
looking for them.
Here are two additional definitions, also straightforward.
Definition 21 collinear points
 words: The set of points P1 , P2 ,

meaning: There exists a line L that passes through all the points.
Definition 22 concurrent lines
 words: The set of lines L1 , L2 ,

, Pk  is collinear.
, 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.
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3.3.3. 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. The question is first raised in the
current chapter, in discussions of finite geometries. But it will come up throughout the course.
3.4. 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 from the set of all points to the set 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 about Fano’s geometry.
 Fano’s Geometry Theorem #1: There are exactly seven points and seven lines.
 Fano’s Geometry Theorem #2: Any two lines intersect exactly once.
 Fano’s Geometry is a model of Incidence Geometry (defined below)
 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 from the set of all points to the set 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 about Young’s geometry.
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



Young’s Geometry Theorem #1: There are exactly nine points and twelve lines.
Young’s Geometry Theorem #2: Two lines parallel to a third line are parallel to each other.
Young’s Geometry is a model of Incidence Geometry (defined below)
Young’s axioms are independent.
3.5. Incidence Geometry
3.5.1. Incidence Relations and Axioms of Incidence
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 Point 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 Geometry 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‖. Such a primitive relation could referred to as the incidence relation.
Axioms such as the two above that described the behavior of the incidence 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 referred to as axioms of incidence. This can be confusing.
3.5.2. The Axiom System for Incidence Geometry
To add to the confusion, the following axiom system is usually called Incidence Geometry.
Axiom System: Incidence Geometry
Primitive Terms:  point
 line
Primitive Relations: relation from the set of all points to the set of all lines, spoken the point
lies on the line
Axioms: IA1: There exist three distinct non-collinear points. (at least three)
IA2: For every pair of distinct points, there exists exactly one line that
passes through both points.
IA3: Every line passes through at least two distinct points.
Notice that the axioms for Incidence Geometry are less specific than any axiom system that we
have seen so far. The number of points is not even specified. Even so, notice that axioms IA2
and IA3 do guarantee that the primitive points and lines in Incidence 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.
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 3.1.1 that in analytic geometry, lines behave this way as a consequence of behavior of
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solutions of systems of linear equations. In Section3.1.2, it was mentioned that in axiomatic
geometry, every aspect of the behavior of points and lines will need to be specified by the
axioms. Well, the axioms of Incidence Geometry do in fact guarantee that lines have the
particular behavior we are discussing. The first of the following five theorems articulate this fact.
You will study the proofs of these theorems in a class drill.
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.
3.6. Models of Incidence Geometry
3.6.1. Abstract versus Concrete Models; Relative versus Absolute Consistency
It is interesting to consider the question of whether or not the Axiom System for Incidence
Geometry is Consistent. That is, whether or not it is possible to find a model. Remember that a
model is a successful interpretation. A very simple interpretation of Incidence Geometry is
possible using the Three Point Geometry:
objects in Incidence Geometry  objects in the Three Point Geometry
points  points
lines  lines
relation in Incidence Geometry  relation in Three Point Geometry
the point lies on the line  the point lies on the line
To determine whether or not the interpretation is successful, we use the interpretation to translate
the axioms of Incidence Geometry into statements in Three Point Geometry, and then consider
whether or not the resulting statements are true in Three Point Geometry.
axioms of Incidence
True?
 statements in Three Point Geometry
Geometry
IA1) There exist three
Statement 1: There exist three distinct true because of
distinct non-collinear

non-collinear points. (at least three)
Three-Point Axiom 1
points. (at least three)
IA2) For every pair of
Statement 2: For every pair of distinct
distinct points, there exists
true because of Three
 points, there exists exactly one line
exactly one line that passes
Point Axiom 2
that passes through both points.
through both points.
IA3) Every line passes
true because of Three
Statement 3: Every line passes
through at least two
Point Geometry

through at least two distinct points.
distinct points.
Theorem 4
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The table above demonstrates that Three Point Geometry can provide a successful interpretation
of Incidence Geometry. That is, Three Point Geometry can be a model of Incidence Geometry.
This demonstrates that Incidence Geometry is consistent. Sort of…
The concept of consistency seemed to be about demonstrating that the words of the axiom
system could be interpreted as actual, concrete things. It is a little unsatisfying that we have only
demonstrated that the words of Incidence Geometry can be interpreted as other words from
another axiom system. That’s a little bit like paying back an I.O.U. with another I.O.U. It would
be more satisfying if we had an interpretation involving actual, concrete things. Here’s one that
uses a picture:
objects in Incidence Geometry  objects in the picture at right
points  dots
lines  line segments
relation in Incidence Geometry  relation in the picture at right
the point lies on the line  the dot touches the line segment
To determine whether or not the interpretation is successful, we use the interpretation to translate
the axioms of Incidence Geometry into statements about the picture, and then consider whether
or not the resulting statements about the picture are true.
axioms of Incidence Geometry
statements about the picture
True?

IA1) There exist three distinct nonStatement 1: There exist three distinct nontrue

collinear points. (at least three)
collinear dots. (at least three)
IA2) For every pair of distinct
Statement 2: For every pair of distinct dots,
points, there exists exactly one line
true
 there exists exactly one line segment that
that passes through both points.
passes through both dots.
IA3) Every line passes through at
Statement 3: Every line segment passes
true

least two distinct points.
through at least two distinct dots.
There is terminology that applies to the above discussion
Definition 23 Abstract Model and Concrete Model
 An abstract model of an axiom system is a model that is, itself, another axiom system.
 A concrete model of an axiom system is a model that uses actual objects and relations.
The Three Point Geometry is an example of an abstract model for Incidence Geometry. The
picture with three dots and three line segments is an example of a concrete model.
Definition 24 Relative Consistency and Absolute Consistency
 An axiom system is called relatively consistent if an abstract model has been
demonstrated.
 An axiom system is called absolutely consistent if a concrete model has been
demonstrated.
It is possible for an axiom system to be both relatively consistent and absolutely consistent. The
fact that Three Point Geometry is a model for Incidence Geometry merely proves that Incidence
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Geometry is Relatively Consistent.We would say that Incidence Geometry is absolutely
consistent because of the model involving the picture with three dots.
In the exercises, you will prove that the Four Point Geometry, Fano’s Geometry, and Young’s
Geometry are also models of incidence geometry.
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3.7. Exercises
The first six exercises deal with the Three Point Geometry that was introduced in Section 3.2.1.
In the reading, it was pointed out that Three Point Geometry is just Axiom System #9 from
the Exercises of Section 2.6, but with some of the wording changed. In those exercises, you
proved a bunch of facts about Axiom System #9. All proofs of facts about Axiom System #9
can be recycled, with their wording changed, into proofs of corresponding facts about the
Three Point Geometry. The first three exercises are of that type.
1) Prove Three Point Geometry Theorem #1.
2) Prove Three Point Geometry Theorem #2.
3) Are the Three Point Geometry axioms independent? Explain.
The remaining three exercises about Three Point Geometry are not simply adaptations of
facts from Axiom System #9. They are new.
4) Prove Three Point Geometry Theorem #4.
5) In the Three Point Geometry, do parallel lines exist? Explain.
6) In the Three Point Geometry, what is the answer to The Big Question? Explain.
The next six exercises deal with the Four Point Geometry that was introduced in Section 3.2.2.
In the reading, it was pointed out that Four Point Geometry is basically just Axiom System
#2 from Section 2.3.2, but with some of the wording changed. All proofs of facts about
Axiom System #2 can be recycled, with their wording changed, into proofs of corresponding
facts about the Four Point Geometry.
7) Prove Four Point Geometry Theorem #1. (Hint: translate the proof of Theorem #1 of
Axiom System #2.)
8) Prove Four Point Geometry Theorem #2. (Hint: translate the proof of Theorem #2 of
Axiom System #2.)
9) Are the Four Point Geometry axioms independent? Explain. (Hint: see your class
notes from Thursday, April 10. There, I discussed the corresponding question about
Axiom System #2.)
The remaining three exercises about Four Point Geometry are not simply adaptations of facts
from Axiom System #2. They are new.
10) Prove that the Four Point Geometry is a model of Incidence Geometry.
11) Prove that in the Four Point Geometry, parallel lines exist.
12) In the Four Point Geometry, what is the answer to The Big Question? Explain.
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The next nine exercises explore Fano’s Geometry, introduced in Section 3.4.
13) Give a model for Fano’s geometry that interprets points as dots and interprets lines as
curves.
14) Give a model for Fano’s geometry that interprets points as letters and interprets lines as
sets.
15) Are Fano’s axioms consistent? Explain.
16) Prove Fano’s Geometry Theorem #1: There are exactly seven points and seven lines.
17) Prove Fano’s Geometry Theorem #2: Any two lines intersect exactly once.
18) Prove that Fano’s Geometry is a model of Incidence Geometry.
19) Prove that Fano’s axioms are independent. (All the steps are easy except for one killer.)
20) Are there parallel lines in Fano’s Geometry? Explain.
21) In Fano’s Geometry, what is the answer to The Big Question? Explain.
The next eight exercises explore Young’s Geometry, introduced in Section 3.4.
22) Give a model for Young’s geometry that interprets points as letters and interprets lines as
sets.
23) Are Young’s axioms consistent? Explain.
24) Prove Young’s Geometry Theorem #1: There are exactly nine points and twelve lines.
25) Prove Young’s Geometry Theorem #2: Two lines parallel to a third line are parallel to
each other. Hint: Do a proof by contradiction, involving axiom #5.
26) Prove that Young’s Geometry is a model of Incidence Geometry.
27) Prove that Young’s axioms are independent.
28) Are there parallel lines in Young’s Geometry? Explain.
29) In Young’s Geometry, what is the answer to The Big Question? Explain.
Recall that incidence geometry was presented in Section 3.5.2.
30) Explain why each of the pictures below could not be an incidence geometry.
picture (a)
picture (b)
picture (c)
picture (d)
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4. Models of Incidence Geometry Whose Sets of Points are Infinite
In the axiom systems for Three Point Geometry, Four Point Geometry, Fano’s Geometry, and
Young’s Geometry, there were axioms that specified a certain number of points. Nothing in the
axioms for Incidence Geometry specifies the number of points and, as a result, there are models
of Incidence Geometry with many different numbers of points. As varied as they may be,
however, each of the models that we studied in the previous chapter had a finite number of
points. It is natural to wonder if Incidence Geometry has an infinite model—one whose set of
points is an infinite set.
4.1. A Model of Incidence Geometry Involving Straight-Line Drawings
Consider the following interpretation of Incidence Geometry as ―straight-line drawings‖.
objects in Incidence Geometry
points
lines
relation in Incidence Geometry
the point lies on the line





objects in straight-line drawings
drawn dots (drawn with a pencil or computer)
drawn lines (drawn with a ruler or computer)
relation in straight-line drawings
the drawn dot touches the drawn line
Is this interpretation successful? That is, is it a model? As always, we consider what happens
when we translate the axioms of Incidence Geometry into statements about straight-line
drawings.
axioms of Incidence Geometry
IA1) There exist three distinct noncollinear points. (at least three)
IA2) For every pair of distinct
points, there exists exactly one line
that passes through both points.
IA3) Every line passes through at
least two distinct points.

statements about straight-line drawings
Statement 1: There exist three distinct non
collinear drawn dots. (at least three)
Statement 2: For every pair of distinct drawn
 dots, there exists exactly one drawn line that
touches both drawn dots.
Statement 3: Every drawn line passes

through at least two distinct drawn dots.
True?
yes
no
yes
There are problems with this interpretation. Whether drawing on paper with a pencil or pen, or
on a chalkboard with chalk, or on a computer screen, a dot will have some breadth, and a line
will have some width as well as length. In extreme cases, the breadth of a point or the width of a
line will cause violations of the Incidence Axioms. For example, in the picture below, drawn line
L is meant to touch drawn dots A, B, and C. Line M is meant to touch drawn dots A and D. But
because of the thickness of the dots and lines, line M also touches dot B. So there are two lines
that touch dots A and B. Thus, Statement 2 is false.
line M
D
A
line L
C
B
Of course the solution to the problem with the above drawing would be to simply reduce the line
width, maybe from 1.5 to 1, and reduce the dot size. But even then, a smaller drawing could that
would still cause Statement 2 to be false.
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As an alternative, we could continue to use actual drawn dots and drawn lines, but think of them
as crude versions of idealized imaginary drawings. In those imaginary drawings, dots would
have no breadth, and lines would have no width. And while we’re at it, in our imaginary
drawings, the flat sheet of paper would continue on forever in all directions. These imaginary,
idealized straight-line drawings would be a model of incidence geometry. And note that the set
of all points in such a drawing is an infinite set.
4.2. A Model of Incidence Geometry Involving the Usual Analytic Geometry
Consider the following interpretation of Incidence Geometry as the usual Analytic Geometry of
the (x,y) plane.
objects in Incidence Geometry  objects in Analytic Geometry
points  (x,y) pairs where x and y are real numbers
sets of the form L   x, y  : ax  by  c , where a, b, and
lines 
c are real number constants and x and y are real variables.
relation in Incidence Geometry  relation in Analytic Geometry
the pair (d,e) is an element of the set  x, y  : ax  by  c
the point lies on the line 
.
Is this interpretation successful? That is, is it a model? As always, we consider what happens
when we translate the axioms of Incidence Geometry into statements about Analytic Geometry.
axioms of Incidence Geometry
IA1) There exist three distinct noncollinear points. (at least three)
IA2) For every pair of distinct
points, there exists exactly one line
that passes through both points.

statements about Analytic Geometry
Statement 1: There exist three distinct pairs, P =
(x1,y1), Q = (x2,y2), R = (x3,y3) such that there is
 no one set L of the form L   x, y  : ax  by  c


such that P  L and Q  L and R  L .
Statement 2: For every pair of distinct pairs P =
(x1,y1) and Q = (x2,y2), there exists exactly one set
 L of the form L   x, y  : ax  by  c such that


True?
yes
yes
P  L and Q  L .
Statement 3: For every set L of the form
L   x, y  : ax  by  c , there exist at least two
IA3) Every line passes through at
yes

least two distinct points.
distinct pairs P = (x1,y1) and Q = (x2,y2), such that
P  L and Q  L .
The three statements are true in Analytic Geometry. (We know the statements are true because of
the behavior of solutions of linear equations in two unknowns x and y.) So the usual Analytic
Geometry of the (x,y) plane is a model of Incidence Geometry. And notice that it is a model
whose set of points is an infinite set.
4.3. A Model of Incidence Geometry Involving the Poincare Disk
Our third interpretation of Incidence Geometry is another Analytic Geometry. That is, points will
be represented in some way by real numbers and lines will be certain kinds of sets of points.
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Before introducing the interpretation, 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.
y (0,1)
Definition 25 Hpoint
 word: Hpoint
 meaning: an  x, y  pair in the interior of the unit
circle
(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 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.
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two intersection points with
two intersection points with
perpendicular tangent lines
non-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.
one intersection point
Definition 26 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.
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. .
objects in Incidence Geometry  objects in the Poincare disk
points  Hpoints
lines  Hlines
relation in Incidence Geometry  relation in the Poincare disk
the point lies on the line  the Hpoint is an element of the Hline
Is this interpretation successful? That is, is it a model? As always, we consider what happens
when we translate the axioms of Incidence Geometry into statements about the Poincare Disk.
axioms of Incidence Geometry
IA1) There exist three distinct noncollinear points. (at least three)

statements about the Poincare Disk
True?
Statement 1: There exist three distinct Hpoints,
 P, Q, and R such that there is no one Hline L such yes
that P  L and Q  L and R  L .
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IA2) For every pair of distinct
points, there exists exactly one line
that passes through both points.
IA3) Every line passes through at
least two distinct points.
Statement 2: For every pair of distinct Hpoints, P
 and Q, there is exactly one Hline L such that
P  L and Q  L .
Statement 3: For every Hline L, there exist at
 least two distinct Hpoints P = (x1,y1) and Q =
(x2,y2), such that P  L and Q  L .
yes
yes
It should come as no surprise that Statements 1 and 3 are true. But what about Statement 2? It
turns out that given two Hpoints A and B, there is exactly one Hline containing both points. (We
would use the symbol AB to denote the Hline, being sure to remember that the meaning of this
symbol in the context of the Poincare disk is different from the meaning of the same symbol in
the context of the usual Analytic Geometry of the plane.) This can be shown by some rather
messy mathematics involving lines and circles. In class, in the exercises, and in the computer lab,
you will explore the Poincare disk by drawing examples of Hpoints and Hlines. 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.
4.4. Other Interpretations of Incidence Geometry
In a class drill, we will investigate a variety of interpretations of Incidence Geometry. Some of
the interpretations will involve the sphere. These, we will investigate using the ―Lenart Sphere‖.
For each interpretation, the goal will be to determine whether or not the interpretation is
successful. That is, to determine whether or not the interpretation is a model.
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4.5. Exercises
Recall that the Poincare disk interpretation of incidence geometry was presented in Section 4.3.
[1] For each picture, draw Hline PQ . The center of the circle is marked with a small cross for
reference.
Q
Q
P
P
Q
P
P
Q
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[2] On the labeled circle below
 Draw Hline AB . Label this Hline L.
 Draw Hline PQ . Label this Hline M.
 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.
[3] On the labeled circle below, draw Hlines AB , BC , and CA .
[4] On the labeled circle below, draw Hlines AB , BC , and CA .
[5] On the labeled circle below, draw Hlines AB , BC , CA , and CD .
C
R
P
Q
B
A
A
B
Exercise [2]
Exercise [3]
C
B
A
B
D
A
C
Exercise [4]
Exercise [5]
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[6] Below are eight interpretations of Incidence Geometry. For each interpretation, do the
following:
i) Determine whether or not the interpretation is a model. Explain your answers.
ii) If the interpretation is a model, are there any parallel lines in the model? Explain.
iii) If the interpretation is a model, what is the answer to THE BIG QUESTION?
Interpretation (a): ordered pairs and straight lines in the interior of the unit circle
objects in Incidence Geometry  objects in Interpretation (a)
point  an (x,y) pair in interior of unit circle
the set L consisting of the portion of an ordinary line that
line 
lies in the interior of the unit circle.
relation in Incidence Geometry  relation in Interpretation (a)
The point lies on the line.  The pair (x,y) is an element of the set L.
Interpretation (b): ordered pairs and straight lines through the origin
objects in Incidence Geometry  objects in Interpretation (b)
point  an (x,y) pair in the plane
line  an ordinary line L through the origin in the plane.
relation in Incidence Geometry  relation in Interpretation (b)
The point lies on the line.  The pair (x,y) is an element of the set L.
Interpretation (c): lines and planes in three-space
objects in Incidence Geometry 
point 
line 
relation in Incidence Geometry 
The point lies on the line. 
objects in Interpretation (c)
an ordinary line M in three-space.
an ordinary plane T in three-space.
relation in Interpretation (c)
The set M is a subset of the set T.
Interpretation (d): lines through the origin and planes through the origin in three-space
objects in Incidence Geometry  objects in Interpretation (d)
point  an ordinary line M through the origin in three-space.
line  an ordinary plane T through the origin in three-space.
relation in Incidence Geometry  relation in Interpretation (c)
The point lies on the line.  The set M is a subset of the set T.
Interpretation (e): Ordered pairs and circles in the plane
objects in Incidence Geometry  objects in Interpretation (e)
point  an (x,y) pair in the plane
an ordinary circle C in the plane
line 
(just the circle, not the interior)
relation in Incidence Geometry  relation in Interpretation (e)
The point lies on the line.  The pair (x,y) is an element of the set C.
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Interpretation (f): Ordered pairs and circles centered at the origin in the plane
objects in Incidence Geometry  objects in Interpretation (f)
point  an (x,y) pair in the plane
an ordinary circle C in the plane, centered at the origin
line 
(just the circle, not the interior)
relation in Incidence Geometry  relation in Interpretation (f)
The point lies on the line.  The pair (x,y) is an element of the set C.
The next two interpretations involve the sphere. By sphere, we mean the surface of a ball of
radius 1. That is,
 The ball is the set B   x, y, z  : x 2  y 2  z 2  1 .

The sphere is the set S   x, y, z  : x 2  y 2  z 2  1 .
Interpretation (g): points and great circles on the sphere
objects in Incidence Geometry  objects in Interpretation (g)
an ordinary point on the sphere
point 
that is, a triple (x,y,z) such that x 2  y 2  z 2  1
a great circle C on the sphere
line 
(that is, a circle of radius 1 on the sphere)
relation in Incidence Geometry  relation in Interpretation (g)
The point lies on the line.  The triple (x,y,z) is an element of the set C.
Interpretation (h): sets of antipodal points and great circles on the sphere
objects in Incidence Geometry  objects in Interpretation (h)
a set consisting of two antipodal points on the sphere
point  (that is, a set P   x, y, z  ,   x,  y,  z  where (x,y,z) is
a point on the sphere)
a great circle C on the sphere
line 
(that is, a circle of radius 1 on the sphere)
relation in Incidence Geometry  relation in Interpretation (h)
The point lies on the line.  The set P is a subset of the set C.
Page 60 of 150
Page 61 of 150
5. Incidence and Betweenness Geometry
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 because our ―straight line‖ drawings and the usual Analytic Geometry of
the x-y plane behave the same way, we would expect that the usual Analytic Geometry would be
a model. But we want those to be the only models of our axiom system. Or rather, we want any
other models of the axiom system to be isomorphic to the Analytic Geometry Model. That is, we
would like our axiom system to be complete.
Complete
Axiom
System
our goal:
model
Straight
Line
Drawings
model
isomorphic
Ordinary Analytic
Geometry of
the x-y plane
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 list of Incidence Geometry 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 take the first
step towards our goal of building a complete axiom set by adding a set of axioms that insure that
any model will have an infinite number of points. (The axioms specify other kinds of behavior,
as well.) Much of the material is follows Marvin Greenberg’s book, Euclidean and NonEuclidean 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.
5.1. 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. (Notice that this terminology was introduced in 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
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Exercise 8).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 27 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
For example, you are undoubtedly familiar with the following ternary relation on the set of real
numbers.
Definition 28 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 introduce a new ternary relation that will also be called The Betweenness Relation, but it
will be a ternary relation on the set of points and it will be an undefined relation.
5.2. 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, among other things, 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 Incidence and
Betweenness Geometry. It is presented on the next page.
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.
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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 stuff: ―passes through‖, ―intersect‖, ―parallel‖, ―concurrent‖, ―collinear‖, ―same
side‖, ―opposite side‖
Axioms: Incidence Axioms
IA1: There exist three distinct non-collinear points. (at least three)
IA2: For every pair of distinct points, there exists exactly one line that
passes through both points.
IA3: Every line passes through at least two distinct points.
Betweenness Axioms
BA1: If A*B*C, then A, B, and C are distinct collinear points, and
C*B*A.
BA2: 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.
BA3: 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.
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.
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 3.5 will be continued as we
continue to build on the axioms of Incidence Geometry in this section and throughout the rest of
this book.
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 3.5.)
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 justify the steps in a proof of this theorem in a class drill.)
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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.
5.3. Line Segments and Rays
Our first three definitions make reference only to concepts from Incidence Geometry.
Definition 29 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. But 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 30 the set of points that lie on a line
 

symbol: AB

meaning: the set of points that lie on line AB
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 31 the plane
 meaning: the set of all points
In this course, we are studying ―plane geometry‖. Our models are the straight-line drawings that
we have been making all of our lives, and the usual analytic geometry of the x-y plane. Both of
these models are 2-dimensional things. The axiom system that we develop will be for plane
geometry. In an axiom system 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 3-dimensional sculptures. In space geometry, points, lines, and planes are all
undefined, and space is defined to be the set of all points. We are not studying the geometry of
space. So to reiterate, in our geometry points and lines are undefined, and the plane 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 32 line segment, endpoints of a line segment
 symbol: AB
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


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 33 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 34 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.
The proof of this theorem is short and gives a taste of the combined use of axioms and
definitions. See if you can justify the steps:
Proof of Theorem 7
1) Let QR be a ray.
2) There exists a points P such that P*Q*R (by ______________________________)
3) Rays QP and QR are opposite rays. (by _________________________________)
End of Proof
Theorem 8 In I&B Geometry, for any two distinct points A and B, AB BA  AB .
(The proof of this theorem is not particularly hard, but it is not particularly illuminating, either.
So we will omit the proof and assume the theorem as given.)
Theorem 9 In I&B Geometry, for any two distinct points A and B, AB
 
BA  AB .
(The proof of this theorem is hard and not particularly illuminating. So we will omit the proof
and assume the theorem as given.)
5.4. Plane Separation
Keep in mind that with the axioms, we are trying to prescribe behavior for our points and lines
(undefined terms) 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
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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 and
definitions in logical, chronological order, we would put these definitions after axiom BA3,
before axiom BA4.
Definition 35 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 36 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. One simple proof of the theorem uses the
technique of proof by contradiction.
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.
(You will justify the steps in a proof of this theorem in the exercises.)
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 37 half plane
 words: half-plane bounded by L, containing point A.
 symbol: symbol: HA
 usage: L is a line and A is a point not on L.
 meaning: the set of points that are on the same side of L as point A.
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.
It is difficult to write a proof of Theorem 11 that shows enough detail to be rigorous but not so
many that it becomes difficult to read. A proof that I have written uses the method of proof by
division into cases. You will justify the steps of the proof in a class drill.
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5.5. 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 assume it as given.)
Theorem 13 In I&B Geometry, if A*B*C and B*C*D, then A*B*D and A*C*D.
(You will justify the proof of this theorem in the exercises.)
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 assume it as 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,
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
segment AC or segment BC . Furthermore, if C does not lie on L, then L does not
intersect both segment AC and segment BC ..
(You will justify a proof of this theorem in a class drill.)
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 .
(This theorem will not be proved in this course. We will assume it as given.)
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 assume it as 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 justify the steps in a proof of this theorem in the exercises.)
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5.6. Angles and Triangles
With the notion of rays and line segments, we are able to define angles and triangles. In this
section, we will define them and explore some of their simpler behavior.
Definition 38 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.
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 39 supplementary angles
 words: supplementary angles
 meaning: two angles that share a common side and whose other sides are opposite rays.
We won’t use 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 a later chapter.
Definition 40 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).
Our first theorem about the behavior of angles states something that is fairly easy to visualize
and not difficult to prove.
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 justify the steps in a proof of this theorem in a class drill.)
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The next theorem about the behavior of angles also states something that is fairly easy to
visualize. But the proof of the theorem is messy and not particularly illuminating, so we will
omit the proof
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.)
Remember that in Incidence and Betweenness Geometry, the betweenness relation for points is a
primitive (undefined) relation. What is interesting is that there is also a betweenness relation for
rays and that it is not a primitive relation: it is defined in terms of other stuff. The definition
follows.
Definition 41 ray between two other rays
 words: ray BD is between BA and BC .
 meaning: Point D is in the interior of ABC .
The next theorem about the behavior of angles is famous enough that it shows up in some
elementary Geometry books.
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 justify the proof of this theorem in the exercises.)
Our final three definitions for this section have to do with triangles.
Definition 42 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 angles ABC , BCA , and CAB are called the angles of the triangle.
o The angle ABC is called angle B when there is no chance of this causing
confusion. Similarly for the other angles.
Definition 43 interior of a triangle
 words: the interior of ABC
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
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 44 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.
Notice that a triangle and the interior of a triangle and the exterior of a triangle are all defined as
sets of points.
The final three theorems of the section have to do with basic facts about the intersection of rays
and lines with triangles. All three theorems are easy to visualize.
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
.
(We won’t prove this theorem in this class, but will take it as a given.)
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.
(We won’t prove this theorem in this class, but will take it as a given.)
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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5.7. Exercises
[1] Recall the statement of 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.
Justify the steps in the follwing proof:
Proof.
1. Let A, B, and C be any points such that A and B are on opposite sides of line L, and B and
C are on the same side of L.
2. Suppose that A and C are not on opposite sides of L. (_______________________)
3. None of the points A, B, or C lie on line L. (___________________________)
4. Points A and C are on the same side of L. (___________________________________)
5. Points A and B are on the same side of L. (___________________________________)
6. But (____________________) contradicts (__________________) Therefore, our
assumption must be wrong. Conclude that A and C must be on opposite sides of L.
End of Proof
[2] Recall the statement of Theorem 13: In I&B Geometry, if A*B*C and B*C*D, then A*B*D
and A*C*D
Justify the steps in the following proof
Step 1: Show that A*C*D. (Note that we will create a line through C)
1. Suppose that A*B*C and B*C*D.
2. A, B, C, and D are distinct collinear points. (by ________________) Let L be the line
that passes through all four points.
3. There exists a point E not on the line L. (by ____________________)
4. point C is an element of segment BD (because___________) and point C also lies on
line EC and. So segment BD intersects line EC .
5. Points B and D are on opposite sides of line EC . (by ______________________)
6. Point C is the only point of intersection of line EC and line L. (by _________________)
7. Point C is not an element of segment AB , because of the fact that C ≠ A and C ≠ B (by
_______) and and the fact that C is not between A & B. (by _______________________)
8. Segment AB does not intersect line EC . (by ______________)
9. Points A and B are on the same side of line EC . (by ___________________________)
10. Points A and D are on opposite sides of line EC . (by ____________________________)
11. Segment AD intersects line EC at a point between A and D. (by
___________________)
12. The point of intersection of segment AD and line EC must be point C. (by _________)
13. A*C*D. (by ____________________________)
Step 2: Use a previous theorem to show that A*B*D
14. A*B*D (by ____________________________)
End of Proof
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[3] Recall the statement of 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.
Justify the steps in the following proof:
1. Suppose that point A lies on line L and point B does not lie on line L, and let P be any
point such that P  AB and P  A .
2. Either A*P*B or P = B or A*B*P. (by ________________________________)
3. Assume that P is on the opposite side of line L from point B. (________________)
4. Segment PB intersects L at a point between P and B. (by ________________________)
5. Point A is an intersection point of line L and ray AB . (by given information)
6. Line L and line AB can’t have any other intersection points. (by ___________________)
7. Point A must be the intersection point of segment PB and line L. (by 4 and 6 and fact that
points P and B lie on line AB .)
8. Point A is between points P and B. That is, P*A*B. (by __________________________)
9. Step 8 contradicts step (___________). So our assumption in step 3 was wrong. Point P
cannot be on the opposite side of line L from point B. Therefore, P must be on the same
side of L as B.
End of Proof
[4] Recall the statement of Theorem 21 (The Crossbar Theorem): In I&B Geometry, if ray AD
is between rays AB and AC then ray AD intersects segment BC .
Justify the steps in the following proof:
1. Suppose ray AD is between rays AB and AC .
2. Let E be a point such that C*A*E (we know such a point exists by _____________)
3. Segment CE . intersects AD at point A, which is between C and E (by __________).
4. Points E and C are on opposite sides of AD . (by ____________________________)
5. Point B is in the interior of DAE (by Theorem _________)
6. Point B is on the same side of AD as E is. (by definition of __________________)
7. Point B is on the opposite side of AD from point C. (by 4, 6, and Theorem ______)
8. Segment BC intersects AD at a point between points B and C. (by ______________)
Let G be the point of intersection of segment BC and line AD . So B*G*C.
9. Point G is in the interior of CAB . (by step ____ and Theorem _______)
10. Point G is on ray AD , and not on the ―opposite ray‖. (by Theorem _______)
11. Ray AD intersects segment BC . (by steps _____ and _______)
End of Proof
[5] Prove Theorem 24: In I&B Geometry, a line cannot be contained in the interior of a triangle.
Hint: Do a proof by contradiction. Suppose that a line L is contained in the interior of a triangle.
Show that it is possible to reach a contradiction of an earlier theorem.
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6. Neutral Geometry I
6.1. 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 some 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 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 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 probably won’t be obvious to the reader 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 need to
include them as axioms.
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 are not the only model of this geometry: the
drawings of Hpoints and Hlines in the Poincare disk are also 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. 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 .
Incidence Axioms
IA1: There exist three distinct non-collinear points. (at least three)
IA2: For every pair of distinct points, there is exactly one line that passes through both points.
IA3: Every line passes through at least two distinct points.
Betweenness Axioms
BA1: If A*B*C, then A, B, and C are distinct collinear points, and C*B*A.
BA2: 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.
BA3: 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.
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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6.2. Line Segment Subtraction and the Ordering of Segments
Notice that numbers have played very little role in our geometry so far. In particular, we have not
introduced the notion of the length of a line segment. In Incidence & Betweenness Geometry, we
found 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 Neutral Geometry, 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. But as was the case with Incidence & Betweenness Geometry, we will find
that it is possible to do things in Neutral Geometry that we might have thought impossible
without the notion of length. We are able to do this because of the Segment Construction Axiom,
CA2, and the Segment Addition Axiom, CA3. These axioms are statements that would
automatically be true in a geometry with a notion of length. But because we have no notion of
length in our geometry, the statements must be made axioms if we want them to be true. What is
remarkable, though, is that there are many other statements about segments that would
automatically be true in a geometry with a notion of length, but we don’t have to make them all
axioms in order for them to be true. It turns out that with just axioms CA1, CA2, and CA3, we
can prove many other statements as theorems.
For example, the following theorem is a statement very similar to the segment addition axiom,
CA3. But this statement does not need to be included in the list of axioms. (That’s why it is here,
listed as a theorem.) The proof is fairly simple. You will provide the justifications in a homework
exercise.
Theorem 25 (segment subtraction) In Neutral Geometry, if A*B*C and D*E*F and AB  DE
and AC  DF , then BC  EF .
The following theorem proves aanother claim that would be obvious for drawn line segments,
but which is a little tricky to prove for abstract segments. You will justify the steps in the proof
in a class drill.
Theorem 26 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 .
The following definition is very important because introduces a way to compare two line
segments without a having a notion of length.
Definition 45 the order relation on the set of line segments
 Symbol: AB  CD
 Spoken: “Segment A,B is less than segment C,D.”
 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.
It turns out that the order relation for line segments has behavior that is entirely analogous to the
behavior of line segment length, even though the order relation does not use the notion of length.
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That behavior is spelled out in the following theorem. You will justify the steps in a proof of this
theorem in a homework exercise.
Theorem 27 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 .
6.3. Triangle Congruence and its Role in the Neutral Geometry Axioms
As mentioned in Section 6.1, 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.
A remark: We won’t prove any theorems about triangles until the next chapter. The current
chapter deals only with theorems about lines, rays, line segments, and angles. It is natural to
wonder, then, why the current section on triangles is even included in this chapter. The reason is
that some of the theorems about lines, rays, line segments, and angles that we prove in this
chapter use triangles and triangle congruence.
6.3.1. 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 46 ―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 .
Page 77 of 150

Machine Diagram:
a
f
input

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
1) Let f :

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 :  .
2) Let S1   A, B, C , D, E and S 2   L, M , N , O, P . Define a function f : S1  S2 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  S2 .
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
Page 78 of 150
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 47 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
1)  A, B, C    D, E, F 
2)
3)
4)
ABC and DEF .
 A, B, C    D, F , E 
 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 48 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.
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
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 .
6.3.2. 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 49 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.”
 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
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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
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 ‖.
6.3.3. 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
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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.
 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.
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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.
6.4. Right Angles
In Section 6.2, 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 supplemental angles and right
angles and to define an order relation on the set of angles.
Remember that the term supplementary angle was introduced in Definition 39, in Section 5.6.
What was interesting about that definition was that it does not say that supplementary angles are
angles whose measures add up to 180. It is worth noting that every angle has a supplementary
angle. You will prove this fact in the exercises.
In the exercises, you will justify the steps in the proof of the following theorem.
Theorem 28 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 .
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. You will justify a proof of the theorem in the exercises.
Definition 50 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.
Theorem 29
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
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measure is 90 ! (We do not even have a notion of angle measure.) The trick is to define right
angles in terms of supplementary angles, as follows.
Definition 51 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 justify a
proof of it in the exercises.
Theorem 30
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 52 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 next two theorems are extremely important in Neutral Geometry. You will justify the steps
of the proofs in class drills. 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.
Theorem 31 In Neutral Geometry, for every line L and every point P not on L there exists a
line through P perpendicular to L.
Theorem 32 In Neutral Geometry, for every line L and every point P on L there exists a unique
line through P perpendicular to L.
6.5. 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. The proofs of the theorems are not terribly
hard, but they would not enhance our understanding very much. (They are in the same spirit of
mind-numbing attention to detail that we found in the proofs of the theorems about segment
subtraction and ordering of line segments.) So we will not discuss the proofs of these theorems,
and will just assume them as given.
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Theorem 33 (angle addition)
Given: Neutral Geometry; angle ABC with point D in the interior; angle
point H in the interior; DBA  HFE ; DBC  HFG
Claim: ABC  EFG
EFG with
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.
Theorem 34 (Angle Subtraction).
Given: Neutral Geometry; angle ABC with point D in the interior; angle
point H in the interior; DBA  HFE ; ABC  EFG
Claim: DBC  HFG
EFG with
The following theorem about rays and angles is analogous to Theorem 26, which was about
segments.
Theorem 35 (Theorem about Rays and Angles) In Neutral Geometry, if ABC  EFG and
ray BD is between rays BA and BC , then there is a unique ray FH between rays FE
and FG such that ABD  EFH .
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. But 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.
Definition 53 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 36 (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 ,
(b): If P  Q and Q  R then P  R .
(c): (transitivity): If P  Q and Q  R then P  R .
P
Q , or
Q
P.
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Now that we have the order relation on the set of angles, it is possible to prove one important
remaining theorem about right angles.
Theorem 37
(Euclid’s 4th postulate). All right angles are congruent to each other.
6.6. The Alternate Interior Angle Theorem and Some Corollaries
In this final section of the chapter, we will study a major theorem called the Alternate Interior
Angle Theorem. It is important because it enables us to prove a number of important theorems,
including one that gives a partial answer to THE BIG QUESTION.
In order to understand the wording of the Alternate Interior Angle Theorem, we need two
definitions.
Definition 54 ―transversal‖
 Words: “Line T is transversal to lines L and M.”
 Meaning: “T intersects L and M in distinct points.”
Definition 55 ―alternate interior angles‖, ―corresponding angles‖
 Usage: Lines L, M, and transversal T are given.
 Labeled points: Let B be the
H T
intersection of lines T and L, and let E
D
be the intersection of lines T and M.
M
E
F
(By definition of transversal, B and E
A
B
C
L
are not the same point.) By the
G
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.
 Meaning:
ABE and FEB is a pair of alternate interior angles.


CBE and DEB is a pair of alternate interior angles.

ABG and DEG is a pair of corresponding angles.
ABH and DEH is a pair of corresponding angles.


CBG and FEG is a pair of corresponding angles.

CBH and FEH is a pair of corresponding angles.
Theorem 38 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 39 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 40
In Neutral Geometry, two distinct lines perpendicular to the same line are parallel.
Remember that Theorem 31 tells us that for any line L and any point P not on L, there exists a
line that passes through P and is perpendicular to L. To say ―there exists a line‖ means ―at least
one line‖. Nothing in the statement or proof of that theorem said anything about the possibility of
there being more than one such line. The following theorem states that there cannot be more than
one line.
Theorem 41 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 31.)
Notice that none of the Neutral Geometry axioms say anything about THE BIG QUESTION.
Even so, the following theorem proves that the answer is ―at least one line‖. This makes it one of
the most important theorems of Neutral Geometry. You will prove it in a class drill.
Theorem 42 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.
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6.7. Exercises
[1] Recall the statement of Theorem 25:
(Segment Subtraction) In Neutral Geometry, if A*B*C and D*E*F and AB  DE and
AC  DF , then BC  EF .
Justify the steps in the following proof. Make a drawing in the indicated step.
Proof
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
[2] Recall the statement of Theorem 27:
(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 .
Justify the steps in the following proof. Make drawings where indicated:
Proof of Claim (a)
1) There exists a unique point G on ray CD such that CG  AB . (by axiom _____)
2) Exactly one of the following statements is true about point G:C*G*D or C*D*G or
G=D. (by definition of ________________)
Case I
3) If C*G*D then AB  CD . (by definition of ___________________________)
Case II
4) If C*D*G, then there exists a unique point H such that A*H*B and AH  CD . (by step
________ and Theorem ______________)
5) In this case, CD  AB . (by definition of ____________________)
Case III
6) If G = D then AB  CD .
End of Proof of Claim (a)
Proof of Claim (b)
1) Suppose that AB  CD and CD  EF .
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2) There exists a point G such that C*G*D and CG  AB . (by definition of __________)
3) There exists a point H such that E*H*F and EH  CG . (by Theorem ______)
4) EH  AB (by steps ____ and ____ and axiom _____)
5) AB  EF . (by steps ___ and ___ and the definition of _________________________)
End of proof of Claim (b)
Proof of Claim (c)
1) Suppose that AB  CD and CD  EF .
2) There exists a point G such that C*G*D and CG  AB . (by definition of __________)
3) There exists a point H such that E*H*F and EH  CD . (by definition of __________)
4) There exists a point I such that E*I*H and EI  CG . (by steps ___ and ___ and
Theorem ____)
5) EI  AB (by steps ___ and ___ and axiom ____)
6) E*I*F (by steps ___ and ___ and Theorem ______)
7) AB  EF .(by steps ___ and ___, and the definition of _________________________)
End of proof of Claim (c)
[3] Recall the statement of Theorem 28: 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 .
Justify the steps in the following proof. Make a drawing where indicated.
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. (_________________________________)
(draw a picture to illustrate)
2.
ABC  EFG . (_________________________________________________________)
3. AC  EG , and A  E . (________________________________________________)
4. AD  EH . (_____________________________________________________________)
5. CAD  GEH .
(_________________________________________________________)
6. CD  GH and D  H . (________________________________________________)
7. CBD  GFH .
(_________________________________________________________)
8.
CBD  GFH (________________________________________________________)
End of proof
[4] The term supplementary angle was introduced in Definition 39, in Section 5.6. Prove that
every angle has a supplementary angle.
[5] Recall the statement of Theorem 28: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 .
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Justify the steps in the following proof. Draw a picture where indicated.
Proof
1. Suppose that if ABC and CBD are supplementary, and EFG and GFH are
supplementary, and ABC  EFG .
2. 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)
3.
ABC  EFG . (____________________________________________________)
4. AC  EG , and A  E . (___________________________________________)
5. AD  EH . (________________________________________________________)
6. CAD  GEH . (____________________________________________________)
7. CD  GH and D  H . (___________________________________________)
8. CBD  GFH . (____________________________________________________)
9.
CBD  GFH (___________________________________________________)
End of proof
[6] Prove Theorem 29, which says that in Neutral Geometry, vertical angles are congruent.
Solution:
[7] Recall the statement of Theorem 30: In Neutral Geometry, any angle congruent to a right
angle is also a right angle.
Justify the steps in the following proof: Make a drawing where indicated.
1) Suppose that angles ABC and EFG are given; that ABC  EFG ; and that
EFG is a right angle.
2) There exists a point D is a point such that A*B*D. (by axiom _____________)
3) Angle ABC , is supplementary to angle CBD . (by definition of ______________)
4) There exists a point H is a point such that E*F*H. (by ____________________)
5) Angle EFG , is supplementary to angle GFH . (by _____________________)
(make a drawing)
6) EFG  GFH (by steps ____ and ____ and definition of _________________)
7) CBD  GFH (by steps ____ and ____ and Theorem )
8) ABC  CBD (by steps ____ and ____ and ____ and axiom ____)
9) Angle ABC is a right angle. (by steps ____ and ____ and the definition of _______)
End of Proof
[8] Prove Theorem 39 (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.
Hint: Use Theorem 38 in your proof.
[9] Prove Theorem 40: In Neutral Geometry, two distinct lines perpendicular to the same line are
parallel.
Hint: Use Theorem 38 or Theorem 39 in your proof.
[10] Recall the statement of Theorem 41 (uniqueness of the perpendicular from a point to a line):
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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.
Justify the steps in the following proof:
1. Suppose that L is a line and P is a point not on L.
2. Assume that there are two different lines that pass through P and are perpendicular to L.
Call the lines M and N.
3. Lines M and N are not parallel. (because ____________________________________)
4. Lines M and N are parallel. (by Theorem ____________________________________)
5. Steps 4 and 5 are a contradiction. Therefore, our assumption in step 3 was wrong. There
cannot be two different lines that pass through P and are perpendicular to L. That means
there must only be the one line, M.
End of Proof
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7. Neutral Geometry II: Triangles
In this section, we will find that many important theorems that are commonly thought of as
Euclidean Geometry theorems can actually be proven to be true in Neutral Geometry.
7.1. The Isosceles Triangle Theorem and Its Converse
In Section 6.3.3, 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 a homework exercise.
Theorem 43 (Angle Side Angle Congruence) (ASA). 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 next theorem is called the Isosceles Triangle Theorem. We should officially define the term
isosceles before stating the theorem.
Definition 56 isosceles triangle
 Words: isosceles triangle
 Meaning: two sides of the triangle are congruent to each other (at least two)
 Additional Terminology: The angles opposite the two congruent sides are called the base
angles.
Here is the theorem:
Theorem 44 (The Isosceles Triangle Theorem)
Given: Neutral Geometry, ABC
Claim: If AB  AC then B  C
Alternate wordings for the statement of the Isosceles Triangle Theorem:
 If two sides are congruent, then the two opposite angles are congruent.
 If the triangle is isoscelese, then the base angles are congruent.
Notice that in the Isosceles Triangle Theorem, the fact that the triangle is isosceles is part of the
hypothesis. So the name is in keeping with the naming convention that we have discussed.
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.
Theorem 45
(Converse of the Isosceles Triangle Theorem)
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Given: Neutral Geometry, ABC
Claim: If B  C then AB  AC (If two angles are congruent, then the two opposite sides are
congruent.)
Proofs of both the Isosceles Triangle Theorem and its converse will be studied in a class drill. As
a summary of the statements of the two theorems, it might be useful to put them together in a
table:
Isosceles Triangle Theorem: If two sides are congruent,
then the two opposite angles are congruent
Converse of the Isosceles Triangle Theorem: If two angles are congruent,
then the two opposite sides are congruent.
In other words, in triangles, congruent angles are always opposite congruent sides. This summary
is easy to remember, and has a convenient acronym, which we will present as the title of a
theorem.
Theorem 46 The CACS Theorem: In Neutral Geometry triangles, Congruent Angles are always
opposite Congruent Sides.
Again, Theorem 46 is not really a new theorem, but rather just a combination of Theorem 44 and
Theorem 45 with a name that is perhaps easier to remember.
Here is a theorem that is presented only because it is useful in the proofs of some other more
important theorems. In other words, it could be called a lemma. The proof is not difficult, but it
is also not particularly illuminating, and so we will omit the proof and assume the theorem as
given.
Theorem 47 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 .
The Triangle Construction Theorem just presented and the Isosceles Triangle Theorem can be
used to prove the Side Side Side Congruence Theorem.
Theorem 48 (The Side Side Side Congruence Theorem) (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.
The proof of the Side Side Side Congruence Theorem takes awhile to digest because there are a
number of cases to consider. You will study the proof in a homework exercise.
7.2. The Exterior Angle Theorem
The next major theorem that we will study is the Exterior Angle Theorem. In order to understand
the wording of the theorem, we need to define some terms.
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Definition 57 ―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 49 The Exterior Angle Theorem
In Neutral Geometry, each of the remote interior angles is less than the exterior angle.
The proof of the Exterior Angle Theorem is rather tricky: it uses the method division into cases
and the method of contradiction. You will study it in a class drill.
The Exterior Angle Theorem plays an important role in the proof of the Angle Angle Side
Congruence Theorem. Remember that the statement of this theorem is the last of the four
congruence conditions (SAS, ASA, SSS, AAS) that were discussed in Section 6.3.3. Like the
proof of the Exterior Angle Theorem, the proof of the Angle Angle Side Theorem uses both the
method division into cases and the method of contradiction. You will study it in a class drill.
Theorem 50 The Angle Angle Side Congruence Theorem (AAS)
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.
In addition to the four congruence conditions introduced Section 6.3.3, there is another famous
congruence theorem that applies only to right triangles. First, some terminology.
Definition 58 ―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.
Here is the theorem:
Theorem 51 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.
You will study a proof of the Hypotenuse Leg Congruence Theorem in the homework.
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7.3. Comparison Theorems
In this section, we will turn our attention to theorems that are not about pairs of objects that are
congruent, but rather are about pairs objects in which one of the objects can be judged less than
the other. Since we are comparing objects, we could call these comparison theorems.
Theorem 52
In Neutral Geometry, in triangle
ABC , if AC  BC then
B
A.
It turns out that the converse of the statement of the above theorem is also a theorem:
Theorem 53
In Neutral Geometry, in triangle
ABC , if
B
A then AC  BC .
Proofs of both these theorems will be studied in a class drill. As a summary of the statements of
the two theorems, it might be useful to put them together in a table, just as we did when studying
the Isoscelese Triangle Theorem and its converse:
Theorem 52: If one side of a triangle is smaller than another side,
then the two opposite angles have the same ordering.
Theorem 53: If one angle of a triangle is smaller than another angle,
then the two opposite sides have the same ordering.
In other words, in triangles, smaller angles are always opposite smaller sides. This summary is
easy to remember, and has a convenient acronym, which we will present as the title of a theorem.
Theorem 54 The SASS Theorem: In Neutral Geometry triangles, Smaller Angles are always
opposite Smaller Sides.
Note that Theorem 54 is not a new theorem, but rather just a combining of Theorem 52 and
Theorem 53.
It actually helps to take the combining of old theorems further. One can combine Theorem 46
(the CACS Theorem), and Theorem 54 (the SASS Theorem) into a single theorem:
Theorem 55 The CACS and SASS Theorem: In triangles of Neutral Geometry, Congruent
Angles are always opposite Congruent Sides, and Smaller angles are always opposite
Smaller Sides.
In summary, Theorem 55 is a combination of the two theorems: Theorem 46 (the CACS
Theorem), and Theorem 54 (the SASS Theorem). But these in turn are combinations of four
theorems: Theorem 44, Theorem 45, Theorem 52 and Theorem 53.
We end this section with the Hinge Theorem. The name 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
Page 95 of 150
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 difficult and would is not particularly illuminating, so we will
omit the proof and assume the Hinge Theorem as given.
7.4. Midpoints and bisectors
We end the chapter with a short presentation of three theorems about midpoints and bisectors.
We most likely will not discuss this section in class, but you will study the proofs in homework
exercises.
Definition 59 ―midpoint‖ of a line segment
 Words: C is a midpoint of segment A, B
 Meaning: A*C*B and CA  CB .
Theorem 57 Existence of Midpoint of a Line Segment
In Neutral Geometry, every segment has exactly one midpoint.
Definition 60 ―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 58
In Neutral Geometry, every line segment has exactly one perpendicular bisector.
Definition 61 ―bisector‖ of an angle
 Words: a bisector of angle ABC
 Meaning: a ray BD between rays BA and BC such that
Theorem 59
DBA  DBC .
In Neutral Geometry, every angle has exactly one bisector.
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7.5. Exercises
[1] Recall the statement of Theorem 43 (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.
Justify the steps in the following proof. Draw a picture where indicated.
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 . (by axiom _________________)
3.
ABC  DGF . (by steps ___ and ___ and axiom _____) (Draw a picture.)
4.
C  DFG . (by step ___ and definition of __________________________________)
5. Ray FE must be the same ray as ray FG . (by axiom ________________)
6. Point E must be the same point as point G. (There can be only one point of intersection of
lines DE and FE , by Theorem _____)
7.
ABC  DEF . (by steps ___ and ___ and ___ and axiom ______________)
End of proof
[2] Recall the statement of Theorem 48 (The Side-Side-Side Congruence Theorem) (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 sides of the other triangle, then all the
remaining corresponding parts are congruent as well, so the triangles are congruent.
Justify the steps in the following proof. Draw pictures where indicated.
We will consider four cases. In the first three cases, triangles ABC and DEF are ―back-toback‖. In the fourth case, they are not.
 case 1: A = D, C = F, and the points B and E are on opposite sides of AC , and points A
and C are on opposite sides of BE .
 case 2: A = D, C = F, and the points B and E are on opposite sides of AC , and point A is
between B and E. That is, B*A*E.
 case 3: A = D, C = F, and the points B and E are on opposite sides of AC , and points A
and C are on the same sides of BE .
 case 4: All other possible configurations.
Proof of Case 1
1) Suppose that triangles ABC and DEF have a correspondence between their parts
such that the three sides of one triangle are congruent to the corresponding sides of the
other triangle, and suppose further that A = D, C = F, and the points B and E are on
opposite sides of BE , and points A and C are on opposite sides of BE . (Draw a picture.)
2) AB  DE (by step ______)
3) ABE  AEB (by step ___ and Theorem ______________________________)
4) BC  EF (by step ___)
5) CBE  CEB (by step ___ and Theorem ______________________________)
6) ABC  DEF (by steps ___ and ___ and Theorem ______________________)
7) ABC  DEF (by steps ___ and ___ and ___ and axiom ___________________)
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End of proof of case 1
Proof of Case 2
1) Suppose that triangles ABC and DEF have a correspondence between their parts
such that the three sides of one triangle are congruent to the corresponding sides of the
other triangle, and suppose further that A = D, C = F, and the points B and E are on
opposite sides of BE , and point A is between B and E. That is, B*A*E. (Draw a picture.)
2) BC  EF (by step ___)
3) CBE  CEB (by step ___ and Theorem ___________________________)
4) BA  ED (by step ___)
5) ABC  DEF (by steps ___ and ___ and ___ and axiom ____________________)
End of proof of case 2
Proof of Case 3
1) Suppose that triangles ABC and DEF have a correspondence between their parts
such that the three sides of one triangle are congruent to the corresponding sides of the
other triangle, and suppose further that A = D, C = F, and the points B and E are on
opposite sides of BE , and points A and C are on the same sides of BE . (Draw a picture.)
2) AB  DE (by step ___)
3) ABE  AEB (by step ___ and Theorem _______________________________)
4) BC  EF (by step ___)
5) CBE  CEB (by step ___ and Theorem _______________________________)
6) ABC  DEF (by steps ___ and ___ and Theorem _______________________)
7) ABC  DEF (by steps ___ and ___ and ___ and axiom _____________________)
End of proof of case 3
Proof of Case 4
Suppose that triangles ABC and DEF have a correspondence between their parts
such that the three sides of one triangle are congruent to the corresponding sides of the
other triangle. One can use Theorem 47, the Triangle Construction Theorem, to build a
new triangle GHI that is congruent to DEF and that is ―back-to-back‖ with ABC .
Then, case 1, 2, or 3 (with the letters D,E,F replaced by G,H,I) would describe the
configuration of triangles ABC and GHI . The results of those cases could be used to
conclude that ABC  GHI . Transitivity would then say that ABC  DEF .
End of proof of case 4
[3] Recall the statement of Theorem 51, 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.
Justify the steps in the following proof. (Draw pictures where indicated.)
1. Let ABC and DEF be triangles such that angles A and D are right angles,
BC  EF , and BA  ED .
Construct a special point G that creates a segment congruent to segment DF .
2. There exists a point H such that H*A*C. (by axiom BA3)
3. There exists a point G on ray AH such that AG  DF . (by axiom CA2, the segment
construction axiom) (Draw a picture.)
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Show that the resulting angle is congruent to EDF .
4. G*A*C (a brief argument of the type we did in Chapter 5 would show this)
5. Angles BAG and BAG are supplemental. (by step 5 and definition of supplemental)
6.
BAG  BAC (by steps 1 and 7 and definition of right angle)
7.
BAC  EDF (by Theorem 37 )
8.
BAG  EDF (by steps 7 and 8 and axiom CA4)
Show that the resulting triangle is congruent to DEF .
9.
ABG  DEF . (steps 1 and 8 and 3 and axiom CA6 (SAS))
Show that the resulting triangle is congruent to ABC .
10. BG  EF . (by step 9 and definition of triangle congruence, that is, CPCTC)
11. BG  BC . (by steps 1 and 10 and axiom CA1)
12. G  C . (by step 11 and Theorem 44, the Isosceles Triangle Theorem)
13. ABG  ABC . (by steps 6 and 12 and 11 and Theorem 50 (AAS))
14. DEF  ABC . (by steps 9 and 13 and the fact that triangle congruence is an
equivalence relation (which follows from the fact that line segment congruence is an
equivalence relation, by CA1, and angle congruence is an equivalence relation, by CA4))
End of Proof
[4] Recall the statement of Theorem 57 (Existence of Midpoint of a Line Segment)
In Neutral Geometry, every segment has exactly one midpoint.
Justify the steps in the following proof. Draw pictures where indicated.
1. Let AB be a line segment.
2. There exists a point C that is not on line AB . (by Theorem ____________________)
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 . (by axiom ___________________________)
4. There exists a unique point D on ray BX such that BD  AC . (by axiom ___________)
5. Points D and X are on the same side of line AB , (by step ___ and Theorem _________)
6. Points D and C are on opposite sides of AB , (by steps ___ and ___ and Theorem _____)
7. Line AB intersects segment CD . (by step ___ and definition of ___________________)
We can call the intersection point E.
8. Point E is between A and B. (This is a very tricky step to justify. We will assume it as
given.) (Draw a picture.)
9.
AEC  BED . (by Theorem _________________________________________)
10. AEC  BED . (by steps ___ and ___ and ___ and Theorem ____________________)
11. AE  BE . (by step ___ and the definition of _________________________________)
12. Point E is the midpoint of segment AB . (by steps ___ and ___ and definition of
______________________)
End of Proof
[5] Prove Theorem 58: In Neutral Geometry, every line segment has exactly one perpendicular
bisector.
[6] Prove Theorem 59: In Neutral Geometry, every angle has exactly one bisector.
Page 99 of 150
8. Measure of Line Segments and Angles
8.1. 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: they are 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 60 Existence of a Length Function for Line Segments
Given: Neutral Geometry and some reference line segment AB .
Claim: There exists a unique function length   : line segments 
properties.

with the following
 
1. length AB  1 .
2. The length function is onto. That is, for every positive real number L, there exists a line
 
segment CD such that length CD  L .
 
 
length  CD   length  EF  if and only if. CD  EF .
length  CE   length  CD   length  DE  if and only if C*D*E.
3. length CD  length EF if and only if CD  EF .
4.
5.
Theorem 61 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. degrees  A  90 if and only if A is a right angle.
2. The degrees function is onto. That is, for every real number 0  m  180 , there exists an
angle A such that degrees  A  m .
3. degrees  A  degrees  B  if and only if
A
B.
4. degrees  A  degrees  B  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 100 of 150
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.
The length definition obviously allows an amount of choice. That is, if one uses a different
reference segment 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 
. 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  and the
codomain of the degrees function is the set  0,   . 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.)
8.2. 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 62 the length function for line segments in straight-line drawings (Euclidean)
Meaning: the function lengthE : Euclideanline segments   defined by the following
formula
 
lengthE AB 
 x1  x2    y1  y2 
2
2
,
where segment AB is a Euclidean line segment with endpoints A   x1 , y1  and
B   x2 , y2  . (This implies that x1, y1, x2,and y2 are real numbers.)
Definition 63 the length function for line segments in the Hyperbolic plane
Meaning: the function lengthH : Hyperbolicline segments   defined by the
following formula
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 lengthE  AR 


lengthE  AS  
,
lengthH  AB   ln 
 lengthE  BR 



length
BS


E


where segment 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 .
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? 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
Page 102 of 150
(You will prove this in the exercises.) Therefore, interchanging the symbols C & D in the
formula would have no effect on the outcome.

For an 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
 
Euclidean distances needed in the formula for lengthH. The results are lengthE AR  5 ,
4
lengthE AS  3 , lengthE BR  1 , and lengthE BS  7 . Plugging these into the formula
4
4
4
for lengthH, we obtain
 lengthE AR



 54 
lengthE AS 

 34 
 57 
 35 
lengthH AB  ln 

ln

 1   ln 
  ln    2.46
 1 3 
 3 
 lengthE BR

 47 
4 



lengthE BS 


Notice that the hyperbolic length of line segment AB is not the same as the Euclidean length.
 
 
 
 
 
 
 
 
 
The Euclidean length of the segment is just lengthE AB  1 .
8.3. 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 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 curvylooking Hline whose equation and missing endpoints are not so hard to determine.
Page 103 of 150
One simple example of an orthogonal circle is simply the unit circle moved to the right.
Recall that the unit circle contains twelve famous points as shown in the figure below.
(0,1)
 1 3
1 3
  ,

 ,

 2 2 
2 2 
 2 2
,


 2 2 
 3 1
, 

 2 2

2 2
,
 

 2 2 

3 1
, 
 
2
2

(-1,0)
(1,0)

3 1
,  
 
2
 2

2
2
,
 

2 
 2
 1
3
  , 

2 
 2
A circle of radius 1 centered at the point




(0,-1)
 3 1
,  

2
 2
2
2
,

2
2 
1
3
 , 

2 
2

2, 0 will have twelve famous points obtained by
2 to the x-coordinates of the points above. The result is shown in the figure below.
2,1
 1
1
3
3
   2,

  2,

2 
2 
 2
2
 2 2
3 2 2 
,
,




2
2 
 2 2 


 3
3
1
1
 2, 
 
  2, 
2
2
 2
2
adding
 1 

2, 0



3
1
 2,  
 
2
 2
 2
2
,


2 
 2
 1
3
   2, 

2 
 2
1 
2, 0

 3
1
 2,  

2
 2


2, 1
3 2
,

2

1
  2, 
2
2

2 
3

2 
Page 104 of 150
 2 2
 2
2
This new circle intersects the unit circle at points 
and 
,
,


 . Furthermore, the
 2 2 
 2
2




two circles are orthogonal, as shown in the figure below.
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:

x  2

x2  y 2  1
2
 y2  1
The point  x, y  lies on the circle of radius 1 centered at


2, 0 .
The point  x, y  lies in the interior of the unit circle.
Three of the twelve famous points on the right circle above qualify as Hpoints. They are are


3
1
3
1
A   
 2,   , B  1  2, 0 , C   
 2,  .
2
2
 2
 2
Since these three points all lie on the same Hline, they are collinear Hpoints. (Of course, they are
not collinear as points of Euclidean Geometry.) The two points of intersection of the two circles
are the missing endpoints of the Hline. They are
1 
 1
 1 1 
R
,
,
, S 
.
2
 2
 2 2


In the exercises, you will compute the hyperbolic lengths of Hyperbolic segments AB ands AC .
8.4. 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.
Page 105 of 150
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.
Theorem 62 In Neutral Geometry, the sum of the measures of any two angles of a triangle is
less than 180.
The Triangle Inequality is well known, and also has a fairly simple proof that you will justify in
the exercises.
Theorem 63 (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.
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. In fact, it is not even true. All that
can be proven is the following very famous theorem.
Theorem 64 (The Saccheri-Legendre Theorem) In Neutral geometry, the angle sum of any
triangle is less than or equal to 180.
The proof of Theorem 64 is difficult; we will accept the theorem as given. But the following
theorem is not very hard to prove. You will justify a proof of it in the exercises.
Theorem 65 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 106 of 150
8.5. Exercises for Chapter 8
u 
v 
v


[1] In the reading, you were told that ln
 ln  u  . Prove this.
 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 8.3, an
example of a curvylooking Hline was
presented. It has
missing endpoints R
and S, and famous
points A, B, and C as
shown in the figure at
right. Compute
 
 AC  .
lengthH AB and
lengthH
 2 2
S  
,

2
2 


3
1
C   
 2, 
2
 2

B  1  2, 0


3
1
A   
 2,  
2
 2
 2
2
R  
,

2 
 2
[4] Let A   0,0  , and let B   x,0  where x is some unknown real number such that 0  x  1.
 
Compute lengthH AB . For consistency, label the missing endpoints R  1,0  and S   1,0 .
(b) Your answer to part (a) will be a symbolic expression that includes the absolute value
symbol. In the next problem, it will be necessary to have this expression simplified further, with
the absolute value symbol eliminated. Determine how to simplify the expression in a way that
the absolute value symbol is eliminated.
(Hint: to eliminate the absolute value symbol, you will have to determine whether the thing
inside it is positive or negative. You can do that by exploiting the fact that 0  x  1. Because of
1 x
that, you know that 1  1  x   2 and 0  1  x   1 . In the fraction
, you are dividing a
1 x
1 x
number that is greater than 1 by a positive number that is less than 1. So the fraction
will
1 x
number greater than 1. What does that tell you about the logarithm of that fraction? More
specifically, will the logarithm be positive, negative or zero? With that information, decide what
to do with the absolute value symbol.)
Page 107 of 150
[5] The goal of this exercise is to prove that the lengthH function is onto. To do that, you must
prove that given any unknown positive number L, there exists a segment AB such that
 
lengthH AB  L . There are lots of such segments. The easiest one to find is the one that has
point A at the origin, that is A   0,0  , and point B somewhere on the positive x-axis, inside the
unit circle. That is, point B of the form B   x, 0  where 0  x  1. Given an unknown L, find the
 
value of x such that 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 in terms of x. Simply solve that equation for x in terms of L.)
[6] Recall the statement of Theorem 62: In Neutral Geometry, the sum of the measures of any
two angles of a triangle is less than 180.
Justify the steps in the following proof. Draw a picture where indicated.
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. (by axiom _______) (Draw a picture.)
3. Angle DBC is an exterior angle of ABC , and A is one of its remote interior angles.
(by steps ___ and ___ and definition _________________________________________)
4.
A  DBC . (by step ___ and Theorem _____________________________________)
5. degrees  A  degrees  DBC  . (by step ___ and Theorem _____________________)
6. degrees  A  degrees  B   degrees  DBC   degrees  B  . (by step ___ and
algebra)
7. Angles DBC and B are supplemental. (by steps ___ and ___ and definition of
______________________________________________________)
8. degrees  DBC   degrees  B   180 . (by step ___ and Theorem _________________)
9. degrees  A  degrees  B   180 . (by steps ___ and ___ and algebra)
End of proof
Page 108 of 150
[7] Recall the statement of Theorem 63: (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.
Justify the steps in the following proof. Draw a picture where indicated.
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 . (Axiom ____ tells us that a point
D exists such that A*B*D. Axiom _________________________________________
tells us that a special point exists on ray BD that will create a segment congruent to
segment BC . With no loss of generality, we can just rename that special point as point
D.) (Draw a picture.)
 
 
3. length BC  length BD . (by step ___ and Theorem ___________________________
4. length  AB   length  BC   length  AB   length  BD  . (by step ___ and algebra)
5. length  AB   length  BD   length  AD  . (by step ___ and Theorem _______________)
6. length  AB   length  BC   length  AD  . (by steps ___ and ___)
7.
ADC  DCB . (by step ___ anTheorem ____________________________________)
8. Ray CB is between rays CA and CD . (Accept this as given.)
9.
DCB  DCA .(by definition ___________________________________________)
10. ADC  DCA . (by steps ___ and ___ and Theorem ___________________________
_______________________________________________________________________)
11. AC  AD . (by Theorem __________________________________________________)
 
 
13. length  AC   length  AB   length  BC  . (by steps ___ and ___ and algebra)
12. length AC  length AD . (by step ___ and Theorem __________________________)
End of proof
[8] Recall the statement of Theorem 65: 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.
Justify the steps in the following Proof. Draw a picture where indicated.
1. Let two angles and a remote exterior angle be given. Without loss of generality, we can
assume that the triangle vertices are labeled so that the given interior angles are A and
B of ABC , with remote exterior angle BCD , where D is a point such that A*C*D.
(Draw a picture.)
2. Angles C and BCD are supplementary. (by step ___ and definition ____________
______________________________________________________________________)
3. degrees  C   degrees  BCD   180 . (by step ___ and Theorem ________________
_______________________________________________________________________)
4. degrees  A  degrees  B   degrees  C   180 . (by Theorem 64, The SaccheriLegendre Theorem)
5. degrees  A  degrees  B   degrees  BCD  . (by steps ___ and ___ and algebra)
End of proof
Page 109 of 150
[10] Three questions about proof structure
(a) 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
(b) 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
(c) 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 110 of 150
Page 111 of 150
9. Building Euclidean Geometry from Neutral Geometry
At the beginning of this course, we were introduced to the notion of an axiom system. We also
learned about three properties that an axiom system may or may not have: consistency,
independence, and completeness. In Chapters 2 and 3, we studied simple axiom systems and
their properties. Then in Chapter 5, we embarked on a program to build an axiom system that
would describe the straight line drawings that we have been making for much of our lives. (The
axiom system would also describe the usual analytic geometry of the x-y plane, in which the
graphs of lines look straight.). One goal was to have an axiom system that is independent—has
no redundancy. Another goal was to have an axiom system that is complete. That is, we wanted
the axiom system to thoroughly specify the geometry, so that there would be no other model
besides our straight-line drawings.
Our first step towards this goal was actually already taken back in Chapter 4, when we studied
Incidence Geometry. We observed that the incidence axioms insured that abstract points and
lines in our axiomatic geometry have some of the ―normal‖ behavior that we expect of points and
lines. But we observed that it was not very difficult for a geometry to satisfy all of the incidence
axioms. As a result, there are many different models. (The axiom system for incidence geometry
is not complete.)
Since then, we have studied Incidence and Betweenness Geometry (in Chapter 5) and Neutral
Geometry (in Chapters 6, 7, and 8). During that time, we have studied only two models: straightline drawings and drawings in the Poincare Disk. But we never really proved that these two
models were actually models. It would have been more correct to just call them interpretations
instead of calling them models. And we never considered the possibility that there might be other
models as well.
But the fact is that the straight-line interpretation and the Poincare Disk interpretation are both
models of Neutral Geometry.
And the two models are not the same: From studying the two types of drawings in the computer
lab, we have seen many instances where the two kinds of drawings have important differences
beyond the obvious difference that they simply look different. For example, in Computer Project
3, we studied the following two statements:
Statement S: If a pair of alternate interior angles is congruent, then lines L and M are parallel
Converse of S: If lines L and M are parallel, then a pair of alternate interior angles is congruent.
We observed that Statement S was true in straight-line drawings and in the Poincare Disk. (That
makes sense, because the statement is actually a theorem of Neutral Geometry.) However, the
Converse of S was true in straight-line drawings but not in the Poincare Disk. Based on this
observation, we would conclude that the straight-line drawings and the Poincare Disk drawings
are non-isomorphic models of Neutral Geometry.
We did not consider the possibility of other models of Neutral Geometry besides these two. But
the fact is that these are the only two models. More correctly, these are the only two types of
Page 112 of 150
models. There are other models of Neutral geometry that look different, but they all behave like
either the straight-line model or the Poincare Disk model.
Most of you had probably not seen Poincare Disk drawings before this course. They look very
different from straight-line drawings and, as we have observed, they have some important
differences in terms of the behavior of objects. But the similarities between the straight-line
drawings and the Poincare Disk drawings are more remarkable. In spite of the fact that the
drawings look very different, there are many instances in which the objects in the drawings
behave in identical ways. The extent to which this is true may best be illustrated by noting that
our list of Neutral Geometry theorems ends with Theorem 65. Every one of those theorems
describes something that is true in both the straight line drawing model and the Poincare Disk
model. It turns out that many important facts about lines and triangles that we might have
previously associated with straight-line drawings are also true in Poincare Disk drawings.
So the Neutral Geometry axiom system is very very rich but in the end is not complete: it does
not thoroughly specify the world of straight-line drawings and, as a result, there are two nonisomorphic models. In this chapter, we turn our attention to the question of what axioms must be
added to the Neutral Geometry in order to turn it into a complete axiom system that does fully
specify the world of straight-line drawings.
9.1. Thirteen Statements that are Logically Equivalent in Neutral Geometry
What is remarkable is that it is possible to turn Neutral Geometry into a complete axiom system
by adding just one more axiom. And what is equally remarkable is that there is a huge list of
candidate statements that will work as the one additional axiom. The reason is that a long list of
statements can be proven to be logically equivalent to each other:
Theorem 66 Thirteen Statements that are Logically Equivalent in Neutral Geometry
Given: The axioms for Neutral Geometry
Claim: The following statements are logically 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) A 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
that passes through P and is parallel to L.
3) Playfair’s Postulate: For every line L and every point P not on L, there is a unique line
that passes through 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) ―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 38, The Alternate
Interior Angle Theorem.)
7) ―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
39,The Corresponding Angle Theorem,)
Page 113 of 150
8) ―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.‖
9) ―Every triangle has angle sum exactly 180.‖
10) ―Given any triangle 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 .‖
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.‖
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. That thing is called Euclidean Geometry. The resulting larger axiom set will be
complete: its only model will be straight-line drawings (and the usual analytic geometry of the xy plane).
On the other hand, if we make the negation of any one of these thirteen statements an axiom (that
is, we declare the negation of the statement 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. That thing is called
Hyperbolic Geometry.The resulting larger axiom set will be complete: its only model will be
Poincare Disk drawings (and the analytic geometry of the Hyperbolic plane).
This ―completion‖ of Euclidean geometry has a fascinating history. For roughly 2000 years,
from the time of Euclid (300bc) until the 1700’s, mathematicians thought that the Neutral
Geometry axioms were complete. This belief was fostered by the fact that there was only one
known model for Neutral Geometry: straight-line drawings. These mathematicians, including
Euclid, himself, thought that Euclid’s Fifth Postulate was not needed in the list of axioms. They
sought to prove that the statement of the Fifth Postulate was always true as a consequence of the
other axioms. That is, they sought to prove that it was reudundant, or not independent.
But such a proof was elusive. Eventually, mathematicians began to wonder if perhaps the
statement of the fifth postulate might be independent, after all. To demonstrate that the statement
was independent, a model would have to be found that satisfied all of the Neutral Geometry
axioms, but in which the statement of the Fifth postulate was false. Such a model was discovered
in the 1800’s.
9.2. What’s up with the answer to THE BIG QUESTION?!?
Notice that there is a difference between statements 2 and 3 on the list
 statement 2: For any line L and for any point P not on L, there is at most one line that
passes through P and is parallel to L.
Page 114 of 150

statement 3: For every line L and every point P not on L, there is a unique line that passes
through P and is parallel to L.
Which of these statements is correct? And why does the theorem say that they are equivalent
when they clearly say different things?
Well, first of all remember that correctness is not the issue here. If we put a statement on the list
of axioms, we are simplay playing a game where we pretend that the statement is true.
But what about the issue that the two statements seem to say different things? The answer to that
is that even though the two statements seem to say different things, in the context of Neutral
Geometry they actually are equivalent. To see why, consider what various statements say about
the answer to THE BIG QUESTION. First, recall the statement of Theorem 42:
Theorem 42 (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.
Now, let n be the number of lines that pass through P and are parallel to L.



Notice that Theorem 42 says that n  1.
Notice that Statement 2 above says that n  1.
Notice that Statement 3 above says that n = 1.
If we add statement 2 to the list of Neutral Geometry axioms, then Theorem 42 will still be true.
Since both Theorem 42 and Statement 2 will have to be true, that will require that n  1 and n  1
. The only way for this to be true is if n = 1. So even though Statement 2 does not say that the
number of parallel lines is exactly 1, the end result will still be that the number of lines is exactly
1.
9.3. Proving Theorems about Equivalent Statements
We won’t prove all of Theorem 66, but we will discuss proof strategies for theorems of this type
and we then prove parts of the theorem.
Recall that we have encountered a few different methods of proving the conditional statement
If P then Q.
A direct proof has the following form.
Proof
1) Suppose that P is true
2)
3)
4) Q is true. (by some justification)
End of proof
Alternately, we could prove the contrapositive statement,
Page 115 of 150
If not Q then not P.
Such a proof would have the following form
Proof
1) Suppose that Q is false
2)
3)
4) P is false. (by some justification)
End of proof
We could use an indirect proof by the method of contradiction. We have used this method to
prove the original conditional statement:
Proof
1) Suppose that P is true
2)Assume that Q is false. (assumption)
3)
4)
5)Reach some contradiction. The assumption must be wrong. Therefore, Q must be true.
End of proof
But we could also use the method of contradiction to prove the contrapositive statement:
Proof
1) Suppose that Q is false
2)Assume that P is true. (assumption)
3)
4)
5)Reach some contradiction. The assumption must be wrong. Therefore, P must be false.
End of proof
In summary, if we wish to prove that P  Q, we may do it by proving that P  Q or by proving
that ~Q  ~P. And we may choose to use the method of contradiction or not.
Now, suppose that we wish to prove that P is logically equivalent to Q. To say that P and Q are
logically equivalent means that P  Q and Q  P. In light of the discussion above, we could
use the following four proof strategies:
Strategy 1:
 Part 1: Prove that P  Q.
 Part 2: Prove that Q  P.
Strategy 2:
 Part 1: Prove that ~Q  ~P.
 Part 2: Prove that Q  P.
Strategy 3:
 Part 1: Prove that P  Q.
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
Part 2: Prove that ~P  ~Q.
Strategy 4:
 Part 1: Prove that ~Q  ~P.
 Part 2: Prove that ~P  ~Q.
Four possible strategies!! It’s actually worse than that. Remember that we can also choose
whether or not to use the method of contradiction for any of the parts. So there are many many
proof strategies to choose from.
Now, what if we want to prove that three statements are logically equivalent? The available
strategies multiply like bunnies. Here are just a few:
(1)
(2)
(1)
(2)
(1)
(2)
(3)
(3)
(3)
Strategy 2
Strategy 1
Strategy 3
I have left out the multitude of choices available such as whether to prove a conditional
statement or its contrapositive, or whether or not to use the method of proof by contradiction. For
each of the arrows in the diagrams above, those choices are still available. What you should
notice is that there are new choices to be made. Notice that Strategy 1 would involve a proof of
the equivalence of statements (2) and (3), but it would not involve a proof of the equivalence of
statements (1) and (3). This is different from Strategy 2. Notice that Strategy 3 is very different
from the first two strategies. What is cool about Strategy 3 is that it would involve a total of three
proofs, whereas the other two strategies would involve four proofs.
Theorem 66 is about the logical equivalence of thirteen statements! Using the method of Strategy
1 or 2 in the above diagram would involve twenty four proofs. Using the method of Strategy 3
would involve thirteen proofs. So it seems obvious that we ought to use Strategey 3. But the
decision is not so simple. For one thing, some statements can be much harder to prove than
others. So the fact that Strategy 3 involves only thirteen proofs does not mean that it would be
the simplest strategy. Also, realize that when using Strategy 3, no equivalences are proven until
the very end, when the circle of thirteen proofs is completed. At that time, all thirteen statements
will have been proven to be complete. We don’t have the time for thirteen proofs. But we would
like to get a taste of the equivalence proofs. For that reason, we will study the following proofs
this week:
(3)
(6)
(9)
9.4. Finally, Euclidean Geometry
With Theorem 66, we come to the end of our study of Neutral Geometry. We will add one of the
thirteen equivalent statements to the list of Neutral Geometry axioms, and the resulting axiom
system will be called Euclidean Geometry. It doesn’t matter which of the thirteen statements we
choose to add, but the most popular one to add is Statement (2), the one that we are call EPP.
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9.5. Exercises
[1] Negations of statements from Theorem 66.
(a) Write Statement (2) and its negation.
(b) Write Statement (4) and its negation.
(c) Write Statement (6) and its negation.
(d) Write Statement (8) and its negation.
(e) Write Statement (9) and its negation.
(f) Write Statement (10) and its negation.
(g) Write Statement (11) and its negation.
(h) Write Statement (13) and its negation.
[2] Some missing parts of the proof of Theorem 66.
(a) Prove that (11)  (12).
(b) Prove that (13)  (11).
[3] Supply one of the missing parts of the proof of Theorem 66. (Not one of the parts that we
proved in class and not one of the parts from exercise [2].
Page 118 of 150
Page 119 of 150
10.Euclidean Geometry I
In the previous chapter we studied Theorem 66, which presented thirteen statements that are
equivalent in Neutral Geometry. We ended the chapter by adding one of the statements to the
Neutral Geometry axiom list to create a longer axiom list. This new, longer axiom list describes
Euclidean Geometry. The statement that we added to the axiom list is Statement (2), the one that
we call EPP. By adding that statement to the axiom list, we declared it to be true. (Remember
that there is no way to prove that Statement (2) is true, because Statement (2) is an independent
statement in Neutral Geometry. That is, there are models of Neutral Geometry where Statement
(2) is true, and there are models of Neutral Geometry where Statement (2) is false.)
Because Statement (2) is true in Euclidean Geometry and Euclidean Geometry is a kind of
Neutral Geometry, Theorem 66 tells us that all of the other twelve statements are also true in
Euclidean Geometry. They are all true, but they need to be proven true, as theorems. That is,
there would be twelve theorems, each proving that one of those remaining twelve statements is
true in Euclidean Geometry.
In the previous chapter, we did not thoroughly present proofs of all of the equivalences
mentioned in Theorem 66. If we had thoroughly proved all the equivalences, then those proofs
could now serve as proofs of the twelve theorems. So we have not yet done the job of proving
that all of those twelve statements are true. We start this chapter by presenting some of those
twelve statements as theorems, with explicit proofs. Following that, we will introduce circles and
prove some theorems about them. Then we will go to Coldstone.
10.1. Some results from the previous chapter presented as theorems
As discussed in class, a full proof of Theorem 66 would be quite long, because we would have to
prove equivalences. We only proved one of the equivalences in class. (We proved (2)  (4).) In
the current section, we will not prove equivalences. We will just prove that a bunch of the
thirteen statements are true as a consequence of statement (2) being true. Here, the statements
will be presented as theorems, and they will be added to our list of Euclidean Geometry
Theorems.
The first theorem that we will prove is the statement that is number (4) on the list of thirteen
statements. We discussed a proof of it in class. You will justify a proof of it in the exercises.
Theorem 67 In Euclidean Geometry, if a line intersects one of two parallel lines, then it also
intersects the other.
Remember the naming convention for theorems: They are named for the situation described in
their hypotheses. So the Alternate Interior Angle Theorem, Theorem 38, tells us something about
congruent alternate interior angles: it tells us that they create parallel lines. It is important to
remember that the Alternate Interior Angle Theorem can never be used to prove that alternate
interior angles are congruent.
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Finally now we come to a theorem that can be used to prove that alternate interior angles are
congruent. It is the Converse of the Alternate Interior Angle Theorem. It was the statement that
was number (6) on the list of thirteen. You will justify a proof of this theorem in the exercises.
Theorem 68 In Euclidean Geometry, 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 the
statement of Theorem 38, The Alternate Interior Angle Theorem.)
The next two theorems could really be called corollaries of Theorem 68, because their proofs are
very short if that theorem is used. The statements are the ones that were numbered (7) and (8) on
the list of thirteen. The first one is very easy to prove, and will not be assigned on homework or
discussed in class. But the second one is a little harder to prove. You will justify a proof of it in
the exercises.
Theorem 69 In Euclidean Geometry, 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 39,The Corresponding Angle Theorem.)
Theorem 70 In Euclidean Geometry, 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.
This next theorem was statement (5) on the list of thirteen statements. Its proof is also very short,
but it uses both Theorem 67 and Theorem 68. You will prove it in the exercises.
Theorem 71 In Euclidean Geometry, if a line is perpendicular to one of two parallel lines, then
it is perpendicular to the other.
Statement (9) on the list of thirteen statements says that every triangle has angle sum exactly
180. I proved this statement in class. The proof is not difficult, but it is worth reviewing. You
will justify a proof of the statement in the exercises.
Theorem 72
In Euclidean Geometry, every triangle has angle sum exactly 180.
The three statements numbered (11), (12), and (13) on the list of thirteen are presented here as
the folloing three theorems. In Chapter 9 Exercise [2], you were asked to verify that (13)  (11)
and (11)  (12). Here, though, the statements are listed as theorems. The ordering of the
theorems means that Theorem 73 will have to be proved without any reference to the subsequent
two thereoms. That means that the proof of Theorem 73 will be very different than the proof of
(13)  (11) that you did in Chapter 9 Exercise [2]. You will justify a proof of Theorem 73 in the
homework.
Theorem 73
In Euclidean Geometry, the opposite sides of a parallelogram are congruent.
Theorem 74
In Euclidean Geometry, the opposite angles of a parallelogram are congruent.
Theorem 75
In Euclidean Geometry, the diagonals of a parallelogram bisect each other.
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10.2. Circles, chords, and diameters and some Neutral Geometry Theorems
In this section, we will define circles and some related things, and we will prove a few theorems
about them.
We start with the definition of the circle. It is noteable that this is the first new definition since
the introduction of Theorem 60, which was the theorem about existence of a length function for
line segments. Now that we have the notion of length, all of our definitions and theorems can use
it. However, it is useful to observe that many definitions and theorems can be formulated without
referring to the notion of length. This is useful because it reminds us that the notion of length is
not essential to Euclidean geometry. And it is useful because it would allow us to postpone the
introduction of length until later, if we wanted to. And most importantly, it is useful to know
about both versions because not all books use the same version.
Definition 64 Circle
 Version of the definition that does not use the notion of distance


o Symbol: Circle A, AB .
o Words: Circle centered at A, with radial segment AB .
o Meaning:The set of points P such that segment AP is congruent to segment AB .


o Meaning in symbols: P : AP  AB .

Version of the definition that does use the notion of length
o Symbol: Circle  A, r  .
o Words: Circle centered at A, with radius r.
o Meaning:The set of points P that are a distance r from point A.

  
o Meaning in symbols: P : length AP  r .
The theorems about circles that we prove in this section will be about the relationship between
circles and certain segments and lines. The segments and lines that we will be interetested in are
presented in the following definition.
Definition 65 Segments and Lines related to circles.
 Given a circle,
o A chord is a line segment whose endpoints are points on a circle.
o A diameter is a chord that also contains the center of the circle.
o A secant is a line that passes through exactly two points of the circle.
o A tangent is a line that passes through exactly one point of the circle.
The theorems in this section can be proven using only the Neutral Geometry axioms. Because of
that, I present them here as a Neutral Geometry Theorems.
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The first two theorems have fairly easy proofs. The theorems are presented here mainly because
they enable us to eliminate some of the clutter from later proofs. (Most books would call these
Lemmas.) We discussed them in class on Monday June 2 and Tuesday June 3.
Theorem 76 In Neutral Geometry, the following are equivalent:
(1) Point P lies on the perpendicular bisector of segment AB .
 
 
(2) Point P is equidistant from A and B. That is, AP  BP and length AP  length BP
Theorem 77
In Neutral Geometry, all diameters of a given circle are congruent.
The next four theorems about circles are more difficult to prove. You will justify their proofs in
class and/or in the exercises.
Theorem 78 In Neutral Geometry, diameters of a circle are the longest chords.
Given a circle, if segment CD is a chord but not a diameter, and segment AB is a diameter, then
 
 
segment CD  AB . That is, length CD  length AB .
Theorem 79 In Neutral Geometry, the following are equivalent:
(1) Diameter AB is perpendicular to non-diameter chord CD .
(2) Diameter AB bisects non-diameter chord CD .
Theorem 80 In Neutral Geometry, the perpendicular bisector of a chord passes through the
center of the circle.
Theorem 81 In Neutral Geometry, the following are equivalent:
(1) Line L is perpendicular to segment AP at point P.


(2) Line L is tangent to Circle A, AP at point P.
10.3. Some Euclidean Geometry Theorems about Circles and Triangles
You might be a little bit puzzled by the fact that each of the theorems that we have so far proven
about circles has been a theorem of Neutral Geometry. That tells us that certain aspects of the
behavior of circles is the same in Euclidean and Hyperbolic Geometries. That should not be a big
surprise. We previously found that many aspects of the behavior of triangles are the same in
Euclidean and Hyperbolic Geometries. We will will now set out to state and prove a famous
statement that is true about circles in Euclidean Geometry but is not true about circles in
Hyperbolic Geometry. Before presenting the big theorem, it is useful to identify two smaller
theorem that can serve as lemmas. These theorems are not famous, but they turn out to be the key
to the proof of some famous theorems.
Theorem 82 (really a Lemma) In Euclidean Geometry, If A, B, and C are non-collinear points,
and line L is perpendicular to line AB , and line M is perpendicular to line BC , then lines
L and M intersect.
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Theorem 83 (really a Lemma) In Euclidean Geometry, If A, B, and C are non-collinear points,
and line L is the perpendicular bisector of segment AB , and line M is the perpendicular
bisector of segment BC , and line N is the perpendicular bisector of segment CA , then
lines L, M, and N are concurrent.
Theorem 84 In Euclidean Geometry, If A, B, and C are non-collinear points, then there exists
exactly one circle that contains all three points.
You will justify proofs of Theorem 82 and Theorem 83 and Theorem 84 in class drills. Theorem
84 is often presented in a different way, in terms of triangles. The term ―circumscribed‖ is often
used. Here is a definition.
Definition 66 Circumscribe


Words: Triangle

Meaning: Points A, B, and C lie on the circle.

ABC is circumscribed by circle Circle P, PQ .
With this terminology, we can present the following theorem, which is really just a re-packaging
of the previous theorem:
Theorem 85
In Euclidean Geometry, every triangle can be circumscribed by exactly one circle.
That is for every triangle


ABC , there exists exactly one circle Circle P, PQ such that
points A, B, and C lie on the circle.
10.4. Could Theorem 84 be a theorem of Neutral Geometry?
As an introduction, consider the following statement and a proof:
Statement S: If the hypotenuse and one leg of one right triangle are congruent to the hypotenuse
and one leg of another right triangle, then the two triangles are congruent.
Proof
Let a1 , b1 , and h1 be the lengths of the two legs and hypotenuse of the first triangle, and let
a2 , b2 , and h2 be the lengths of the two legs and hypotenuse of the second triangle. We are
told that the hypotenuses are congruent, so h1 and h2 must actually be the same number, that
we can just call h. We are also told that one leg of the first triangle is congruent to one leg of
the second triangle. We can assume that it is the sides labeled a1 and a2 that are congruent, so
a1 and a2 must actually be the same number, that we can just call a.
The Pythagorean Theorem of Euclidean Geometry tells us that the three sides of a right
triangle must satisfy the equation a 2  b 2  h 2 , where a and b are the lengths of the two legs
and h is the length of the hypotenuse.
Applying the Pythagorean Theorem to the first triangle, we have a 2  b12  h 2 . Solving this
equation for b1, we obtain the equation b1  h 2  a 2 .
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Applying the Pythagorean Theorem to the second triangle, we have a 2  b2 2  h 2 . Solving
this equation for b2, we obtain the equation b2  h 2  a 2 .
We see that b1 and b2 must actually be the same number. So the second legs of the two
triangles are also congruent. Therefore, the triangles are congruent by Side-Side-Side.
End of Proof
The proof is perfectly valid, and it shows that Statement S is true in Euclidean Geometry. But it
is a famous fact that the Pythagorean Theorem is not valid in Hyperbolic Geometry. Based on
that fact, one might reasonably suspect that Statement S is not true in Hyperbolic Geometry, and
so it cannot be a theorem of Neutral Geometry.
But notice that S is the statement of Theorem 51, the Hypotenuse-Leg Congruence Theorem of
Neutral Geometry. We have already proven that statement S is true in Neutral Geometry.
What’s wrong? Nothing is wrong. But this example illustrates the idea that one should not
assume that because a proof of some statement cites a theorem of Euclidean Geometry, the
statement is only true in Euclidean Geometry. Maybe there is some other proof possible that only
uses the axioms and theorems of Neutral Geometry.
In light of this discussion, notice that Theorem 82 is listed as a theorem of Euclidean Geometry.
You might think that the reason for that is that one of the steps in the proof is justified by a
theorem of Euclidean Geometry. But that is not the reason. Just because you find a proof of
Theorem 82 that uses a theorem of Euclidean geometry, you should not assume that Theorem 82
has to be listed as a Euclidean geometry theorem. You should wonder if maybe the statement of
Theorem 82 is actually true in Neutral Geometry, and I was just not clever enough to come up
with a proof that used only the Neutral Geometry axioms and theorems.
We should consider the statement of Theorem 82 and try to determine if it is also true or false in
Hyperbolic Geometry. It is useful to write down the statement of the theorem with explicit
quantifiers, and then also write down the negation of the theorem, with explicit quantifiers.
Statement S: For all points A, B, and C, and for all lines L and M, if A, B, and C are noncollinear , and line L is perpendicular to line AB , and line M is perpendicular to
line BC , then lines L and M intersect.
Statement ~S: There exist points A, B, and C, and lines L and M such that A, B, and C are noncollinear, and line L is perpendicular to line AB , and line M is perpendicular to
line BC , and lines L and M do not intersect.
Notice that statement S is a universal statement. To prove that statement S is true, we would need
to come up with a general proof.
If we want to prove that statement S is false, then we would need to do it by proving that the
negation of S is true. But the negation, ~S, is an existential statement. To prove that ~S is true,
we would need to come up with an example.
Page 125 of 150
I will present an example that can prove that statement ~S is true.
Consider the picture at right showing three
circles all of radius 1. The bottom left
circle is centered at the (0,0), the top circle


is centered at 0, 2 , and the circle on the
right is centered at


2, 0 . The top and
right circles are orthogonal to the bottom
left circle. So the top and right circle can be
used as a source of Hlines that contain
some known points.
line L
The picture at right shows the
standard unit circle, along with
the two Hlines obtained from the
top and right circles of the
previous picture. We can call the
lines L and M. One known Hpoint
on each line is labelled. Those are
the points P and Q.

P  0, 1  2

 1 1 
missing endpoint 
,

 2 2
line M

Q  1  2, 0

B = (0,0)
Notice that Hlines L and M share a missing endpoint. Remember, though, that the missing
endpoint is a point on the unit circle, not in the interior. So the missing endpoint is not an Hpoint.
That means that Hlines L and M do not intersect. Notice also that Hline L is perpendicular to the
y-axis and Hline M is perpendicular to the x-axis.
Page 126 of 150
By axioms BA3 and CA2, we know that
there is some point A such that A*P*B and
such that AP  BP . So point P is the
midpoint of segment AB and Hline L is the
perpendicular bisector of segment AB .
Similarly there is some point C such that
C*Q and such that CQ  BQ . So point Q is
the midpoint of segment CB and Hline M is
the perpendicular bisector of segment CB .
The points A, B, C, P, Q are shown at right.
line L
point A
point P
missing endpoint
line M
point C
point Q
point B
This picture is a remarkable thing. It was created without the aid of the NonEuclid program,
using only some translations of the unit circle and some references to the axioms of Neutral
Geometry. It shows that there exist Hpoints A, B, and C, and Hlines L and M such that A, B, and
C are non-collinear, and line L is perpendicular to line AB , and line M is perpendicular to line
BC , and lines L and M do not intersect. So the picture serves as an example that shows that
Statement ~S is true in Hyperbolic Geometry. In other words, it illustrates the fact that Statement
S is not true in Hyperbolic Geometry. This tells us that Theorem 82 cannot be a theorem of
Neutral Geometry. It is only true in Euclidean geometry.
Notice that the three non-collinear Hpoints A, B, and C define a triangle. So the same picture
(with those three points connected by line segments) could serve as an example of a triangle in
Hyperbolic Geometry that cannot be circumscribed!
Page 127 of 150
10.5. Exercises
The first six exercises have to do with proofs of theorems that were originally on the list of
thirteen equivalent statements. Now, instead of trying to prove that the statements are equivalent
to each other, we are trying to prove that each is true asa theorem of Euclidean Geometry.
[1] Recall the statement of Theorem 67:
In Euclidean Geometry, if a line intersects one of two parallel lines, then it also intersects the
other.
Justify the steps in the following proof (we discussed this proof in class). Make a drawing.
1. Suppose that lines L and M are parallel and line T intersects line M. Denote by P the point
of intersection of lines T and M, and observe that line M is a line that passes through P
and is parallel to L, and also observe that line T is a line that passes through P. (Make a
drawing)
2. There is at most one line that passes through P and are parallel to L. (by _____________
___________________________________________, so line T cannot be parallel to L.
3. Line T intersects line L. (by ________________________________________________
______________________________________________________________________)
End of proof
Page 128 of 150
[2] Recall the statement of Theorem 68:
In Euclidean Geometry, 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 the statement of Theorem
38, The Alternate Interior Angle Theorem.)
Justify the steps in the following proof. Make a drawing where indicated.
1. Suppose that L and M are parallel lines cut by a transversal T.
2. There exist points A, B, C on line L and points D, E, F on line M such that point B is the
point of intersection of lines L and T, and point E is the point of intersection of lines M
and T, and A*B*C, and D*E*F, and such that points A and D are on the same side of line
T. (This step can be justified by labelling the intersection points and by invoking the
betweenness axioms.) (Make a drawing) With this labelling scheme, angles ABE and
FEB are alternate interior angles. (Our goal is to show that these two angles are
congruent.)
3. There exists a ray EG such that points G and A are on opposite sides of line T and such
that GEB  ABE . (by axiom ________________________) (Add to your drawing)
4. Line EG is parallel to L. (by step ___ and Theorem ___________________________
_____________________________________________________________________)
5. Line EG must be the same line as line M. (because ___________________________
_____________________________________________________________________
_____________________________________________________________________)
6. Rays EF and EG are the same ray. (By step 2, 3, and 5, we know that E, F, and G are
collinear and that points F and G are on the same side of line T. Theorem __________
_____________________________________________________________________
can be used to justify that the rays are the same.) So angles GEB and FEB are the
same angle.
7.
FEB  ABE (by seps ___ and ___)
End of proof
Page 129 of 150
[3] Recall the statement of Theorem 70:
In Euclidean Geometry, 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.
Justify the steps in the following proof. Make a drawing where indicated.
1. Suppose that L and M are parallel lines cut by a transversal T.
2. There exist points A, B, C on line L and points D, E, F on line M such that point B is the
point of intersection of lines L and T, and point E is the point of intersection of lines M
and T, and A*B*C, and D*E*F, and such that points A and D are on the same side of line
T. (This step can be justified by labelling the intersection points and by invoking the
betweenness axioms.) (Make a drawing) With this labelling scheme, angles ABE and
DEB are a pair of angles on the same side of the transversal. (Our goal is to show that
the measures of these two angles add up to 180.)
3. Angles DEB and FEB are supplementary angles. (by step ___ and definition of
________________________________________________)
4. degrees  DEB   degrees  FEB   180 (by step ___ and Theorem _____________
____________________________________________________________________)
5. Angles ABE and FEB are alternate interior angles. (by _____________________
_____________________________________________________________________)
6.
ABE  FEB (by steps ___ and ___ and Theorem ___________________________
______________________________________________________________________)
7. degrees  ABE   degrees  FEB  (by step ___ and Theorem ___________________
______________________________________________________________________)
8. degrees  DEB   degrees  AEB   180 (by steps ___ and ___ and algebra)
End of proof
[4] Prove Theorem 71: In Euclidean Geometry, if a line is perpendicular to one of two parallel
lines, then it is perpendicular to the other. Hint: the proof can be done in just three lines if you
use two recent theorems.
Page 130 of 150
[5] Recall the statement of Theorem 72:
In Euclidean Geometry, every triangle has angle sum exactly 180.
Justify the steps in the following proof. (We discussed the proof in class). Make a drawing
1. Let ABC be given. (make a drawing, and try to make the triangle lopsided)
2. There exists a line L that passes through C and is parallel to line AB . (by Theorem ____
____________________________________________________) (add to your drawing)
3. There exist points D and E on line L such that D*C*E and such that points D and A are
on the same side of line BC . (This step can be justified by betweenness axioms and
theorems.) (add to your drawing)
4. Angles DCB and BCE are supplementary. (by steps ___ and ___ and definition of
______________________________.)
5. degrees  DCB   degrees  BCE   180 (by step ___ and Theorem _______________
_______________________________________________________________________)
6. degrees  DCB   degrees  DCA  degrees  ACB  (by Theorem ______________)
7. degrees  DCA  degrees  ACB   degrees  BCE   180 (by steps ___ and ___ and
algebra)
1. Angles DCA and CAB are alternate interior angles. (by ______________________
______________________________________________________________________)
2.
DCA  CAB (by steps ___ and ___ and Theorem ___________________________
_______________________________________________________________________)
3. degrees  DCA  degrees  CAB  (by step ___ and Theorem ____)
4. Angles ECB and CBA are alternate interior angles. (by ______________________
______________________________________________________________________)
5.
ECB  CBA (by steps ___ and ___ and Theorem ___________________________
______________________________________________________________________)
6. degrees  ECB   degrees  CBA (by step ___ and Theorem ______________)
7. degrees  CAB   degrees  ACB   degrees  CBA  180 (by steps ___ and ___ and
___ and algebra)
End of proof
Page 131 of 150
[6] Recall the statement of Theorem 73:
In Euclidean Geometry, the opposite sides of a parallelogram are congruent.
Justify the steps in the following proof. Make a drawing where indicated. And supply the second
half of the proof.
1. Let
ABCD be a parallelogram. (so AB CD and AD BC .) (Make a drawing)
Part 1: Prove that DA  BC by using diagonal AC .
2. Angles DCA and BAC are alternate interior angles. (by _________________
________________________________________________) (add to your drawing)
3.
DCA  BAC (by steps ___ and ___ and Theorem _______________________
__________________________________________________________________)
4. Angles DAC and BCA are alternate interior angles. (by ___________________
___________________________________________________________________)
5.
DAC  BCA (by steps ___ and ___ and Theorem ________________________)
6. AC  AC (by _______________________________________________________)
7. DAC  BCA (by steps ___ and ___ and ___ and __________________________)
8. DA  BC (by step _ and the definition of __________________________________)
Part 2: Prove that AB  CD by using steps analogous to steps 2 – 8, but using diagonal BD .
End of Proof
Page 132 of 150
The next three exercises deal with justifications of proofs of Neutral Geometry Theorems about
Circles.
[7] Recall the statement of Theorem 78:
In Neutral Geometry, diameters of a circle are the longest chords.
Given a circle, if segment CD is a chord but not a diameter, and segment AB is a
 
 
diameter, then segment CD  AB . That is, length CD  length AB .
Justify the steps in the following proof. Make a drawing where indicated.
1. Suppose that segment CD is a chord but not a diameter, and segment AB is a diameter.


2. Let point F be the center of the circle. We can denote the circle by Circle F , FC . Let
point G be the second intersection of line CF with the circle. Then segment CG is a
diameter. (Make a drawing showing the circle and chord CD and diameter CG .)
Part 1: show that CD  CG .
3. FD  FG (by _________________________________________) (add to your drawing)
4.
FGD  FDG (by ____________________________________________________)
5. Observe that point F is in the interior of angle ADG , so that ray DF is between rays
DC and DG .
6.
FDG  CDG (by step ___ and definition of ________________________________)
7.
FGD  CDG (by steps ___ and ___ and Theorem ___________________________)
8. segment CD  CG . (by step ___ and Theorem _________________________________
_______________________________________________________________________)
Part 2: show that CD  AB .
9. CG  AB (by steps ___ and ___ and Theorem _________________________________
_______________________________________________________________________)
10. CD  AB (by steps ___ and ___ and Theorem __________________________________
_______________________________________________________________________)
End of proof
Page 133 of 150
[8] Recall the statement of Theorem 79:
In Neutral Geometry, the following are equivalent:
(1) Diameter AB is perpendicular to non-diameter chord CD .
(2) Diameter AB bisects non-diameter chord CD .
Justify the steps in the following proof. Make drawings where indicated.
Part 1: Prove that (2)  (1)
1. Suppose that diameter AB bisects non-diameter chord CD . Let E be the center of the
circle and let F be the point of intersection of AB and CD . (Make a drawing)
2. CF  DF (by definition of ____________________) (add to your drawing)
3. EF  EF (by ___________________________________________________)
4. EC  ED (by ____________________________________________________)
5. CFE  DFE (by steps ___ and ___ and ___ and Theorem _________________
__________________________________________________________________)
6.
CFE  DFE (by step ___ and the definition of _________________________
__________________________________________________________________)
7. Angles CFE and DFE are supplementary. (by ________________________)
8. Angles CFE and DFE are right angles. (by steps ___ and ___ and definition of ___
______________________________________________________________________)
9. AB is perpendicular to CD . (by step ___ and definition of perpendicular)
Part 2: Prove that (1)  (2)
1. Suppose that diameter AB is perpendicular to non-diameter chord CD . Let E be the
center of the circle and let F be the point of intersection of AB and CD . (We must show
that F is the midpoint of segment CD .) (make a drawing)
2. Assume that F is not the midpoint of segment CD . (assumption)
3. Segment CD has a midpoint that we can label G. (by Theorem _________________)
4. Line AG bisects segment CD . (by step ___ and definition of _________________)
5. Line AG contains a diameter HJ that also bisects segment CD . (by steps 4 and 3 and
definition of diameter)
6. Diameter HJ is perpendicular to segment CD . (by step 5 and Part 1 of this proof)
7. Line AG is perpendicular to line CD . (by steps 5 and 6)
8. Line AB is perpendicular to line CD . (by step 1)
9. Lines AG and AB are parallel. (by step ___ and Theorem _______________________
______________________________________________________________________)
10. Step 9 contradicts the fact that lines AG and AB intersect at point A. Therefore, our
assumption in step 2 was wrong. It cannot be that F is not the midpoint of segment CD .
Therefore, F is the midpoint of segment CD .
End of proof
Page 134 of 150
Recall the statement of Theorem 80:In Neutral Geometry, the perpendicular bisector of a chord
passes through the center of the circle.
Justify the steps in the following proof. Make a drawing where indicated.


1. Suppose that segment CD is a chord of Circle A, AB .
2. Segment CD has a perpendicular bisector that we can call line L. (by Theorem ______
______________________________________________________________) We must
show that line L passes through the center of the circle.
3. Line L is perpendicular to segment CD at a point E that is the midpoint of segment CD .
(by definition of perpendicular bisector)
4. Either segment CD is a diameter chord or segment CD is a non-diameter chord.
Case 1: Segment CD is a diameter chord.
5. Suppose that segment CD is a diameter chord. (make a drawing)
6. Then the midpoint E of segment CD is actually the center of the circle, so line L passes
through the center of the circle.
Case 2: Segment CD is a non-diameter chord
7. Suppose that segment CD is a non-diameter chord. (make a new drawing)
8. Line AE intersects the circle at two points that we can call F and G, creating diameter
segment FG . (add to your drawing)
9. Diameter FG bisects non-diameter chord CD . (because diameter FG is part of line
AE that passes through point E.)
10. Diameter FG is perpendicular to non-diameter chord CD . (by step ___ and Theorem
__________________________________________________________________)
11. Line AE is a perpendicular bisector of segment CD . (by steps ___ and ____ and
definition of _____________________________________________)
12. Line L and line AE must be the same line. (because segment CD has only one
perpendicular bisector, by Theorem ____________________________) So line L passes
through the center of the circle.
End of proof
Page 135 of 150
11.For Reference: Axioms, Definitions, and Theorems of Euclidean Geometry
11.1. The Axioms of Euclidean Geometry
Primitive Relations:
 incidence relation, written ―the point lies on the line.‖
 betweenness relation for points, written ―B is between A and C‖, denoted A*B*C.
 congruence of segments, written ―segment AB is congruent to segment CD‖, AB  CD .
 congruence of angles, written ―angleABC is congruent to angleDEF‖, ABC  DEF
Incidence Axioms
IA1: There exist three distinct non-collinear points. (at least three)
IA2: For every pair of distinct points, there is exactly one line that passes through both points.
IA3: Every line passes through at least two distinct points.
Betweenness Axioms
BA1: If A*B*C, then A, B, and C are distinct collinear points, and C*B*A.
BA2: 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.
BA3: 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.
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) 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.
The Euclidean Parallel postulate (EPP): For any line L and for any point P not on L, there is at
most one line that passes through P and is parallel to L.
Page 136 of 150
11.2. The Definitions of Euclidean Geometry
Definition 18 passes through
 words: Line L passes through point P.
 meaning: Point P lies on line L.
Definition 19 intersecting lines
 words: Line L intersects line M.
 meaning: There exists a point P that lies on both lines. (at least one point)
Definition 20 parallel lines
 words: Line L is parallel to line M.
 symbol: L M

meaning: Line L does not intersect line M.
Definition 21 collinear points
 words: The set of points P1 , P2 ,

meaning: There exists a line L that passes through all the points.
Definition 22 concurrent lines
 words: The set of lines L1 , L2 ,

, Pk  is collinear.
, Lk  is concurrent.
meaning: There exists a point P that lies on all the lines.
Definition 23 Abstract Model and Concrete Model
 An abstract model of an axiom system is a model that is, itself, another axiom system.
 A concrete model of an axiom system is a model that uses actual objects and relations.
Definition 24 Relative Consistency and Absolute Consistency
 An axiom system is called relatively consistent if an abstract model has been
demonstrated.
 An axiom system is called absolutely consistent if a concrete model has been
demonstrated.
Definition 25 Hpoint
 word: Hpoint
 meaning: an (x,y) pair in the interior of the unit circle
Definition 26 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 27 Ternary Relation on a Set
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



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
Definition 28 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 29 symbol for a line
 symbol: AB
 meaning: the (unique) line that passes through points A and B
Definition 30 the set of points that lie on a line
 

symbol: AB

meaning: the set of points that lie on line AB
Definition 31 the plane
 meaning: the set of all points
Definition 32 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 33 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 34 opposite rays
 words: BA and BC are opposite rays
 meaning: A*B*C
Definition 35 same side
Page 138 of 150



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 36 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 37 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
Definition 38 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 39 supplementary angles
 words: supplementary angles
 meaning: two angles that share a common side and whose other sides are opposite rays.
Definition 40 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 41 ray between two other rays
 words: ray BD is between BA and BC .
 meaning: Point D is in the interior of ABC .
Definition 42 triangle
 symbol: ABC
 spoken: triangle A, B, C
 usage: A, B, and C are non-collinear
 meaning: the set AB BC CA
 additional terminology:
Page 139 of 150
o
o
o
o
o
The points A, B, C are called the vertices of the triangle.
The segments AB, BC, CA are called the sides of the triangle.
Side BC is said to be opposite vertex A. Similarly for the other sides.
The angles ABC , BCA , and CAB are called the angles of the triangle.
The angle ABC is called angle B when there is no chance of this causing
confusion. Similarly for the other angles.
Definition 43 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 44 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.
Definition 45 the order relation on the set of line segments
 Symbol: AB  CD
 Spoken: “Segment A,B is less than segment C,D.”
 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 46 ―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.
Page 140 of 150
Definition 47 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 48 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 49 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.”
 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.
Definition 50 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 51 right angle
 words: right angle
 Meaning: An angle that is congruent to its supplementary angle
Definition 52 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.
Definition 53 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 .
Page 141 of 150

Remark: The order relation is a binary relation on the set of angles.
Definition 54 transversal
 Words: “Line T is transversal to lines L and M.”
 Meaning: “T intersects L and M in distinct points.”
Definition 55 transversal―alternate interior angles‖; ―corresponding angles‖
 Usage: Lines L, M, and transversal T are given.
 Labeled points: Let B be the
H T
intersection of lines T and L, and let E
D
be the intersection of lines T and M.
M
E
F
(By definition of transversal, B and E
A
B
C
L
are not the same point.) By the
G
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.
 Meaning:
ABE and FEB is a pair of alternate interior angles.


CBE and DEB is a pair of alternate interior angles.

ABG and DEG is a pair of corresponding angles.
ABH and DEH is a pair of corresponding angles.


CBG and FEG is a pair of corresponding angles.

CBH and FEH is a pair of corresponding angles.
Definition 56 ―isosceles triangle‖
 Words: isosceles triangle
 Meaning: two sides of the triangle are congruent to each other (at least two)
 Additional Terminology: The angles opposite the two congruent sides are called the base
angles.
Definition 57 ―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.
Definition 58 ―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 59 ―midpoint‖ of a line segment
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

Words: C is a midpoint of segment A, B
Meaning: A*C*B and CA  CB .
Definition 60 ―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 61 ―bisector‖ of an angle
 Words: a bisector of angle ABC
 Meaning: a ray BD between rays BA and BC such that
DBA  DBC .
Definition 62 the length function for line segments in straight-line drawings (Euclidean)
Meaning: the function lengthE : Euclideanline segments   defined by the following
formula
 
lengthE AB 
 x1  x2    y1  y2 
2
2
,
where segment AB is a Euclidean line segment with endpoints A   x1 , y1  and
B   x2 , y2  . (This implies that x1, y1, x2,and y2 are real numbers.)
Definition 63 the length function for line segments in the Hyperbolic plane
Meaning: the function lengthH : Hyperbolicline segments   defined by the
following formula
 lengthE  AR 


lengthE  AS  
,
lengthH  AB   ln 
 lengthE  BR 



length
BS


E


where segment 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 .
Definition 64: Circle
 Version of the definition that does not use the notion of distance


o Symbol: Circle A, AB .
o Words: Circle centered at A, with radial segment AB .
o Meaning:The set of points P such that segment AP is congruent to segment AB .


o Meaning in symbols: P : AP  AB .

Version of the definition that does use the notion of length
o Symbol: Circle  A, r  .
o Words: Circle centered at A, with radius r.
Page 143 of 150
o Meaning:The set of points P that are a distance r from point A.

  
o Meaning in symbols: P : length AP  r .
Definition 65: Segments and lines related to a circle
 Given a circle,
o A chord is a line segment whose endpoints are points on a circle.
o A diameter is a chord that also contains the center of the circle.
o A secant is a line that passes through exactly two points of the circle.
o A tangent is a line that passes through exactly one point of the circle.
Definition 66: Circumscribe

Words: Triangle

Meaning: Points A, B, and C lie on the circle.


ABC is circumscribed by circle Circle P, PQ .
11.3. The Theorems of Euclidean 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.
Page 144 of 150
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 segment
AC or segment BC . Furthermore, if C does not lie on L, then L does not intersect both segment
AC and segment 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 .
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 (segment subtraction): In Neutral Geometry, if A*B*C and D*E*F and AB  DE
and AC  DF , then BC  EF .
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Theorem 26: 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 27: 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 .
Theorem 28: 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 29: In Neutral Geometry, vertical angles are congruent.
Theorem 30: In Neutral Geometry, any angle congruent to a right angle is also a right angle.
Theorem 31: In Neutral Geometry, for every line L and every point P not on L there exists a line
through P perpendicular to L.
Theorem 32: In Neutral Geometry, for every line L and every point P on L there exists a unique
line through P perpendicular to L.
Theorem 33 (angle addition):
Given: Neutral Geometry; angle ABC with point D in the interior; angle
point H in the interior; DBA  HFE ; DBC  HFG
Claim: ABC  EFG
Theorem 34 (angle subtraction):
Given: Neutral Geometry; angle ABC with point D in the interior; angle
point H in the interior; DBA  HFE ; ABC  EFG
Claim: DBC  HFG
EFG with
EFG with
Theorem 35 (Theorem about Rays and Angles): In Neutral Geometry, if ABC  EFG and
ray BD is between rays BA and BC , then there is a unique ray FH between rays FE and FG
such that ABD  EFH .
Theorem 36 (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 ,
(b): If P  Q and Q  R then P  R .
(c): (transitivity): If P  Q and Q  R then P  R .
P
Q , or
Theorem 37 (Euclid’s 4th postulate): All right angles are congruent to each other.
Q
P.
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Theorem 38 (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 39 (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 40: In Neutral Geometry, two distinct lines perpendicular to the same line are parallel.
Theorem 41 (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.
Theorem 42 (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 43 (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.
Theorem 44 (The Isosceles Triangle Theorem):
Given: Neutral Geometry, ABC
Claim: If AB  AC then B  C . (If two sides are congruent, then the two opposite angles are
congruent.)
Theorem 45 (The Converse of the Isosceles Triangle Theorem)
Given: Neutral Geometry, ABC
Claim: If B  C then AB  AC (If two angles are congruent, then the two opposite sides are
congruent.)
Theorem 46 (The CACS Theorem): In Neutral Geometry triangles, Congruent Angles are always
opposite Congruent Sides.
Theorem 47 (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 48 (The Side-Side-Side Congruence Theorem) (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 sides of the other triangle, then all the remaining corresponding
parts are congruent as well, so the triangles are congruent.
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Theorem 49 (The Exterior Angle Theorem): In Neutral Geometry, each of the remote interior
angles is less than the exterior angle.
Theorem 50 (The Angle Angle Side Angle Congruenc Theorem) (AAS): 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 51 (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 52: In Neutral Geometry triangle
ABC , if AC  BC then
Theorem 53: In Neutral Geometry triangle
ABC , if
B
B
A.
A then AC  BC .
Theorem 54 (The SASS Theorem): In Neutral Geometry triangles, Smaller Angles are always
opposite Smaller Sides. (Not a new theorem, but rather just a combining of Theorem 52 and
Theorem 53.)
Theorem 55: The CACS and SASS Theorem: In triangles of Neutral Geometry, Congruent Angles
are always opposite Congruent Sides, and Smaller angles are always opposite Smaller Sides.
(Not a new theorem, but rather just a combination of the two theorems: Theorem 46 (the CACS
Theorem), and Theorem 54 (the SASS Theorem). But these in turn are combinations of four
theorems: Theorem 44, Theorem 45, Theorem 52 and Theorem 53.)
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: In Neutral Geometry, every segment has exactly one midpoint.
Theorem 58: In Neutral Geometry, every line segment has exactly one perpendicular bisector.
Theorem 59: In Neutral Geometry, every angle has exactly one bisector.
Theorem 60 Existence of a Length Function for Line Segments.
Given: Neutral Geometry and some reference line segment AB .
Claim: There exists a unique function length   : line segments 
properties.

with the following
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 
1. length AB  1 .
2. The length function is onto. That is, for every positive real number L, there exists a line
 
segment CD such that length CD  L .
 
 
length  CD   length  EF  if and only if. CD  EF .
length  CE   length  CD   length  DE  if and only if C*D*E.
3. length CD  length EF if and only if CD  EF .
4.
5.
Theorem 61 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. degrees  A  90 if and only if A is a right angle.
2. The degrees function is onto. That is, for every real number 0  m  180 , there exists an
angle A such that degrees  A  m .
3. degrees  A  degrees  B  if and only if
A
B.
4. degrees  A  degrees  B  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 .
Theorem 62 In Neutral Geometry, the sum of the measures of any two angles of a triangle is less
than 180.
Theorem 63 (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 64 (The Saccheri-Legendre Theorem) In Neutral geometry, the angle sum of any
triangle is less than or equal to 180.
Theorem 65 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.
Theorem 66: Thirteen Statements that are Logically Equivalent in Neutral Geometry
Given: The axioms for Neutral Geometry
Claim: The following statements are logically 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.‖
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2) A 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
that passes through P and is parallel to L.
3) Playfair’s Postulate: For every line L and every point P not on L, there is a unique line
that passes through 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) ―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 38, The Alternate
Interior Angle Theorem.)
7) ―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
39,The Corresponding Angle Theorem,)
8) ―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.‖
9) ―Every triangle has angle sum exactly 180.‖
10) ―Given any triangle 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 .‖
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.‖
Theorem 67: In Euclidean Geometry, if a line intersects one of two parallel lines, then it also
intersects the other.
Theorem 68: In Euclidean Geometry, 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 the statement
of Theorem 38, The Alternate Interior Angle Theorem.)
Theorem 69: In Euclidean Geometry, 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 39,The Corresponding Angle Theorem.)
Theorem 70: In Euclidean Geometry, 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.
Theorem 71: In Euclidean Geometry, if a line is perpendicular to one of two parallel lines, then it
is perpendicular to the other.
Theorem 72: In Euclidean Geometry, every triangle has angle sum exactly 180.
Theorem 73: In Euclidean Geometry, the opposite sides of a parallelogram are congruent.
Theorem 74: In Euclidean Geometry, the opposite angles of a parallelogram are congruent.
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Theorem 75: In Euclidean Geometry, the diagonals of a parallelogram bisect each other.
Theorem 76: In Neutral Geometry, the following are equivalent:
(1) Point P lies on the perpendicular bisector of segment AB .
 
 
(2) Point P is equidistant from A and B. That is, AP  BP and length AP  length BP
Theorem 77: In Neutral Geometry, all diameters of a given circle are congruent.
Theorem 78: In Neutral Geometry, diameters of a circle are the longest chords.
Given a circle, if segment CD is a chord but not a diameter, and segment AB is a
 
 
diameter, then segment CD  AB . That is, length CD  length AB .
Theorem 79: In Neutral Geometry, the following are equivalent:
(1) Diameter AB is perpendicular to non-diameter chord CD .
(2) Diameter AB bisects non-diameter chord CD .
Theorem 80: In Neutral Geometry, the perpendicular bisector of a chord passes through the
center of the circle.
Theorem 81: In Neutral Geometry, the following are equivalent:
(1) Line L is perpendicular to segment AP at point P.


(2) Line L is tangent to Circle A, AP at point P.
Theorem 82: In Euclidean Geometry, If A, B, and C are non-collinear points, and line L is
perpendicular to line AB , and line M is perpendicular to line BC , then lines L and M intersect.
Theorem 83: In Euclidean Geometry, If A, B, and C are non-collinear points, and line L is the
perpendicular bisector of segment AB , and line M is the perpendicular bisector of segment BC ,
and line N is the perpendicular bisector of segment CA , then lines L, M, and N are concurrent.
Theorem 84: In Euclidean Geometry, If A, B, and C are non-collinear points, then there exists
exactly one circle that contains all three points.
Theorem 85: In Euclidean Geometry, every triangle can be circumscribed by exactly one circle.
That is for every triangle ABC , there exists exactly one circle that contains points A, B, and C.