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Chapter 1
Geometry and axiomatic
Method
1.1
Origin of Geometry
The word geometry has its roots in the Greek word “geometrein”, which
means earth measuring. Before the time of recorded history, geometry originated out of practical necessity; it was the science of measuring land. Many
ancient civilizations (Babylonian, Hindu, Chinese, and Egyptian) possessed geometric informations. The first geometrical considerations had their origin in
simple observations stemming from human ability to recognize physical form
and to compare shapes and sizes. There were many circumstances in which
primitive people were forced to take on geometric topics, although it may not
have been recognized as such. For instance, man had to learn with situations involving distance, bounding their land, and constructing walls and homes. These
types of situations were directly related to the geometric concepts of vertical,
parallel, and perpendicular.
The geometry of the ancient days was actually just a collection of rules concerning lengths, angles, areas, and volumes which were found through experimentation, observation of analogies, guessing, and sometimes even intuition to meet
some practical need in surveying, construction, astronomy, and various crafts.
The Babylonian from 2000 to 1600 B.C. were familiar with the general
rules for computing the area of rectangle, the area of right isosceles triangles,
and the volume of a rectangular parallelepiped. The circumference of a circle
was taken as three times the diameter and the area as one twelfth the square of
the circumference. Both formulas correct for π = 3. Also, the volume of a right
circular cylinder was then obtained by finding the product of the base and the
altitude. While the volume of a truncated square pyramid appears incorrectly
as a product of the altitude and half the sum of the bases. They also knew that
corresponding sides of two similar right triangles are proportional. Pythagorean
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CHAPTER 1. GEOMETRY AND AXIOMATIC METHOD
theorem was also known.
The geometry in ancient Egypt (3000 B.C.– 600 B.C.) was arising out of
the need of survey land. Because of the Nile frequently flooded, washing away
land, property had to be frequently surveyed for taxation. Egyptian surveyors
were highly skilled in geometry. Also the precise construction of the pyramids
indicate the use of geometry.
The mathematics of Egyptian geometry is documented by examples of rules
for determining areas and volumes of common plane and solid objects. They
appear to be based on trial and error results and observations.
Egyptian knew that the area of any triangle is 21 the product of the base times
the height. The Egyptian had the correct formula for the volume of a square
1
pyramid which is ha2 , where h is the height of the pyramid and a is the length
3
of the side of its square base. Even more remarkably, they had the formula for
the volume of a truncated square pyramid which is
1
h(a2 + ab + b2 ),
3
where h is the height and a, b are length of sides of the two square bases.
There is no documentary evidence that ancient Egyptian were ware of Pythagorean
theorem, early Egyptians surveyors realized that the triangle having sides of
lengths 3, 4 and 5 units is right triangle. They had the incorrect formula for
the area of a quadrilateral with successive sides of lengths a, b, c and d to be
(a + c)(b + d)
.
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This formula is true for rectangles.
Very likely, mathematical achievements similar to those of ancient Babylonians and Egyptians also occurred in ancient India and China, but we know
very little indeed with any degree of certainty about those achievements. The
ancient Egyptians recorded their work on stones and papyrus, the latter fortunately resisting the ages, because of Egypt dry climate. And the Babylonian
used everlasting baked clay tablets.
In contrast to the use of these median, the early indians and chinese used very
perishable writing materials like bark bast and bamboo.
The Greeks (600 B.C.– 400 A.D.) worked to transform geometry into something much different than the former geometry of their predecessors. The Greeks
insisted that geometric facts must be established, not by empirical procedures,
but by deductive reasoning, geometrical conclusions must be arrived at by logical demonstration rather than by trial and error experimentations. In short,
Greeks transformed the empirical geometry into deductive or systematic geometry.
Virtually there exist no first-hand sources of early Greek geometry. Hence,
1.1. ORIGIN OF GEOMETRY
3
the following is based on manuscripts written hundreds of years after the early
Greek geometry had been developed. According to these manuscripts, Thales
of Miletus was the one who began early Greek geometry in the sixth century
B. C. He insisted that, geometric statements must be established by deductive
reasoning rather than by trial and error. He was familiar with the computations recorded from Egyptian and Babylonian mathematics, and he developed
his logical geometry by determining which results were correct.
The next mentioned great Greek geometer is one who quite possibly studied
under Thales of Miletus. This geometer is Pythagoras. He lived around 570–490
B. C., he founded the Pythagorean school, which was committed to the study of
philosophy, mathematics, and natural science. The systematization of geometry
was further developed during this time. The members of this school developed
the properties of parallel lines to prove that the sum of the angles of any triangle
is equal to two right angles. They also worked with proportion to study similar
figures. We all think of the Pythagorean Theorem when we think of Pythagoras,
however it is important to note that this theorem was used (although it may
not have been proved) before his time.
Hippocrates of Chios was one of these students at the Pythagorean school.
It is suggested that he was the first to attempt a logical presentation of geometry in the form of a single chain of propositions. He is credited to writing the
first “Elements ”. Although his book has been lost, it covered most of Books
of Euclid’s Element, which appeared about century later. Hippocrates was the
first to show that the ratio of the areas of two circles equals the ratio of the
squares of the radii of the circles .
Although Plato around 427 to 327 B. C. did not make any major mathematical discoveries himself, he did emphasize the idea of proof. He insisted
on accuracy, which helped pave the way for Euclid. Around 300 B. C. Euclid collected the theorems of Pythagoras, Hippocrates, and others into a work
called “The Elements ”. Euclid is the most widely read author in the history of
mankind. The teaching of geometry has been dominated by Euclid’s approach
to the subject. In fact, Euclid’s axiomatic method is the prototype for all pure
mathematics. By pure, it is meant that all statements can be verified through
reasoning of demonstrations, no physical experiments are necessary.
The “Elements”was divided into thirteen books and contained 465 propositions from plane and solid geometry and from number theory. His genius was
not so much in creating new mathematics but rather in the presentation of old
mathematics in a clear, logical and organized manner. He provided us with an
axiomatic development of the subject. The Elements begins with 23 definitions,
5 postulates and 5 common notions or general axioms. From these he proved
his first proposition. All subsequent results were obtained from a blend of his
definitions, postulates, axioms and previously proven propositions.
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1.2
Axiomatic Method
Greeks devised the axiomatic method, and modern science still uses this
method. The axiomatic method is based on a system of deductive reasoning.
In a deductive system, statements used in an argument must be derived from
some previous statement or statements. These previous statements must themselves be derived from even earlier statements. Clearly this cannot be continued
indefinitely, nor should one resort to logical circularity by deriving statement A
from statement B and statement B from statement A.
The Greek found a way out of this difficulty, by assuming the truth of
a collection of initial statements whose truth are acceptable without justification, and then proceeding, by purely deductive reasoning, to derive all the other
statements of the discourse. Euclid called these prior statements postulates or
axioms1 .
The prior and the derived statements involve special terms. These terms need
to be defined. Since these should be defined by means of other terms, one is
faced with the problem of circularity. Thus, to avoid circularity of definition,
where term y is defined by term x, and then term x is defined by means of term
y, one is forced to set down at the very start a collection of basic terms whose
intended meanings should be made clear to the reader. All subsequent terms
must then be defined by means of terms already introduced. This method, created by Greek, is called axiomatic method.
Starting from a base of undefined terms and agreed upon axioms, we can
define other terms and use our axioms to argue the truth of other statements.
These other statements are called the theorems of the system. Also, there must
be agreement on how and when one statement follows logically from another,
that means agreement on certain rules of reasoning. Thus deductive system or
axiom system consists of the following components:
1. Undefined terms
2. Axioms (or postulates)
3. Defined terms
4. Theorems
5. Rules of logic.
The axioms; which are statements to be accepted that are true without
justification, they state the basic properties of undefined terms and should be
simple in structure and few in number.
1 Early Greeks made a distinction between axioms and postulates. Evidence exists, that
Euclid made a distinction that an axiom (common notion) is an assumption common to all
sciences and that postulate is an assumption peculiar to the particular science being studied.
In modern times no distinction is made between the two; an axiom or a postulate is an assumed
statement.
1.2. AXIOMATIC METHOD
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In an axiom system a proof of a specific statement is a sequence of statements,
together with a justification for each statement, ending up with desired conclusion. In a given axiom system, the only statements we call theorems are those
statements for which a proof has been given. By the way, all mathematical theorem are conditional statements. In some cases, a theorem may not be written
in a conditional form, but it can be reformulated as a conditional statement.
For example:
Base angles of an isosceles triangle are congruent.
It can be reformulated as:
If a triangle has two congruent sides then
the angles opposite to those sides are congruent.
It is common in mathematics to reserve the word theorem for a result
that has a special importance and significance in the subject and central in
the development of the theory, while other statements of less impact are called
propositions. Now, if a theorem is not really interesting (according to author) in itself, or is a result that is not related to the subject at hand, but
is only stated because it makes the proof of some other theorem more interesting, then it is called lemma. Also, if a theorem follows almost immediately
as a result of another theorem with very short proof, then it is called corollary.
Now, I will remind you with some basic rules of reasoning, that you
already know.
1. The following are types of justifications allowed for statements in a proof:
(a) By hypothesis.
(b) By axiom.
(c) By a previously proven theorem.
(d) By definition.
(e) By a previous established step in the same proof.
(f) By rule of logic.
2. Proof by Contradiction:
To prove P ⇒ Q, assume the negation of statement Q and deduce an
absurd statement, using the hypothesis statement P if needed in the deduction.
3. Negation
(a) ∼ (∼ P ) is equivalent to2 P .
(b) ∼ (P ⇒ Q) is equivalent to P ∧ ∼ Q
2 Two
statements are called equivalent if and only if they have the same truth values.
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(c) ∼ (P ∧ Q) is equivalent to ∼ P ∨ ∼ Q
(d) ∼ (P ∨ Q) is equivalent to ∼ P ∧ ∼ Q
(e) ∼ (∀x, P (x)) is equivalent to the statement ∃x, ∼ P (x)
(f) ∼ (∃x, P (x)) is equivalent to the statement ∀x, ∼ P (x)
4. If P ⇒ Q and P are steps in a proof then Q is a justifiable step in the
proof.
5. Contrapositive: ∼ Q ⇒∼ P is equivalent to the statement P ⇒ Q.
6. For every statements P , Q and R, the following statements are true:
(a) ((P ⇒ Q) ∧ (Q ⇒ R)) ⇒ (P ⇒ R).
(b) (P ∧ Q) ⇒ P
(c) (P ∧ Q) ⇒ Q.
7. Law of excluded middle: For every statement P , P ∨ ∼ P is a valid
step in a proof.
8. Proof by cases:
If ((P1 ⇒ Q) ∧ (P2 ⇒ Q) ∧ · · · ∧ (Pn ⇒ Q)) ⇒ ((P1 ∨ P2 ∨ · · · ∨ Pn ) ⇒ Q).
That means, if P1 ∨ P2 ∨ · · · ∨ Pn is a valid step in a proof and Q is carried
out from each statement Pi . Then Q is a valid step in the proof.
1.3
Examples and Properties of Axiom System
We will start with the following example:
Example 1.3.1 Consider the undefined terms; ant, path, and has. The following is the axiom set.
Axiom 1. Every ant has at least two paths.
Axiom 2. Every path has at least two ants.
Axiom 3. There exists at least one ant.
Note that has indicates the relationship between ant and path. We can prove
two theorems from these axioms.
Theorem 1. There exists at least one path.
Proof: By Axiom 3, there exists an ant. Now since each ant must have at least
two paths by Axiom 1, there exists at least one path. Theorem 2. The minimum number of paths is two.
Proof: By Axiom 3, there exists an ant, call it A1 . Then by Axiom 1, A1 must
1.3. EXAMPLES AND PROPERTIES OF AXIOM SYSTEM
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have two paths call them P1 and P2 . Hence, there are at least two paths. Example 1.3.2 Undefined terms: point, line, has. The set of axioms are:
Axiom 1. There exists exactly four lines
Axiom 2. Any two distinct lines has exactly one point in common.
Axiom 3. Each point is on exactly two lines.
We can prove two theorems in this axiom system:
Theorem 1. There is exactly 6 points.
Proof. By axiom 1, there are four lines, say l,m, n, and t. By axiom 2, l and
n has unique point in common, say B, l and t has unique point in common, say
C, m and n has unique point in common, say D, m and t has unique point in
common, say E, and n and t has unique point in common, say F . Thus, there
is at least 6 points. Now, suppose that, G is another point in the system. Then
by axiom 3, G is on exactly two lines, say l, and m. Then by axiom 2, G = A.
Thus, there is exactly 6 points.
Theorem 2. Each line has exactly 3 points on it.
Proof. Exercise. Example 1.3.3 A group G consists of a set of undefined objects called elements
and a binary operation o that relates two elements to a third one. The axioms
are
Axiom 1. For all x, y ∈ G, x ◦ y ∈ G
Axiom 2. For all x, y ∈ G, (x ◦ y) ◦ z = x ◦ (y ◦ z)
Axiom 3. There is an element in G, say e, such that x ◦ e = e ◦ x = x,
∀x ∈ G.
Axiom 4. For every x ∈ G, there is an element y ∈ G such that x ◦ y = e =
y ◦ x.
Some theorems in this system are:
Theorem 1. A group G has exactly one identity.
Proof. Exercise. Theorem 2. If x, y, z in a group G and x ◦ z = y ◦ z then x = y.
Proof. Exercise. 8
CHAPTER 1. GEOMETRY AND AXIOMATIC METHOD
Example 1.3.4 Given the undefined terms: student, class, and the relation
belongs to. The set of axioms are:
Axiom 1. There exists exactly three distinct students in this system.
Axiom 2. Any two distinct students belong to exactly one class.
Axiom 3. Not all students belong to the same class.
Axiom 4. Any two distinct classes contain at least one student in common.
Some theorems that could be proved from these axioms are:
Theorem 1. Two distinct classes contain exactly one student.
Proof: By Axiom 4, two distinct classes contain at least one student in common. If two distinct classes has two students in common, this contradicts Axiom
2. Thus two distinct classes contain exactly one student in common. Theorem 2. There are exactly three classes.
Proof: By Axiom 1, there are exactly three distinct students, say, S1 , S2 , S3 .
By Axiom 2, two distinct students belong to exactly one class. Thus
S1 , S2 belong to class, say, C1 ,
S1 , S3 belong to class ,say, C2 ,
S2 , S3 belong to class , say, C3 .
Now, C1 , C2 , and C3 are distinct by Axiom 3. Theorem 3. Each class has exactly two students.
Proof: Exercise. Now, We will look at the axiom system structure itself and its properties.
It is important to point out that, in an axiom system, it does not matter what
the terms represent. The only thing that matters is how the terms are related
to each other. In the last example, we can re-label the two terms students
and classes by Bllab and Blloob respectively. We could as well have used the
following axiom set:
Axiom 1. There are exactly three Bllabs.
Axiom 2. Two distinct Bllabs belong to exactly one Blloob.
Axiom 3. Not all of the Bllabs belong to the same Blloob.
Axiom 4. Two different Blloobs have at least one Bllab in common.
By changing the labels in the theorems of the example students and classes, we
would get equivalent theorems about Bllabs and Blloobs. Thus we saw that,
changing the particular meaning of the undefined terms in a system, have no
1.3. EXAMPLES AND PROPERTIES OF AXIOM SYSTEM
9
real change on the structure of the system. Giving the undefined terms a specific
meaning is called an interpretation of the axiom system.
Suppose in the example of students and classes, we replace students and classes
by points and lines respectively. Then, our axioms would be as follows:
Axiom 1. There are exactly three distinct points.
Axiom 2. Two distinct points belong to one and only one line.
Axiom 3. Not all of the points belong to the same line.
Axiom 4. Two distinct lines have at least one point in common.
The theorems above now say that there are exactly three lines, each pair of
lines intersects in exactly one point, and each line has exactly two points. This
axiom system is called three points geometry.
Suppose that we interpret line to be one of the three segments shown in the
following figure, and point to be one of the end points of the three segments, as
shown bellow.
Now, we can see that our axioms still make correct sense with this interpretation.
This interpretation of our original axiom system is called a model of the system.
Thus, a model of an axiom system is an interpretation of the undefined terms
in the system such that all axioms in the system are true statements in this
interpretation.
Other model for three point geometry is given as follows:
Points are the letters, A, B, C.
Lines are{A, B } , {A, C}, {B, C}
and the relation is belong to.
Note that, axiom 1 is true in this model, axiom 2 is also true, axiom 3 is also
true, and finally axiom 4 is true.
A main property of any model of an axiom system, is that all theorems of the system are correct statements in the model. This is because
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logical consequences of correct statements are themselves correct.
Now, we are going to study three important properties of axiom system,
which are consistency, independence, and completeness:
I. Consistency:
An Axiom system is called consistent if no contradiction can be concluded in the system; that means no two statements(two axioms, axiom
and theorem, or two theorems) contradicts each other.
Now, consider the following axiom system:
Axiom 1. There are exactly three points.
Axiom 2. There are at most two points.
It is clear that Axiom 2 contradicts Axiom 1, it would be impossible to
logically deduce theorem in this system as we start with a fundamental
contradiction. This system is not consistent.
It should be clear, that it is essential for an axiom system to be consistent, since a system in which both a statement and its negation can
be proved, is worthless. This is a highly desirable property of any axiom
system. We don’t want to waste our time proving theorems from an axiom
system, that one day may lead to contradiction.
To determine whether an axiom system is consistent, we would have
to examine every possible pair of axioms and/or theorems in the system,
which is not possible in many cases.
Instead, models are used for establishing consistency. If we can find a
model for the given axiom system by using interpretations, that are objects and relations from the real world, we say, we have established absolute consistency. In this case, statements corresponding to contradictory theorems would lead to contradictory statements in the model, but
contradiction in real world are supposedly impossible.
For example, to show that the three points geometry is absolute consistent, we consider the concrete model given in the following figure:
•
HHH
HH
•
•
Where points are the dots in the figure and the segments are the lines.
On the other hand, if the interpretations assigned are taken from
1.3. EXAMPLES AND PROPERTIES OF AXIOM SYSTEM
11
another axiom system, we have only tested consistency relative to the
consistency of the second system, that is; the system we are testing is
consistent only if the system within which the interpretation are assigned
is consistent. In this case, we say we have established relative consistency of the first axiom system.
Thus, we say that axiom system A is relative consistent, if we can find
a model for system A, that is embedded in another system B, and if we
know that system B is consistent, then system A must itself be consistent.
For, if there are two statements in system A that were contradictory, then
this would be a contradiction in system B.
For example, if you believe in the consistency of the theory of real numbers, then you must accept the consistency of Hilbert’s Axiom System,
that is to be given in the next chapter. Because all axioms will hold in
the Cartesian plane.
Often, relative consistency is all we can hope for, as Goedel has shown
that there is no internal proof of consistency for a system that involves
infinite sets.
II. Independence:
A statement is called independent of a given axiom system, if it
cannot be proved or disproved from the axioms in the system.
Example 1.3.5 Consider the following axiom system:
Undefined terms: point, line and belong to.
Axiom 1. There are exactly four lines.
Axiom 2. Any two distinct lines has exactly one point in common.
Axiom 3. Each point is on exactly two line.
Axiom 4. There is exactly six points.
In example 1.3.2, we have seen, that Axiom 4 could be proved from
the first three axioms. Thus, Axiom 4 is not independent from the first
three axioms. This axiom system is just as consistent as the one in the
example, it is not as economical as it could be. A system with numerous
axioms is difficult to remember and confusing to use as basis for proving
theorems. So, it is always desirable, that each axiom in the axiom system
is independent from the other axioms.
To show that an axiom, say X, is independent from the other axioms
in an axiom system, we have to construct two models: one for the system
including axiom X, and another model for the system of the other axioms
and the negation of statement X. If such models exist, then axiom X is
independent from the other axioms.
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CHAPTER 1. GEOMETRY AND AXIOMATIC METHOD
An axiom system, in which each axiom is independent from the other
axioms, is called independent axiom system. Thus the previous example is clearly not independent axiom system. While example 1.3.2 is
independent system.
To show that the system in example 1.3.2 is independent, first we show
that axiom 1 is independent from the other axioms. Consider the following figures:
We see that figure 1 is a model in which all the three axioms hold, figure
2 is a model for which ∼(Axiom 1), Axiom 2, and Axiom 3 hold.
Now, to show that Axiom 2 is independent from the others. Consider
figure 3, which is a model, in which Axiom 1, ∼(Axiom 2), and Axiom 3
hold. Thus, considering the model in figure 1 and the model in figure 3,
we get the independence of Axiom 2.
Third, we show that Axiom 3 is independent from the other axioms. Figure 4 is a model, in which Axiom 1, Axiom 2, and ∼(Axiom 3) hold. Thus,
considering the models in figure 1 and figure 4, we get the independence
of Axiom 3.
Example 1.3.6 Incidence Axiom System:
Undefined terms: point, line, and lie on.
Axiom 1. Two distinct points lie on one and only one line.
Axiom 2. For every line, there is at least two points lie on it.
Axiom 3. There exists three distinct points with the property that no
line contains all three of them.
To show that Axiom 1 is independent from the other axioms, we first construct a model, in which all axioms are true, name it Model 1:
1.3. EXAMPLES AND PROPERTIES OF AXIOM SYSTEM
13
Model 1:
Points are the letters A, B, C.
Lines are the sets {A, B}, {A, C }, {B, C}.
Relation is belong to.
It is easily seen, that Axiom 1, Axiom 2, and Axiom 3 all are true statements in model 1.
Now, we consider a model, call it model 2, in which Axiom 2, and Axiom
3 are true, but ∼(Axiom 1) is true.
Model 2:
Points are A, B, C, D.
Lines are {A, B}, {A, C, D} and {B, C, D}.
It is easily seen, that Axiom 1 is false, Axiom 2 and Axiom 3 are true
statements in Model 2.
Exercise
1. Prove that axiom 2 is independent from Axiom 1 and Axiom 3.
2. Prove that Axiom 3 is independent from Axiom 1 and Axiom 2.
Thus, we conclude that, this system is independent.
II. Completeness
A consistent system is called complete if every statement involving undefined and defined terms in the system can either be proved or
disproved from the axioms. That is, it is not possible to add a new independent statement to the system.
As an example consider the following system for points, and lines:
Axiom 1. Two distinct points belong to one and only one line.
Axiom 2. For every line, there exists at least two points on it.
Now, consider the following statement “There is at least three points, such
that no line pass through all of them ”. In example 1.3.6 we have seen, that
this statement is independent from axiom 1 and axiom 2. That means,
it cannot be proved or disproved in this given axiom system. Hence, this
system is not complete.
Two models of an axiom system are called isomorphic provided that,
there exists one to one correspondence between the interpretation of each
set of undefined terms, such that any relationship between the objects
corresponding to undefined terms in one model, are preserved under correspondence by the second model.
If every two models of an axiom system are isomorphic, then the system is called categorical axiom system.
The axiom system of three points geometry is categorical. For if we label
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the three points A, B, and C, then we know from the two theorems in the
system, that there must be exactly three lines and each line has exactly
two points. The three lines can be symbolized by pairs of points: (A,B),
(A,C), (B,C).
Suppose we have another model of three points; L, M, N. Then we could
construct one to one correspondence between these points and the points
A, B, C, so that the lines in the two models could be put into one to one
correspondence.
Example 1.3.7 The incidence axiom system is not categorical. For:
We construct two models for the system, that are not isomorphic.
First Model
Points: the letters A, B, C, D.
Lines: The sets: {A, B}, {A, C}, {A, D}, {B, C}, {B, D}, {C, D}.
Incidence: Belong to .
Second Model
Points: the letters L, M, N .
Lines: The sets: {L, M }, {L, N }, {M, N }.
Incidence: Belong to .
It is clear that, the two models are not isomorphic, since no correspondence
between the set of points in the two models.
Theorem 1.3.1 Every categorical system is complete.
Proof. We will use contradiction. Let A be a categorical axiom system,
which is not complete. Then there is a statement, say S, which cannot
be proved or disproved in the system, that is S is independent from the
system A. Hence, we can find two models for the axiom system A; the
first in which the statement S is true, and the other in which statement S
is false. Clearly, the two models are not isomorphic. This contradicts the
assumption that our system is categorical. 1.4
Euclid’s Axiom System
Euclid’s monumental achievement in the “Elements ”was to single out
a few simple axioms, and then deduce from them 465 propositions, many
are complicated and not at all intuitively obvious.
The first volume of the “Elements ”begins with a list of 23 definitions,
followed by Euclid’s axiom set. Euclid mistakenly defined all of the terms
that are used. For example he defined a point as that which has no
1.4. EUCLID’S AXIOM SYSTEM
15
part. And a line as breadthless length. These definitions are not very
informative, so it is better, to consider them as undefined term. Similarly,
Euclid defined straight line to be that which lies evenly with points on
itself. This definition is not very useful, to understand it, you must already
have the image of the line. So it is better to consider it as undefined term.
There is also other flaws in the axiom set given by Euclid, which will be
mentioned in the next chapter.
Euclid divided his axioms into two categories:
I. Common Notions
He listed basic assumptions that they are not geometric in nature
and applied to all of mathematics, they are:
1. Things that are equal to the same thing are also equal to one
another.
2. If equals be added to equals, the wholes are equal.
3. If equals be subtracted from equals, the remainders are equal.
4. Things that coincide with one another are equal to one another.
5. The whole is greater than the part.
II. Postulates
We keep Euclid’s terminology and refer to the five axioms, that specifically deal with the geometric basis of his exposition as postulates.
Euclid’s Postulate I. To draw a straight line from any point to
any point. That means, two points determine a unique line.
Euclid’s Postulate II. To produce a finite straight line continuously in a straight line. That means, any line segment may be extended to a line segment that contains it.
Euclid’s Postulate III. To describe a circle with any centre and
radius. That means, given a specified point that will serve as a center and a specified distance to serve as radius, a unique circle can be
drawn.
Euclid’s Postulate IV. All right angles are equal3 to each other.
That means, all right angles are congruent.
Euclid’s Postulate V. If a straight line falling on two straight lines
makes the interior angles on the same side less than two right angles,
the two straight lines, if produced indefinitely, meet on that side on
which are the angles less than the two right angles.
3 In set theory, two sets A and B are said to be equal if every member of A is a member
of B, and vice versa. Thus, equal means identical. Euclid used the word equal in a different
sense, he meant by two equal segments, that the two segments have the same length. And he
meant by two equal angles, that the two angles have the same number of degrees. So to avoid
confusion, it is better to use the word congruent instead of the word equal in these two cases.
16
CHAPTER 1. GEOMETRY AND AXIOMATIC METHOD
The final postulate seems very different from the first four. Euclid
himself waited until Proposition 29 of Book I of his Elements before using the fifth postulate as a justification step in the proof of a theorem.
From the time of Euclid, the fifth postulate axiomatic status has been
questioned. An axiom should be a statement so obvious that it can be
accepted without proof. The first four of Euclid’s postulates are simple statements about the construction of figures, statements that resonate
with our practical experience. We can draw circles, lines, and can measure
angles. However, we cannot follow a line indefinitely to see if it intersects
another line.
Many mathematicians tried to find simpler postulate, one that is more
intuitively believable, to replace Euclid’s fifth postulate, and then prove
the postulate as a theorem.
One of the attempts to find a simpler postulate for Euclid’s fifth was that
of John Playfair(1748–1819):
Playfair’s Postulate: Given a line and a point not on the line, it is
possible to construct one and only one line through the given point parallel to the line.
This statement is certainly easier to read and understand when compared with Euclid’s fifth postulate. In chapter 3, we will show that, this
statement is logically equivalent to Euclid’s fifth postulate. Thus, it does
not simplify Euclid’s system at all.
Other mathematicians tried to prove that Euclid’s fifth postulate was
actually a theorem that could be derived from the first four postulates.
All attempts to derive it from the first four turned out to be unsuccessful
because the so called proofs always entailed a hidden assumption that was
unjustifiable.
1.5
Euclidean Constructions
Euclid’s first three postulates state that lines and circles can be drawn, and line
segments extended. His first proposition states:
Given any segment, there is an equilateral triangle having the given segment as
one of its sides.
Euclid proved this result using a geometric construction; he considered two
circles centred at the endpoints of the segment and having this segment as their
radii. The third vertex of the triangle is one of the points determined by the
intersection of these two circles. See the following figure.
1.5. EUCLIDEAN CONSTRUCTIONS
17
This proof is an example that the concept of construction was part of Euclid’s axiom system and was used as a method of proving propositions.
The Euclid’s first four postulates do not imply that the two circles intersect each
other, therefore there is a gap in his proof. In geometry, diagrams and figures
are very helpful for understanding proofs, and illustrating theorem. But there
is a danger that a diagram may suggest a fallacious argument. A diagram may
be slightly inaccurate or it may represent only a special case. However, every
conclusion must be theoretically justified by the axioms or by the results that
they imply.
Now, we turn to the actual construction of geometric figures with only a straightedge and compass. A straightedge is a ruler without marks, and it is used for
the constructions specified in the first and the second postulate of Euclid. a
compass is used to draw circles. These constructions fascinated mathematicians
from ancient Greece until the nineteenth century, when all classical geometric
constructions were finally solved. In Euclids time he and others were able to do
many constructions including: constructing an equilateral triangle, a perpendicular line to a given line, and a midpoint of a line segment. Throughout history,
mathematicians were able to construct almost all geometric figures following
the postulates laid down by Euclid. The three problems, which were most well
known for not being able to be solved, were:
[I] Trisecting an angle:
Given an angle, construct an angle whose measure is one third of the
original angle.
[II] Doubling a cube:
Given the length of the side of a cube, construct a cube whose area is
double the area of the original cube.
[III] Squaring a circle:
Given a circle, construct a square that has exactly the same area as the
circle.
This problem were unsolvable for over two thousand years with the use of a
compass and a straight edge alone. In 1882, Lindeman proved that this is an
impossible constructions. And Wantzel proved in 1837, that doubling a cube
18
CHAPTER 1. GEOMETRY AND AXIOMATIC METHOD
and trisecting an angle cannot be constructed by only straight edge and a compass.
In the following we will give some examples of euclidean constructions:
[1] Copy a line segment:
−−→
−−→
Given a line segment AB, and a ray CD, we want to find a point E ∈ CD
such that AB ∼
= CE.
First, Place the ends of the compass at the points A and B to determine
the length of AB.
Second, with the compass fixed, place one end at the point C and draw a
−−→
circular arc intersecting CD of radius AB.
Third, define the name of intersection point in the previous step as E.
Hence, we get the desired segment CE.
[2] Copy an angle:
−−→
Given an angle ∠BAC and a ray DE, construct an angle ∠EDF that is
congruent to ∠BAC.
First, construct a circle γ with centre A and radius AB with a compass.
−−→
This circle intersects BC, at a point, say P .
Second, with the compass fixed draw a circle δ with centre D and radius
−−→
AB. This circle intersects DE, say at a point Q.
Third, fix the compass with the length BP , then construct a circle with
radius P B and centre Q. This circle intersects the circle δ at a point, say
F . Then ∠BAC ∼
= ∠EDF .
[3] Angle Bisection:
−−→
Given an angle ∠BAC, we construct a ray AD such that ∠BAD ∼
=
∠CAD.
First, construct a circle with centre A and any radius. This circle intersects the sides of the angle ∠BAC, at two points, say P and Q.
Second, place the ends of a compass at the points P and Q. Construct a
circle δ with centre P and radius P Q, and another circle γ with centre Q
1.5. EUCLIDEAN CONSTRUCTIONS
19
and radius P Q.
Third, the two circles γ and δ intersect at two points, take one of them,
name it D. We obtain that ∠CAD ∼
= ∠BAD.
[4] Perpendicular Line:
Given any line l, and any point P on l, we construct a line through P and
perpendicular to l.
First, construct a circle with centre A and any radius. The circle intersects
l in two points, say A and B.
Second, construct a circle γ with centre A and radius AB.
Third, construct another circle δ with centre B and radius the same as
the one in the second step. The circles γ and δ intersect at two points,
←→
take one of them, call it Q. And then the line P Q is perpendicular to the
line l.
[5] Drop a Perpendicular to a Line:
Given any line l and any point A not on l, we construct a perpendicular
to l through the point A.
First, let B be any point on l. Construct a circle with centre A and radius
20
CHAPTER 1. GEOMETRY AND AXIOMATIC METHOD
BA. This circle intersects line l at another point, say C.
−−→
Second, use construction of bisector of an angle to draw the bisector AD
←→
of the angle ∠BAC. Then, the line AD is the perpendicular to l through
the point A.
[6] Parallel to a Line:
Given any line l and any point P not on l, we construct a line through P
and parallel to l.
−→
First, let A be any point on the line l, construct the line P A with a straight
edge.
Second, construct a circle γ with centre A and radius r > AP . This circle
←→
intersects l at a point say B, and intersects the line P A at a point, say Q.
Third, construct an angle congruent to the angle ∠P AB. Consider the
−−→
←→
−→
ray P Q and the side of the line P Q. We construct a ray P R, such that
←
→
∠QP R ∼
= ∠P AB. Then the line P R is the parallel to l through P .
1.6. EXERCISES
21
[7] Square Root:
Given a segment OI of unit length, and√a segment AB of any length a.
Then we construct a segment of length a.
←→
First, copy the segment OI on the line AB, such that OI ∼
= BC, where C
−−→
is on the opposite ray of BA.
Second, Find the midpoint of the segment AC, call it M .
Third, construct a circle γ with centre M and radius AM . This circle pass
through C.
←→
Fourth, construct a line l through B, that is perpendicular to the line AC.
This line intersects γ at two points, consider one of them, and call it D.
Let c be the length of the segment BD.
Fifth, the triangles 4ABD and 4DBC are similar. Hence
c
a
= .
1
c
Thus a = c2 . See the following figure.
1.6
EXERCISES
[1] Show that the following system is not complete.
Axiom 1 There exists five points.
Axiom 2 Each line is a subset of those five points.
Axiom 3 There exists exactly two lines.
Axiom 4 Each line contains at least two points.
[2] Consider a set of undefined elements to be S, and the undefined relation R
on the elements of S which satisfies the following axioms:
22
CHAPTER 1. GEOMETRY AND AXIOMATIC METHOD
Axiom 1 If a ∈ S then aRa.
Axiom 2 If a, b ∈ S and aRb then bRa.
Axiom 3 If a, b, c ∈ S, aRb, and bRc then aRc.
i. Construct a model of the system.
ii. Is the axiom system independent? Explain.
[3] Consider the following axiom system:
Axiom 1 There exists five points.
Axiom 2 Each two distinct points have exactly one line on both of them.
Axiom 3 Each line contains exactly two points.
i. Construct a model of the system.
ii. Show that there exists exactly 10 lines.
iii. Show that each point has exactly 4 lines through it.
[4] Volume II of Euclid’s Element is concerned with geometric algebra; algebra
based on geometric figures. For example, the distributive law of algebra
a(b + c) = ab + ac can be interpreted as the area of rectangle with sides a
and b + c is equal to the sum of areas of the rectangle with sides a and b
and the rectangle with sides a and c. See the figure:
Show that the algebraic identity (a+b)2 = a2 +2ab+b2 can be established
geometrically.
[5] Consider the following axiom system: undefined terms are point, line and
incident with. The axioms are:
Axiom 1. There exists at least one line.
Axiom 2. Every line has exactly three points on it.
Axiom 3. Not all points are on the same line.
Axiom 4. For any two distinct points, there exists exactly one line on
both of them.
1.6. EXERCISES
23
Axiom 5. Each two lines have at least one point on both of them.
i. Construct a model of this axiom system.
ii. Determine whether axiom 5 is independent of the other axiom
or not. Explain.
iii. Is this axiom system complete? Explain.
iv. Is it categorical? Explain.
[6] Construct a model for the following axiom system. Undefined terms are
point, line, and lie on. And the axioms are:
Axiom 1. If a, b are two distinct points, there exists at least one line
containing both of them.
Axiom 2. If a, b are distinct points, then there exists at most one line
containing both of them.
Axiom 3. If l, m are any two distinct lines, then there exists at least
one point which lies on both l and m.
Axiom 4. Every line contains at least 3 distinct points.
Axiom 5. Not all points lie on the same line.
Axiom 6. No line contains more than 3 distinct points.
Axiom 7. There is at least one line.
[7] Construct an axiom system that is not categorical.
[8] Consider the incidence axiom system. Determine which of the following
interpretations of the undefined terms is a model of the system:
i. Points are dots on a sheet of paper, lines are circles drawn on the sheet
of paper, incidence means that the dot lie on the circle.
ii. Points are lines in the Euclidean 3-space, lines are planes in the space,
incidence is the usual relation of a line lying on a plane.
iii. Same as in ii., but the lines and the planes all pass through the origin.
iv. Consider a given circle γ in the Euclidean plane. Interpret points as
all points inside γ, and lines as all open chords of γ, and incidence
as lie on.
[9] Let us add to the incidence axiom system the following axioms:
Axiom 4: The Euclidean parallel postulate.
Axiom 5: The existence of only a finite number of points.
Axiom 5: There exists two lines l and m such that the number of points
lying on m is different from the number of points lying on l.
Show that this system is not consistent.
24
CHAPTER 1. GEOMETRY AND AXIOMATIC METHOD
[10] With a straight edge and a compass, construct a tangent of a circle at a
point on P on the circle.
[11] With a straight edge and a compass, construct a square with a given side
a segment AB.
[12] With a straight edge and a compass, construct a hexagon.
[13] With a straight edge and a compass, construct a midpoint of a segment.
√
(1 + 5)
[14] The number
was called by Greeks the Golden Ratio, and a
2
rectangle whose sides are in this ratio is called a golden rectangle. With
a straight edge and a compass, construct a golden rectangle.
Chapter 2
Hilbert’s Axiom System
In the late of 19th century and in the 20th century, mathematicians, who work
with the field foundations of mathematics, realized that Euclid’s system is not
complete and pointed out flaws again and again in Euclid’s assumptions. Euclid
assumed assumptions, that had not been stated. For example, he neglected to
state his assumptions, that points and lines do exist, that not all points are
collinear, and that every line has at least two points lying on it.
Some of Euclid’s proof are based on reasoning from diagrams. To make these
proofs rigorous, a much larger system of explicit axioms is needed. For example,
the need for a statement about concept of betweenness, just because diagram
shows a bisector of an angle intersecting the opposite side in a triangle “between”two endpoints, is it true that there is a point between the two endpoints?
There is a need to state that if a straight line enters a triangle at a vertex,
then it must intersect the opposite side. That means, a line divide the plane
into two half disjoint planes. Also, there is a need for a statement about infinite
length of a line, does a line stops at endpoints?
Hence, many mathematicians undertook the quest for a more rigorous and complete foundation for Euclidean geometry. One of the earliest and most notable
attempts are from the German mathematician David Hilbert (1862-1943). His
system of Euclidean geometry built on the work of Pasch, who published the
first rigorous treatise on geometry in 1882. Hilbert’s presentation of Euclidean
geometry in 1899 satisfactorily addresses most of the problems found in the Elements of Euclid, the axioms are stated in away that resembles Euclid’s axiom
set, making them quite intuitive.
Other treatments of Euclidean geometry were done by mathematicians like Veblen (1904), Huntington (1913), Forder (1927), and Birkhoff (1932). Birkhoff
published an Euclidean axiom set with different flavour than those of Hilbert’s
and Euclid’s.
25
26
CHAPTER 2. HILBERT’S AXIOM SYSTEM
Birkhoff remarkably concise axiom set contains only four statements and features measurements as unifying concept. The powerful first axiom puts the
points of any line in Euclidean geometry in one-to-one correspondence with the
real numbers. In this way, all properties of real numbers are associated with
the points of any line. Also Birkhoff axioms are designed so that the Euclidean
parallel axiom can be proved as a theorem.
In this chapter, we will present the Hilbert’s Axiom system for plane Euclidean
geometry, which is presented in 1899 in his book Foundations of Geometry.
Hilbert’s axioms are perhaps the more intuitive and are certainly the closest in
spirit to Euclid’s. Hilbert clarified Euclid’s definitions and filled in the gaps in
some Euclid’s proofs.
We begin by the undefined terms that are basis for defining all other terms in
plane Euclidean geometry. They are:
point
line
incidence with (lie on or pass through)
between
congruent.
Hilbert’s axioms are divided into five groups: incidence axioms, betweenness
axioms, congruence axioms, continuity axioms, and parallelism axiom.
2.1
Incidence Axioms
We said that, Euclid neglected to state his assumptions that points and lines
exist, that not all points are collinear, and that every line has at least two points
lying on it. Hilbert made these assumptions explicitly by adding two more axioms of incidence to Euclid’s first postulate.
Undefined terms: Points, lines, and incident with.
The axioms of incidence are:
I.1 Given two distinct points P and Q there is a unique line incident with P
←→
and Q. This line will be denoted by P Q.
I.2 For every line l there exist at least two distinct points incident with l.
I.3 There exist three distinct points with the property that no line is incident
with all three of them.
Definition 2.1.1 Three points or more are called collinear if there exists a
line which is incident with all of them.
Two lines l and m are called parallel if they do not have a point in common.
Two or more distinct lines with a common point are called concurrent.
In order to make some statements easier to write, if a point P is incident to a
line l , we will write P ∈ l. Here are some results that can be proved from the
axioms above.
2.1. INCIDENCE AXIOMS
27
Proposition 2.1.1 The incidence axioms imply:
i. If l and m are two distinct lines that are not parallel, then they have a unique
point in common.
ii. There exists three distinct lines that are not concurrent.
iii. For every line there is at least one point not incident with it.
iv. For every point there is at least one line not incident with it.
v. For every point P there exist at least two lines through P.
Proof.
i. Assume that l and m are two distinct lines that are not parallel. Hence, l
and m have at least one point in common. To prove that only a unique
common point exist, we use contradiction; that is, we assume that the
points P and Q belong to both l and m and P 6= Q. Axiom I.1 implies
that l = m, and this contradicts the assumption that l 6= m.
ii. It follows from axiom I.3 that there are three points P, Q, and R that are
not collinear. Using axiom I.1, P and Q determine a unique line, say
l, P and R determine a unique line, say m, and Q and R determine a
unique line say n. The three lines are distinct, otherwise P, Q, and R are
collinear. Now, suppose that the three lines are concurrent, that is, they
have a point in common, say A. Then A ∈ l ∩ m. Thus, by part i A = P .
Also, A ∈ m ∩ n and part i implies that A = R. Then, P = R. This is
clearly a contradiction, since we have assumed that P 6= R.
iii. Given any line l. By axiom I.3 there are 3 distinct non-collinear points, say
P, Q, and R. If all the three points belong to l, we get a contradiction to
the assumption, that they are non-collinear. Thus, not all of them belongs
to line l, that is, at least one of them does not belong to l
iv. Given any point P . By part ii there are three distinct non-concurrent lines,
say l, m and n. Either P ∈ (l ∩m∩n) or P ∈
/ (l ∩m∩n). If P ∈ (l ∩m∩n).
Then l, m, and n are concurrent, thus we get a contradiction. Hence, at
least one of these lines is not incident with P .
v. Given a point P . By part iv there is a line, say l, that is not incident with
P . By part iii there is a point, say A, that is not incident with line l.
Now, by axiom I.1 points P and A determine a unique line, say m. Then
l 6= m, otherwise P ∈ l, which is a contradiction. Thus the line l and m
are incident with P . We will call this axiom system incidence geometry. Now, we describe some
models for incidence geometry:
28
CHAPTER 2. HILBERT’S AXIOM SYSTEM
Model 1:
points are the nodes in the following figure, lines are the line segments,
and incident with is being endpoint of the line segment.
Model 2:
See the following figure:
Model 3:
The cartesian plane is a traditional model for incidence geometry. Where
the points defined to be the ordered pairs of real numbers, and lines to be
the collection of all points ax + by + c = 0 for some choice of real numbers
a, b, c.
Note that: These three models for incidence geometry are not isomorphic, and
hence the incidence geometry is not categorical.
The surface of any sphere, where points are defined to be the set of all points
on the sphere, and the lines to be all great circles 1 on the sphere.
This is not a model for incidence geometry, because any two antipodal points
1 A great circle of sphere is a circle on the surface of the the sphere, obtained by the
intersection of a plane through the center of the sphere and the surface of the sphere
2.2. BETWEENNESS AXIOMS
29
on the sphere are on an infinite number of great circles. Explain, why? This
violates axiom I.1.
Consider the following parallelism properties:
Euclidean Parallel Axiom: For every line l and for every point P ∈
/ l there
is exactly one line through P and parallel to l.
Hyperbolic Parallel Axiom: For every line l and every point P ∈
/ l there
are at least two lines through P and parallel to l.
Elliptic Parallel Axiom: No parallel lines exist.
Exercise: Show that each one of the previous three statements is independent
from the incidence geometry.
Thus, incidence geometry is not complete. In example 1.3.6 in chapter 1, we
show that this axiom system is independent.
2.2
Betweenness Axioms
Euclid never mentioned the notion “betweenness”. But tacitly assumed certain
facts about it that are obvious in diagrams. In fact, quite few of Euclid’s proof
are based on reasoning from diagrams. To make these proofs rigorous, a much
larger system of explicit axioms is needed.
Hilbert suggested the following axioms on betweenness:
B.1 If A ∗ B ∗ C then A, B, and C are three distinct points all lying on the
same line, and C ∗ B ∗ A.
Note that (C ∗ B ∗ A) means that, B between A and C, is the same as B
between C and A.
B.2 Given any two distinct points B and D, there exists points A, C and
←→
E lying on BD such that A ∗ B ∗ D, B ∗ C ∗ D and B ∗ D ∗ E.
This axiom ensures that there are points between B and D and that the line
←→
BD does not end at either B or D.
B.3 If A, B, and C are three distinct points lying on he same line, then one
and only one of the points is between the other two.
This axiom ensures that a line is not circular; if the points where on a circle,
we would then have to say that each is between the other two.
Definition 2.2.1 We define a segment AB as the set of all points between
A and B, together with the points A and B which are called the endpoints of
segment AB.
30
CHAPTER 2. HILBERT’S AXIOM SYSTEM
−−→
We define the ray AB as the set of all points on the segment AB together with
all points C such that A ∗ B ∗ C. The point A is called the vertex of the ray
−−→
−−→
AB and we say that ray AB emanates from A.
It follows from axiom B.2 that given any two points A, B on a line l there exists
infinitely many points of l between A and B. Axiom B.2 also implies that there
−−→
are infinitely many points lying on AB that are not on the segment AB.
−−→
−−→
Definition 2.2.2 If A ∗ B ∗ C, then we say that BC is opposite ray to BA.
That is two rays are opposite rays if and only if they are distinct, part of the
same line, and emanating from the same point.
Note that axiom B.2 guarantees that every ray has opposite ray. The next result
is consequence of the three axioms of betweenness.
Proposition 2.2.1 For any two points A and B we have:
−−→ −−→ ←→
i. AB ∪ BA = AB.
−−→ −−→
ii. AB ∩ BA = AB
Proof.
−−→ −−→ ←→
−−→ ←→
−−→ ←→
i. First, we have to show that, AB ∪ BA ⊂ AB. Since AB ⊂ AB and BA ⊂ AB
−−→ −−→ ←→
we have that AB ∪ BA ⊂ AB.
←→
−−→
Second, let P be a point lying on AB. If P = A or P = B, then P ∈ AB, and
−−→ −−→
hence, P ∈ AB ∩ BA.
If P 6= A, and P 6= B, then from B.3 we get that one and only one of the
following cases:
−−→
−−→
Case 1: A ∗ P ∗ B and hence P ∈ AB, then P ∈ AB and P ∈ BA.
−−→
Case 2: P ∗ A ∗ B and hence P ∈ BA.
−−→
Case 3: A ∗ B ∗ P and hence P ∈ AB.
−−→
−−→
−−→ −−→
In each case P belongs to either AB, or BA. Hence, P ∈ BA ∪ AB.
ii. Exercise. Definition 2.2.3 Let l be a line and A and B be two distinct points not on
l. We say that A and B on the same side of l if the segment AB does not
intersect the line l. But if segment AB intersects the line l then we say that
points A and B are on opposite sides of l.
B.4 Given any line l and three distinct points A, B, and C not lying on l, we
have:
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 line l.
2.2. BETWEENNESS AXIOMS
31
Corollary 2.2.1 If A and B are on opposite sides of l and B and C are on the
same side of l, then A and C are on opposite sides of l.
Definition 2.2.4 Let l be a line and A a point not lying on l. The set of all
points that are on the same side of l as A is called the half-plane bounded by
l containing A. We denote it by Hl,A .
Note that B.4 implies that if A and B are on the same side of l then Hl,A = Hl,B .
Theorem 2.2.1 (Plane separation property) Every line bounds exactly two
half-planes and these half-planes are disjoint.
Proof: Given any line l. Axiom I.2 implies that there is a point O on l. And
proposition 2.1.1 implies that there is a point A not on l. From B.2 there is
a point B such that A ∗ O ∗ B. Therefore A and B are on opposite sides of l.
Thus l has at least two half-planes Hl,A and Hl,B . First, we have to prove that,
Hl,A ∩ Hl,B = Φ
If C ∈ Hl,A ∩ Hl,B , then C is on same side of l as A, and C is on same side of
l as B. Then by B.4, A and B are on the same side of l, contradicting that A
and B are on opposite side of l. Thus the two half-planes are disjoint.
Second, we have to prove that, if P is any point not lying on l then either
P ∈ Hl,A or P ∈ Hl,B . Let P ∈
/ l and P ∈
/ Hl,A , then P and A are on opposite
sides of l. Since A and B are on opposite sides of l then by B.4, B and P are
on same side of l. That means P ∈ Hl,B . We next apply the plane separation property to study betweenness among four
points.
Proposition 2.2.2 Given A ∗ B ∗ C and A ∗ C ∗ D. Then
i. B ∗ C ∗ D.
ii. A ∗ B ∗ D.
Proof: i. We prove that B ∗ C ∗ D.
Given A∗B ∗C and A∗C ∗D. By B.1, A, B, and C are distinct and collinear, say
lie on the line l, also A, C, and D are distinct and collinear, say lie on the line
n. By I.1, since A and C lie on both l and n, then l = n. Hence, A, B, C, and
D all lie on l. Now, we need to show that, B 6= D. If not, that is, B = D, then
from the hypothesis, we have both A ∗ B ∗ C and A ∗ C ∗ B, which contradicts
B.3. Thus, A, B, C, and D are distinct and collinear, all lie on l.
Now, by proposition 2.1.1, there exist a point E not on l. Consider the line
←→
←→
EC, AD meets this line in C, hence A, and D are in opposite sides of EC. We
←→
←→
claim that, A and B are in opposite sides of EC, then AB meets EC at a point
←→
M between A and B. By proposition 2.1.1, l and EC meets in a unique point,
which is C. Thus C = M . Then, we have, A ∗ B ∗ C and A ∗ C ∗ B, which
←→
contradicts B.3. Hence A and B are on the same sides of EC. Since A and B
←→
←→
are on the same sides of EC, and A and D are on opposite sides of EC, then
32
CHAPTER 2. HILBERT’S AXIOM SYSTEM
←→
←→
by corollary 1, B and D are on opposite sides of EC. Thus BD intersects EC
at a point between B and D, and this point must be C.
ii. Now, we will prove A ∗ B ∗ D.
Given A ∗ B ∗ C and A ∗ C ∗ D, then from the first part, A, B, C and D are four
distinct collinear points let l be the line containing them.
By proposition 2.1.1 there is a point F ∈
/ l. From I.1, the points B and F determine a unique line, say m. The segment AC meet m at B and by assumption
A ∗ B ∗ C then A and C are in opposite sides of m.
We will use contradiction to prove that C and D are in same side of m. Thus,
assume that C and D are in opposite side of m, then CD intersects m at some
point G between C and D. But l and m are distinct and can have only a unique
point in common, thus G = B. Hence C ∗ B ∗ D, contradicting that B ∗ C ∗ D
in the first part. And so C and D are on same side of m.
From corollary 1 and that A and C are on opposite side and C and D are in on
same side of m, then A and D are in opposite side of m. Thus AD intersects
line m at a point between A and D, this point must be B by proposition 2.1.1.
hence A ∗ B ∗ D. Corollary 2.2.2 Given A ∗ B ∗ C and B ∗ C ∗ D. Then A ∗ B ∗ D and A ∗ C ∗ D.
Theorem 2.2.2 (Line separation property) If C ∗ A ∗ B and l is the line
−−→
through them. Then for every point P =
6 A lying on l, then either P ∈ AB, or
−→
P ∈ AC.
Proof. Given C ∗ A ∗ B and l the line through them. Let P ∈ l and P 6= A,
−−→
−−→
−−→
then either P ∈ AB or P ∈
/ AB. If P ∈ AB, then we are done.
−→
If not, then P ∗ A ∗ B. If P = C then P ∈ AC.
If P 6= C, then by B.3 either P ∗ B ∗ C or B ∗ P ∗ C or B ∗ C ∗ P .
If B ∗ P ∗ C then combining this with P ∗ A ∗ B and proposition 2.2.2 we have
−→
A ∗ P ∗ C then P ∈ AC.
If B ∗ C ∗ P then combining this with C ∗ A ∗ B, we get A ∗ C ∗ P , and hence
−→
P ∈ AC .
If P ∗ B ∗ C then combining this with C ∗ A ∗ B, we get P ∗ B ∗ A a contradiction.
−→
Therefore, either A ∗ C ∗ P or A ∗ P ∗ C, which means that P ∈ AC. The next theorem state a visually obvious property that Pasch discovered Euclid
to be using without proof.
Theorem 2.2.3 (Pasch’s Theorem) If A, B, and C are distinct non-collinear
points and l is any line intersecting AB at a point between A and B, then l also
intersects either AC or BC. If C does not lie on l, then l does not intersects
both AC and BC.
2.2. BETWEENNESS AXIOMS
33
Proof. Let A, B, and C are distinct non-collinear points and l is any line intersecting AB at a point between A and B. Then A and B are in opposite sides
of l. If C ∈ l then l intersects both AC and BC. If C ∈
/ l we obtain from plane
separation property that C ∈ Hl,A or C ∈ Hl,B and not in both of them. If
C ∈ Hl,A then AC intersects l. And if C ∈ Hl,B then BC intersects l. Intuitively, this theorem says that, if a line intersects a triangle in one side,
then it must intersect another side.
Proposition 2.2.3 Given A ∗ B ∗ C. Then
i. AC = AB ∪ BC and B is the only point common to segments AB and BC.
−−→ −→
ii. AB = AC
−−→
−−→
iii. B is the only common point to rays BA and BC.
Proof. i. First, we show that AC ⊂ AB ∪ BC. Let P ∈ AC. Then either
P ∈ AB or P ∈
/ AB.
If P ∈ AB, then since AB ⊂ AB ∪ BC, P ∈ AB ∪ BC.
If P ∈
/ AB then P ∗ A ∗ B or A ∗ B ∗ P . Now, since P ∈ AC then by definition
of segments either P = C or A ∗ P ∗ C.
If P = C then P ∈ BC ⊂ AB ∪ BC.
If A ∗ P ∗ C, then from P ∈
/ AB, either A ∗ B ∗ P or B ∗ A ∗ P . Thus, if
A ∗ B ∗ P , combine it with A ∗ P ∗ C we get by proposition 2.2.2, B ∗ P ∗ C, that
is P ∈ BC ⊂ AB ∪ BC.
If B ∗ A ∗ P , then combine it with A ∗ P ∗ C, by proposition 2.2.2, we get that
B ∗ P ∗ C, then P ∈ BC ⊂ AB ∪ BC.
Second, we show that AB ∪ BC ⊂ AC. Let P ∈ AB ∪ BC, then either P ∈ AB
or P ∈ BC.
If P ∈ AB, then either P = A or P = B or A ∗ P ∗ B. If P = A then P ∈ AC.
If P = B and since A ∗ B ∗ C then P ∈ AC. If A ∗ P ∗ B with A ∗ B ∗ C we get
A ∗ P ∗ C then P ∈ AC.
Now, if P ∈ BC, then either P = B or P = C or B ∗ P ∗ C. If P = C then
P ∈ AC. If P = B with A ∗ B ∗ C then P ∈ AC. Finally, if B ∗ P ∗ C, then
with A ∗ B ∗ C we get A ∗ P ∗ C and hence P ∈ AC. Thus AB ∪ BC ⊂ AC.
To show that only B is the common to AB and BC, use contradiction. Let
E 6= B and E ∈ AB ∩ BC. Then clearly E 6= A and E 6= C, so A ∗ E ∗ B and
B ∗ E ∗ C. Consider A ∗ E ∗ B with A ∗ B ∗ C then from proposition 2.2.2, we
have E ∗ B ∗ C contradicting that B ∗ E ∗ C.
−−→
−→
−−→
ii. First, we will show that, AB ⊂ AC. We use contradiction, let P ∈ AB
−→
−→
and P ∈
/ AC. Then P ∈
/ AC implies that P ∗ A ∗ C. Combine A ∗ B ∗ C with
−−→
P ∗ A ∗ C by proposition 2.2.2, we get P ∗ A ∗ B which contradicts that P ∈ AB.
−−→ −→
−→ −−→
Thus, AB ⊂ AC. Similarly we show that, AC ⊂ AB.
−−→ −−→
iii. Now, we show that BA ∩ BC = B. Use contradiction, assume that P 6= B
−−→ −−→
−−→
and P ∈ BA ∩ BC. Then clearly P 6= A and P 6= C. Now, P ∈ BA implies
34
CHAPTER 2. HILBERT’S AXIOM SYSTEM
that, B ∗ P ∗ A or B ∗ A ∗ P . If B ∗ P ∗ A, with A ∗ B ∗ C we get P ∗ B ∗ C, then
−−→
P ∈
/ BC, hence we get a contradiction.
−−→
If B ∗ A ∗ P with A ∗ B ∗ C we get also P ∗ B ∗ C, which leads to that P ∈
/ BC,
and hence we get a contradiction. Definition 2.2.5 An angle with vertex A is a point A together with two non−−→
−→
opposite rays AB and AC emanating from A. The two rays are called sides of
the angle. Such an angle will be denoted by ∠BAC or ∠CAB. If no confusion
occur, we will denote it by ∠A.
Definition 2.2.6 Given an angle ∠BAC, define a point D to be an interior
←→
point of ∠BAC if D is on the same side of AC as B and D is also on same
←→
side of AB as C. That is D is in the intersection of two half planes.
←→
Proposition 2.2.4 Given an angle ∠CAB and a point D lying on line BC.
Then D is in the interior of ∠CAB if and only if B ∗ D ∗ C.
←→
Proof. Given that D ∈ BC
←→
(=⇒). Let D be in the interior of ∠BAC. Since D ∈ BC, then from B.3, either
B ∗ D ∗ C or D ∗ B ∗ C or B ∗ C ∗ D. If D ∗ B ∗ C, then D and C are in
←→
opposite sides of any line intersecting line DC at B. In particular, D and C are
←→
in opposite sides of line AB contradicting that D is an interior point.
←→
If B ∗ C ∗ D, then D and B are in opposite side of any line intersecting line BD
←→
at C. In particular, D and B are in opposite sides of the line AC contradicting
that D is an interior point. Thus, B ∗ D ∗ C hold.
(⇐=) Conversely, assume B ∗ D ∗ C, if D is not in the interior of the angle
←→
∠BAC. Then either D and B are not on same sides of AC or D and C are not
←→
on same sides of AB.
←→
←→
If D and B are in opposite sides of AC then DB intersect AC at a point say F
←→ ←→
←→
←→
such that D ∗ F ∗ B. Then DB = DC. Thus the two lines AC and DB have
←→
←→
two points in common F and C. First, the two lines DB and AC are clearly
−−→
distinct lines, otherwise A, B and C be would be collinear, and hence AB and
−→
AC are opposite rays, contradicting that the two rays are sides of an angle and
hence non-opposite. Then by proposition 2.1.1 C = F , but this also contradicts
the assumption that B ∗ D ∗ C.
←→
The case that D and C are in opposite sides of line AC leads to similar contradiction. Then we conclude that D is an interior point of ∠BAC. .
Proposition 2.2.5 Let D be interior point of the angle ∠CAB. Then
−−→
i. Every point E 6= A and on AD is an interior point of ∠BAC.
−−→
ii. No point on the opposite ray to AD is in the interior of ∠CAB.
2.2. BETWEENNESS AXIOMS
35
iii. If C ∗ A ∗ E then B is in the interior of ∠DAE.
Proof. Given an angle ∠CAB and point D in the interior of ∠CAB.
−−→
i. Let E 6= A and E ∈ AD. If E = D, then by assumption, E is interior point
of ∠CAB. If E 6= D, we use contradiction. Assume that E is not interior point
←→
of ∠CAB. Then either E and B are in opposite sides of AC or E and C are in
←→
opposite sides of line AB.
←→
If E and B are in opposite sides of line AC, since B and D are on same side of
←→
line AC then by corollary to B.4 the points E and D are in opposite sides of
←→
←→
line AC. Then ED intersects AC at a point say F such that E ∗ F ∗ D. But
−−→
←→
←→
DE ⊂ AD. Then AC and AD have two points in common, A and F , from
proposition 2.1.1 i. and that the two lines are distinct we get a contradiction
−−→
unless F = A. But then E ∗ A ∗ D which contradicts that E ∈ AD.
−−→
ii. Let G be any point on the opposite ray of AD. Then G ∗ A ∗ D and G
←→
←→
and D are in opposite sides of AC since GD intersects AC at A. Now, D is in←→
terior of ∠CAB then B and D are on same side of AC. Then by corollary to B.4
←→
we have G and B are in opposite sides of AC. Hence, G is not interior to ∠CAB.
iii. Assume that C ∗ A ∗ E. To show that B interior point of ∠DAE. First, D
←→ ←→
and B are on same side of AC = EA since D interior of ∠CAB.
←→
Second, we need to show that: E and B are on same side of AD. If not, then
←→
EB intersects line AD at a point, say F, and then F is an interior point of
−→
∠BAE by proposition 2.2.4. Now, we claim that D ∈ AF . Since F is on same
←→
←→
side of AE as B and B is on same side of AE as D, then by B.4 F and D are on
←→
−→
same side of AE. Hence, D ∈ AF . Then D is also an interior point of ∠BAE.
←→
Then D and E are on same side of AB. And from that D is interior to ∠CAB,
←→
we have D and C are on same side of AB. Thus by B.4 we get that E and C are
←→
on same side of AB. This is a contradiction since C ∗ A ∗ E and CE intersects
←→
AB at A. −−→
−→
−−→ −−→
−→
Definition 2.2.7 Ray AD is between rays AC and AB if AB and AC are not
opposite rays and D is interior to ∠BAC.
Note that by previous proposition this definition does not depend on the choice
of point D.
Theorem 2.2.4 Crossbar Theorem
−−→
−→
−−→
−−→
If AD is between AC and AB then AD intersect segment BC.
−−→
−→
−−→
Proof. Given that AD is between AC and AB. Use contradiction, assume
−−→
←→
that AD does not intersect BC. By B.2, there exists a point F ∈ AD such that
−−→ −→
F ∗ A ∗ D, and AD ∩ AF = {A}. By proposition 2.2.4, every point on BC except
−→
B and C is interior point of ∠CAB. And by proposition 2.2.5, no point on AF
−→
is interior to ∠CAB. Thus BC does not intersect AF . And by assumption,
36
CHAPTER 2. HILBERT’S AXIOM SYSTEM
−−→
←→
BC does not intersect AD. Then BC does not intersect the line AD. Hence,
←→
B and C are in the same side of AD. Let E be such that C ∗ A ∗ E, Since by
last proposition B is interior point of ∠DAE. Then B and E are in the same
←→
←→
side of AD. Since B and C are on same side of AD, then by the B.4, C and E
←→
are in the same side of AD, which is a contradiction. 2.3
Congruence Axioms
The last of our undefined terms is congruent. It will refer to either a relation
between segments or a relation between angles. Also, we use congruence to refer
to a relation between triangles, we will define it as follows:
Definition 2.3.1 A triangle 4ABC consists of three non-collinear points A, B
and C and the line segments AB, AC, and BC. Where A, B and C are called
the vertices of the triangle and AB, BC, and AC are called the sides of the
triangle.
Two triangles are said to be congruent if there is a one to one correspondence between their vertices so that the corresponding sides are congruent and
corresponding angles are congruent.
Now we state the axioms of congruence:
C.1 If A and B are two distinct points and if C is any point, then for each given
ray r emanating from C there is a unique point D on r such AB ∼
= CD.
Intuitively, this axioms says that, one can move the segment AB so that
it lies on the ray r with A superimposed on C, and B superimposed on
D.
C.2 If AB ∼
= CD and AB ∼
= EF then CD ∼
= EF . Moreover, every segment is
congruent to itself.
This axiom replaces the Euclid’s first and fourth common notions, since
it says that segments congruent to the same segment are congruent, and
segments that coincide are congruent.
We conclude from this axiom that: If AB ∼
= CD then CD ∼
= AB.
C.3 (Segment Addition)If A ∗ B ∗ C, E ∗ F ∗ G, AB ∼
= EF , and BC ∼
= F G,
∼
then AC = EG.
This axiom replaces the third common notion of Euclid, since it says that,
if congruent segments are added to congruent segments, then the sums
are congruent. Here adding means juxtaposing segments a long the same
line. Using C.1 and C.2, we can lay off a copy of a given segment AB on
the ray r, n times, to get a new segment n.AB.
2.3. CONGRUENCE AXIOMS
37
−−→
←→
C.4 Given ∠BAC and any ray EF , then on a given side of the line EF there
−−→
is a unique ray EG such that ∠BAC ∼
= ∠F EG.
This axiom say that, angle can be laid off on a given side of a given
ray in a unique way.
C.5 If ∠A ∼
= ∠B and ∠A ∼
= ∠C, then ∠B ∼
= ∠C. Moreover, every angle is
congruent to itself.
We conclude from this axiom that: If ∠A ∼
= ∠B then ∠B ∼
= ∠A.
C.6 (SAS) If two sides and the included angle of one triangle are congruent
respectively to two sides and the included angle of another triangle, then
the two triangles are congruent.
Corollary 2.3.1 Given 4ABC and segment DE ∼
= AB, there is a unique point
←→
∼
F on a given side of line DE such that 4ABC = 4DEF .
∼ AB. Then by C.4 there is a
Proof. Given any 4ABC and segment DE =
−−→
←→
unique ray DF on a given side of the line DE such that ∠BAC ∼
= ∠F DE.
−−→
Then by C.1 on the ray DF choose F the unique point such that AC ∼
= DF .
Hence by SAS, 4ABC ∼
= 4DEF . Proposition 2.3.1 (ASA) Given 4ABC and 4DEF with AB ∼
= DE, ∠CAB ∼
=
∼
∼
∠F DE, and ∠CBA = ∠F ED, then 4ABC = 4DEF .
Proof. Given 4ABC and 4DEF with AB ∼
= DE, ∠CAB ∼
= ∠F DE, and
−→
∠CBA ∼
∠F
ED.
By
C.1
there
is
a
unique
point
G
on
ray
AC, such that
=
∼
∼
AG ∼
DF
.
Then
by
SAS,
4ABG
4DEF
,
and
hence
∠ABG
=
=
= ∠DEF .
But by assumption, ∠DEF ∼
∠ABC.
Then
by
uniqueness
of
C.4,
we have
=
−−→ −−→
←→ ←→
BG = BC. We claim that G = C. If not, that is G 6= C then G ∈ AC ∩ BC,
←→ ←→
then by proposition 2.1.1 i, AC = BC. But this contradicts that A, B, and C
are non-collinear. Definition 2.3.2 A triangle is called isosceles if two of its sides are congruent.
Proposition 2.3.2 Given 4ABC, then:
i. If 4ABC is an isosceles triangle then ∠B ∼
= ∠C.
ii. If ∠B ∼
= ∠C then the triangle is isosceles.
Proof. i. Given a triangle 4ABC in which AB ∼
= AC. Consider the triangles
4ABC and 4ACB. Then by C.4, ∠A ∼
= ∠A, by hypothesis AB ∼
= AC, and by
C.2, we have AC ∼
= AB, then by SAS, 4ABC ∼
= 4ACB. Hence, ∠B ∼
= ∠C.
38
CHAPTER 2. HILBERT’S AXIOM SYSTEM
ii. Given a triangle 4ABC in which ∠B ∼
= ∠C. Consider the triangles 4ABC
and 4ACB. Since ∠B ∼
= ∠C, then by C.4, ∠C ∼
= ∠B. And BC ∼
= BC by C.2,
∼
∼
then by ASA, 4ABC = 4ACB and hence AB = AC by definition of congruent
triangles. Proposition 2.3.3 (Segment Subtraction) If A ∗ B ∗ C and D ∗ E ∗ F ,
AB ∼
= DE, and AC ∼
= DF then BC ∼
= EF .
Proof. Given A ∗ B ∗ C and D ∗ E ∗ F , AB ∼
= DE, and AC ∼
= DF . Apply C.1
−−→
−−→
to BC and ray EF . Then there is unique point G ∈ EF such that BC ∼
= EG.
Then by C.3 AC ∼
= DG. But AC ∼
= DF . This contradicts C.1 unless G = F .
Thus BC ∼
= EF . Proposition 2.3.4 Given AC ∼
= DF , then for any point B between A and C,
there is a unique point E between D and F such that AB ∼
= DE.
Proof. Given AC ∼
= DF and given B such that A ∗ B ∗ C. Then by C.1, there
−−→
is a unique point E on the ray DF such that AB ∼
= DE. Then either E = F
or D ∗ E ∗ F or D ∗ F ∗ E.
−→
If E = F , then B and C are two distinct points on AC such that AC ∼
= DF ∼
=
AB contradicting the uniqueness of C.1.
−→
If D ∗ F ∗ E, then by C.1, there is a point G on the opposite ray to ray CA such
that EF ∼
= CG. Then AG ∼
= DE by C.3. Thus there are two distinct points B
−→
and G on AC such that AG ∼
= DE ∼
= AB contradicting the uniqueness in C.1.
∼
Thus, D ∗ E ∗ F and AB = DE. Definition 2.3.3 By AB < CD or CD > AB we mean that there exists a
point E between C and D such that AB ∼
= CE.
Proposition 2.3.5 (Segment Ordering)
2.3. CONGRUENCE AXIOMS
39
i. (Trichotomy) For any two segments AB and CD, exactly one of the
followings hold: AB < CD, AB > CD, or AB ∼
= CD.
ii. If AB < CD and CD ∼
= EF , then AB < EF .
iii. If AB > CD and CD ∼
= EF , then AB > EF .
iv.(Transitivity) If AB < CD and CD < EF , then AB < EF .
Proof.i. Given any two segments AB and CD. By C.1, considering AB and
−−→
−−→
the ray CD there exists unique point on CD, say G, such that AB ∼
= CG. Then
by definition of the ray, either D = G, C ∗ D ∗ G,or C ∗ G ∗ D.
If D = G then AB ∼
= CD.
If C ∗ G ∗ D, then AB < CD.
If C ∗ D ∗ G, then by proposition 2.3.4 and since AB ∼
= CG, there exists unique
point I, A ∗ I ∗ B such that AI ∼
= CD, and hence AB > CD.
ii. Given AB < CD and CD ∼
= EF , then there exists G between C and
D such that AB ∼
= CG. Then by proposition 2.3.4 with CD ∼
= EF , there is
unique point J, between E and F such that CG ∼
= EJ. By C.2, AB ∼
= EJ.
Then EF > AB.
iii. Exercise.
iv. Given AB < CD and CD < EF . Then by definition of greater segment, there exists I such that C ∗ I ∗ D and AB ∼
= CI, also there is a point
G such that E ∗ G ∗ F and CD ∼
= EG. Consider CD, the point I and EG, by
proposition 2.3.4 there exists a point , say H such E ∗ H ∗ G and EH ∼
= CI.
By C.2, AB ∼
= EH. From E ∗ G ∗ F and E ∗ H ∗ G we get E ∗ H ∗ F . From
E ∗ H ∗ F and AB ∼
= EH, we get that AB < EF . Definition 2.3.4 Two angles are called supplementary angles, if they have
the same vertex, have a common side, and the other sides are opposite rays.
An angle is called right angle, if it is congruent to its supplementary angle.
Two angles are called vertical angles if they have the same vertex and the
sides are opposite rays.
Two lines meet at a point, say P are called perpendicular, if the angles consists of P and one side is a ray in the first line emanating from P , and the other
side is a ray in the second line emanating from P , are right angles.
Proposition 2.3.6
i. Supplements of congruent angles are congruent.
ii.Vertical angles are congruent to each other.
iii An angle congruent to a right angle is right.
40
CHAPTER 2. HILBERT’S AXIOM SYSTEM
Proof. i. Given ∠CAD ∼
= ∠GF H. We show that, their supplementary angles
are congruent, that is, ∠BAD ∼
= ∠EF H. See the figure.
By C.1, choose G such that F G ∼
= AC, E such that EF ∼
= AB, and H such
∼
that F H = AD. By SAS, 4DAC ∼
= 4HF G. By segment addition BC ∼
= EG.
Then, since ∠DCA ∼
= ∠HGF and DC ∼
= HG, so by SAS, 4DBC ∼
= 4HEG.
By congruence of triangles, HE ∼
= DB and ∠DBC ∼
= ∠HEG. Then by SAS,
∼
4DBA = 4HEF . Finally, by congruence of triangles ∠DAB ∼
= ∠HF E.
ii. Consider the following figure. ∠1 is supplementary to ∠2 and also supplementary to ∠4. By C.5 ∠1 ∼
= ∠1, then by part i ∠2 ∼
= ∠4.
iii. Let ∠1 be a right angle and ∠2 is any angle congruent to ∠1. See the figure.
Since ∠1 is right angle, then it is congruent to its supplementary angle ∠3.
2.3. CONGRUENCE AXIOMS
41
Then by i, the supplementary angle of ∠2, which is ∠4, is congruent to angle
∠3. That is, ∠4 ∼
= ∠3 ∼
= ∠1 ∼
= ∠2. By C.5, ∠4 ∼
= ∠2. Hence, ∠2 is right. Proposition 2.3.7 For every line l and every point P , there exists a line
through P that is perpendicular to l.
Proof. Given any line l, and P any point. Then, either P ∈ l or P ∈
/ l.
Case 1: let P ∈
/ l. By I.2, there are two points, say A and B on l. By C.4, on
−→
the opposite side of l from P , there exists a ray AG such that ∠GAB ∼
= ∠P AB.
−
→
By C.1, there is a point P 0 on AG such that AP 0 ∼
= AP . Now P P 0 intersects l
at a point Q. Either Q = A or Q 6= A.
If Q = A, since ∠P QB ∼
= ∠P 0 QB and they are supplementary angles, then
←−→
they are right angles. Hence, P P 0 is perpendicular to l.
If Q =
6 A, then by SAS, 4P QA ∼
= 4P 0 QA. Hence, ∠P QA ∼
= ∠P 0 QA. Thus
←−→0
P P is perpendicular to l.
Case 2: Let P ∈ l. By proposition 2.1.1 there is a point not on l, say H.
by case 1, we can drop a perpendicular from H to l, call it m. Let l and m
intersects at a point G. Then either P = G or P 6= G.
If P = G, then we are done.
If P 6= G. See the figure.
42
CHAPTER 2. HILBERT’S AXIOM SYSTEM
By C.4, consider angle ∠HGP and any ray part of l emanating from P and a
given side of l, there is a unique ray emanating from P , which produce an angle
congruent to the right angle ∠HGP . −−→
−−→
−→
Proposition 2.3.8 (Angle Addition) Given AD lies between AB and AC
−−→
−−→
−−→
and ray EG lies between EF and EH, ∠CAD ∼
= ∠GEH, and ∠DAB ∼
= ∠GEF .
∼
Then ∠CAB = ∠F EH.
−−→
−−→
−→
−−→
−−→
Proof. Given AD lies between AB and AC and ray EG lies between EF and
−−→
∼ ∠GEH, and ∠DAB =
∼ ∠GEF . See the figure
EH, ∠CAD =
−−→
Then by Crossbar theorem, AD intersects BC at a point, say M , such that
B ∗ M ∗ C. By C.1, choose F , G and H, such that AB ∼
= EF , AC ∼
= EH, and
AM ∼
EG.
=
Then, by SAS, 4M AC ∼
= 4GEH, and 4BAM ∼
= 4F EG. Then ∠HGE ∼
=
∠CM A and ∠F GE ∼
∠BM
A.
But
∠CM
A
and
∠BM A are supplementary
=
angles. By proposition 2.3.6 i, ∠BM A is congruent to the supplementary angle
2.3. CONGRUENCE AXIOMS
43
of ∠HGE, and since ∠F GE ∼
= ∠BM A. Thus by uniqueness of C.4, the supplementary angle of ∠HGE is ∠F GE. Hence H ∗ G ∗ F .
Since ∠EHF ∼
= ∠ACB, then by SAS, 4ABC ∼
= 4EHF . And by definition of
congruent triangles, ∠BAC ∼
= ∠F EH. −−→
−−→
−→
Proposition 2.3.9 (Angle Subtraction) Given AD lies between AB and AC
−−→
−−→
−−→
and ray EG lies between EF and EH, ∠CAB ∼
= ∠F EH, and ∠DAB ∼
= ∠GEF .
Then ∠CAD ∼
= ∠HEG.
Proof. Consider the following figure, in which ∠CAB ∼
= ∠HEF and ∠BAD ∼
=
∠F EG.
−−→
←→
Now, considering the angle ∠CAD, the ray EG, and the same side of EG as H.
−−→
By C.4, there is a ray EK, such that ∠CAD ∼
= ∠GEK. Thus by hypothesis and
angle addition, ∠KEF ∼
= ∠CAB. By assumption, we have ∠F EH ∼
= ∠CAB.
−−→ −−→
Thus by uniqueness in C.4, EK = EF . Hence, ∠CAD ∼
= ∠HEG. −−→
Definition 2.3.5 By ∠BAC < ∠DEF , we mean that, there is a ray EG be−−→
−−→
tween EF and ED such that ∠BAC ∼
= ∠GEF .
Proposition 2.3.10 (Ordering of Angles)
i.(Trichotomy) Given any two angles ∠A and ∠B. Then only one of the
following is true: ∠A < ∠B, ∠A > ∠B, or ∠A ∼
= ∠B.
ii. If ∠A < ∠B and ∠B ∼
= ∠C, then ∠A < ∠C.
iii. If ∠A > ∠B and ∠B ∼
= ∠C, then ∠A > ∠C.
iv. If ∠A < ∠B and ∠B < ∠C then ∠A < ∠C.
44
CHAPTER 2. HILBERT’S AXIOM SYSTEM
Proof.
i. Given any two angle ∠DAC and angle ∠GBE. Consider ∠DAC and ray
−−→
←→
BE, by C.4 on the same side of line BE as point G, there is a unique ray, say
←→
−−→
BH, such that ∠CAD ∼
= ∠HBE. Then either H ∈ BG or H is on the same
←→
side of line BG as point E or in the opposite side.
←→
←→
Case 1. If H ∈ BG. Since H is on the same side of BE as point G, then H
−−→
−−→
−−→ −−→
does not belong to the opposite ray of BG, that is H ∈ BG, then BH = BG.
Thus ∠CAD ∼
= ∠GBE.
←→
Case 2. If H is on the same side of line BG as point E, and from that H
←→
is on the same side of line BE as G. Then H is interior point of the angle
−−→
−−→
−−→
∠GBE. Hence BH lies between rays BE and BG. Thus, ∠EBG > ∠DAC.
←→
Case 3. If H is on the opposite side of line BG as point E, then EH in←→
tersects line BG at a point, say K and H ∗ K ∗ E. Then either K belong to the
−−→
−−→
ray BG or it belongs to the opposite ray of BG.
−−→
Claim: K belongs to the ray of BG.
−−→
If not, that is K belongs to the opposite ray of BG. Then G and K are in
←→
←→
opposite sides of line BE, but G and H are on same side of line BE, then by
←→
B.4, H and K are in opposite sides of line BE. Which contradicts that K is
interior point of the angle ∠HBE, since H ∗ K ∗ E. Thus K belongs to the ray
−−→
of BG.
Choose D and C, such that AD ∼
= BH and AC ∼
= BE. Since ∠CAD ∼
= ∠HBE,
∼
then by SAS 4ADC = 4BHE. See the figure
Then, by definition of congruent triangles CD ∼
= HE. So, by proposition 2.3.4,
there is a unique point, say Y , D ∗ Y ∗ C, such that CY ∼
= EK. By SAS,
∼
4AY C ∼
4BKE.
Then
∠Y
AC
∠GBE,
and
since
Y
interior
to ∠DAC by
=
=
proposition 2.2.4. Then ∠DAC > ∠GBE.
2.3. CONGRUENCE AXIOMS
45
ii. Assume that ∠EAD < ∠GBF and ∠GBF ∼
= ∠HCI. See the figure.
−−→
−−→
−−→
Then there is a ray, say BK between BG and BF such that ∠KBF ∼
= ∠EAD.
−−→
Also, BK intersects GF at a point, say J such that G ∗ J ∗ F . Now, Choose
I, H such that BG ∼
= CH and BF ∼
= CI. Then by SAS, 4GBF ∼
= 4HCI.
Then ∠GF B ∼
∠HIC.
Then,
apply
proposition 2.3.4 on HI and G ∗ J ∗ F , we
=
get that there is a unique point L on HI, such that F J ∼
= LI. Then by SAS, we
∼
get that 4JBF ∼
4LCI.
Hence,
∠JBF
∠LCI.
By
C.2, ∠EAD ∼
=
=
= ∠LCI.
That is, ∠EAD < ∠HCI.
iii. Exercise.
iv. Assume that ∠EAD < ∠GBF and ∠GBF < ∠HCI. Then there is a ray,
−−→
−−→
−−→
−−→
say BK between BG and BF such that ∠KBF ∼
= ∠EAD. Also, BK intersects
GF at a point, say J such that G ∗ J ∗ F . Since ∠GBF < ∠HCI, then there
−→
−−→
−→
is a ray, say CL between CH and CI such that ∠GBF ∼
= ∠LCI Now, Choose
I, L such that BG ∼
= CL and BF ∼
= CI. Then by SAS, 4GBF ∼
= 4LCI. Then
∠GF B ∼
= ∠LIC. Then, apply proposition 2.3.4 on LI and G ∗ J ∗ F , there
is a unique point M on LI, such that F J ∼
= M I. Then by SAS, we get that
4JBF ∼
= 4M CI. Hence, ∠M CI ∼
= ∠JBF and ∠JBF ∼
= ∠EAD. Then by
C.4, ∠EAD ∼
= ∠M CI.
Claim: M is interior to ∠HCI. Then we conclude that ∠EAD < ∠HCI. To
prove the claim, first, since M interior to ∠LCI, M and L are on the same side
←
→
←
→
of CI, but also, L and H are on same side of CI, then by B.4, M and H are
←
→
on same side of the line CI.
←→
Second, to show that M and I are on same side of the line CH. Use contra←→
diction, let M and I be on opposite sides of the line CH. Since L is interior to
←→
∠HCI then L and I are on the same side of the line CH. Then by corollary
←→
of B.4, L and M are in opposite sides of the line CH. Then M L intersects the
←→
line CH at a point, say P , such that M ∗ P ∗ L. From proposition 2.2.1 and
←→
that L ∗ M ∗ I, we obtain L ∗ P ∗ I. Thus LI intersects the line CH at P , which
46
CHAPTER 2. HILBERT’S AXIOM SYSTEM
contradicts that, L is interior to the angle ∠HCI. Proposition 2.3.11 (SSS) Given 4ABC and 4DEF . If AB ∼
= DE, BC ∼
=
∼
∼
EF , and AC = DF , then 4ABC = 4DEF .
Proof. Given 4ABC and 4DEF , AB ∼
= DE, BC ∼
= EF , and AC ∼
= DF .
←→
By corollary of SAS, on the opposite side of DE as F , there is a unique point,
say M such that 4ABC and 4DEM are congruent. Then BC ∼
= EM and
CA ∼
= DM , by C.2 EF ∼
= EM and F D ∼
= DM .
←→
←→
Since F and M are in opposite sides of DE, F M intersects the line DE at a
point say P . Then either P = D, P = E, P ∗ D ∗ E, D ∗ E ∗ P , or D ∗ P ∗ E.
Case 1: If P = D, see the following figure.
Then by proposition 2.3.2 ∠EF M ∼
= ∠EM F . That is ∠DF E ∼
= ∠DM E. But
∼
∠DM E = ∠ACB. Then by C.4, we get ∠ACB ∼
= ∠DF E. Hence, by SAS
4ABC ∼
= 4DEF .
Case 2: If P = E, the proof is similar to case 1.
Case 3. If P ∗ D ∗ E, see the following figure.
2.3. CONGRUENCE AXIOMS
47
Then by proposition 2.3.2, ∠P F E ∼
= ∠P M E, and ∠P F D ∼
= ∠P M D. Thus by
∼
angle subtraction ∠DF E = ∠DM E. But ∠DM E ∼
= ∠ACB. Then by C.5, we
get ∠ACB ∼
= ∠DF E. By SAS, 4ABC ∼
= 4DEF .
Case 4: If D ∗ E ∗ P , the proof is similar to case 3.
Case 5: If D ∗ P ∗ E then F is in the interior of ∠DM E, and M in the interior
of ∠DF E.
48
CHAPTER 2. HILBERT’S AXIOM SYSTEM
By proposition 2.3.2, in the triangle 4F EM , we get ∠EF M ∼
= ∠EM F , and
in the triangle 4F DM , we get ∠DF M ∼
= ∠DM F . Then by angle addition,
we get that ∠DF E ∼
= ∠DM E. But ∠DM E ∼
= ∠ACB. The by C.4, we get
∼
∠ACB = ∠DF E. Hence, by SAS, 4ABC ∼
= 4DEF . Proposition 2.3.12 (Euclid’s Fourth Postulate) All right angles are congruent to each other.
Proof. Given the following two pairs of right angles, ∠BAD ∼
= ∠CAD and
∠F EH ∼
= ∠GEH. We will use contradiction. Assume that ∠BAD is not congruent to ∠F EH. Then by trichotomy either ∠F EH < ∠BAD or ∠F EH >
∠BAD.
−→
It is enough to consider the case ∠F EH < ∠BAD. Then there is a ray AJ
−−→
−−→
between AB and AD such that ∠BAJ ∼
= ∠F EH.
By proposition 2.3.6, ∠CAJ ∼
= ∠GEH. Then by C.5 and that ∠GEH ∼
=
∠F EH, we get ∠CAJ ∼
∠BAJ.
=
Since ∠BAJ < ∠BAD ∼
= ∠CAD then by ordering of angles ∠BAJ < ∠CAD.
2.4. AXIOMS OF CONTINUITY
49
We show D is an interior point to ∠CAJ. First, D, J are on same side of the
←→ ←→
line AB = AC, since J is interior to ∠BAD.
←
→
Second, C, B are on opposite side of AJ, since C ∗ A ∗ B, and D, B are in oppo←
→
←→
site sides of AJ. Then C and D are on the same side of AB, since D is interior
point of ∠CAJ. Then ∠CAD < ∠CAJ, and ∠CAJ ∼
= ∠BAG < ∠CAD, which
is a contradiction. 2.4
Axioms of continuity
The axioms of continuity are the axioms which give us our correspondence between the real line and an Euclidean line. These are necessary to guarantee that
our geometry “Euclidean Plane Geometry”is complete.
Dedekind
Suppose that the set of all points on a line lP
is the disjoint
P Axiom:
P
union 1 ∪ 2Pof two non-empty subsets, such that no points of 1 is between
two points of 2 and vice versa. Then P
there is a unique
P point O lying on l
such that P1 ∗ O ∗ P2 if and only if P1 ∈ 1 and P2 ∈ 2 . That is one of the
subsets is equal to a ray of l with vertex O and the other subset is equal to the
complement.
P
P
A pair of subsets 1 and 2 with the properties of Dedekind’s axiom is called
a Dedekind cut of the line.
Dedekind’s axiom is a sort of converse to the line separation property, which
says that any point O on l separates all points on l into those to the left of O,
and those to the right of O. In particular, the set of all points on l is the union
of the two rays of l emanating from O. Dedekind’s axiom says that, conversely
any separation of points on l into left and right is produced by a unique point
O.
Corollary 2.4.1 We can define a Dedekind cut on a ray r the same way a
Dedekind cut is defined for a line. Then the conclusion of Dedekind’s axiom
hold for the ray r.
Proof. Let r be
ray. Suppose that the set of all points on the ray r is P
the
P anyP
disjoint union 1 ∪ 2 ofPtwo non-empty subsets, such that no points of 1
is between two points
P of 2 and vice versa. One of the subsets contains the
vertex A of r, say, P1 . Then enlarge this set so as to include the P
ray opposite
P
∗
∗
to r. Call this set 1 Let l be the line containing ray r. Thus, 1 and 2
form a Dedekind
cut of l, since no point in the opposite ray of r lies between two
P
points of 2 , which is subset of the ray r, and vice versa. Then by Dedekind’s
axiom, there is a unique point O on l such that one of the subsets is equal to
the ray of l with vertex O and the other
Psubset
T P is equal to the complement. The
point O belongs to ray r. Otherwise, 1
2 would be non-empty. .
Corollary 2.4.2 We can define a Dedekind cut on a segment AB the same way
a Dedekind cut is defined for a line. Then the conclusion of Dedekind’s axiom
50
CHAPTER 2. HILBERT’S AXIOM SYSTEM
hold for the segment AB.
Proof. Let AB be any segment.
that the set of all points on the
P Suppose
P
segment AB P
is the disjoint union 1 ∪ P
of
two non-empty subsets, such that
2
no points of 1 is betweenP
two points of 2 and vice versa. One of the subsets
P
contains the point A, say, P
1 . The point
P B must be in the other subset, 2 .
For if both A and B are in 1 . Then 2 would be empty, contradicting the
assumption.
P
−−→
Then enlarge the subset
1 so as to include the ray opposite to AB. And
P
−−→
enlarge the subset 2 so as to include P
the ray opposite
to BA. Let l be the
P
line containing the segment AB. Thus, 1 and 2 form a Dedekind cut of l.
Then by Dedekind’s axiom, there is a unique point O on l such that one of the
subsets is equal to the ray of
Pl with vertex
P O and the other subset is equal to
the complement. Since A ∈ 1 and B ∈ 2 . Then A ∗ O ∗ B. .
Dedekinds axiom ensure that a line has no holes in it, in the sense that for
any point O on a line l and any positive real number x there exists unique points
P−x and Px on l such that P−x ∗ O ∗ Px and the segments P−x O and OPx both
have the length x.
Without Dedekind’s axiom there is no guarantee that there is a segment
√ of
length π or of length e or of certain other non-constructible lengths as 2. It
is Dedekind’s axiom that allows us to make correspondence of the line in our
geometry and the real line. It is with Dedekind’s axiom that we are able to
introduce the coordinate system and do geometry analytically, as Descartes and
Fermat discovered in the seventeenth century. This coordinate system enables
us to prove that our axioms for Euclidean geometry are categorical. That is
this system has a unique model up to isomorphism, namely, the usual Cartesian
coordinate plane of all ordered pairs of real numbers.
To see why we need Dedekind’s axiom, consider the manner in which Euclid
construct a perpendicular to a given line at a given point only with a straight
edge and compass:
Given any line l, and any point O on l. First, using the point O as a center,
draw a circle of positive radius. The circle intersects the line l in two points,
say A and B. At each of these points, we then construct a circle of larger radius
and these two circles intersect in two points, say P and Q. Drawing the line
between the two points of intersections gives a line perpendicular to the given
line l at the given point O. See the figure
There are two gaps in this proof:
2.4. AXIOMS OF CONTINUITY
51
1. Why does a line intersects the circle with center O at all?
2. Why do the two circles then intersect?
The first of these gaps can be filled by assuming the Elementary Continuity
Principle, and the second gap can be filled by assuming the Circular Continuity
Principle. Both principles are consequences of Dedekind’s axiom.
Definition 2.4.1 i. Given any two distinct points R and O, then a circle
with center O and radius OR is defined to be the set of all points P such
that OP ∼
= OR.
ii. We define a point X to be inside the circle with center O and radius OR
if OX < OR. A point Y is outside the circle if OY > OR.
Proposition 2.4.1 (Elementary Continuity Principle) Let γ be a circle
with center O and radius OR. If A is a point inside γ, and B is a point outside
γ. Then the segment AB intersects the circle γ.
Proof. We will prove it later.
Proposition 2.4.2 (Circular Continuity Principle) If a circle γ has one
point inside and one point outside another circle γ 0 , then the two circles intersect
in two points.
Proof. We will prove it later.
The next statement is not about continuity but rather about measurements.
Archimedes recognized that a new axiom was needed. It is listed here because
it is a consequence of Dedekind’s axiom. It is needed so that we can assign
a positive real number as a length of an arbitrary segment, as we will see in
chapter 3.
Proposition 2.4.3 (Archimedes’ Axiom)
If CD is any segment, A any point, and r any ray with vertex A. Then for
every point B 6= A on r there is an integer n such that when CD is laid off n
times on r starting at A, a point E is reached such that n.CD ∼
= AE and either
B = E or A ∗ B ∗ E.
Here we use C.1 to begin laying off CD on r starting from A, obtaining a point
A1 on r such that AA1 ∼
= CD. We define 1.CD to be AA1 . Let r1 be the ray
emanating from A1 and contained in r. By C.1, we obtain a unique point A2 on
52
CHAPTER 2. HILBERT’S AXIOM SYSTEM
r1 such that A1 A2 ∼
= CD. And we define 2.CD to be AA2 . Proceeding in the
same manner, we can define by induction on n, the segment n.CD to be AAn .
For example, if AB were π units long and CD of one unit length, then we would
have to lay off CD at least four times to get to a point E, such that A ∗ B ∗ E.
The intuitive content of Archimedes’ Axiom is that if we arbitrarily choose
one segment CD as a unit of length, then every other segment has finite length
with respect to this unit. In terms of the axiom, the length of AB with respect
to CD as unit is at most n units. Another way to look at it to choose AB as
unit length. Then the axiom says that, no other segment can be infinitesimally
small with respect to this unit. That is, the length of CD with respect to AB
1
units.
as unit is at least
n
Proof of Archimedes Axiom.
Given aP
segment CD and a point A on line l, with a ray r of l emanating from
A. Let 1 consists of A and all points P
B on r reached by laying P
off copies of
segment CD on r starting from
A.
Let
be
the
complement
of
in r.
2
1P
P
First, we have to show that 2 is empty. We use contradiction. If 2 6= ∅.
Then, let us show that,
P we have a Dedekind cut on r.
P
Let P and Q be in 2 such that A ∗ P ∗ Q. We must show that P Q ⊂ 2 .
Let B be a point such that P ∗ B ∗ Q. Suppose that B could be reached, that
is there is n and a point E such that n.CD ∼
= AE, and A ∗ B ∗ E. Then apply
proposition 2.2.3 on A ∗ P ∗ Q and P ∗ B ∗ Q, we get A ∗ P ∗ B. Then apply
the proposition on A ∗ P ∗ B and A ∗ B ∗ E, we get A ∗ P ∗ E. ThusPP is also
reachedPby the same n and the same E, contradicting to that P ∈ 2 . Thus
P Q ⊂ 2 . So we have a Dedekind cut on theP
ray r. Let O P
be the unique point
P
on r such
that
P
∗
O
∗
P
if
and
only
if
P
∈
and
P
∈
1
2
1
2
1
2 . Either O ∈
1
P
or O ∈ P 2 .
If O ∈ 1 . Then there is n such that O can be reached by laying off n copies
of segment CD on r P
starting from A. By laying off one more copy of CD, we
can reach a point in 2 , which is impossible.
P
P
If O ∈
2 . Lay off a copy of CD on the ray opposite to
2 starting at
O, obtaining a point P . Then P must lie on r. For, if not, AO < CD, then
there is a point M , such that C ∗M ∗D, and AO ∼
= CM . By applying C.1 on the
−→
opposite ray to OA, there is a point G such that OG ∼
G and
= M D, then A ∗ O ∗ P
AG ∼
be
reached
from
A,
contradiction
to
that
O
∈
= 1.CD. Then O can
2.
P
Since P ∈ r, then P ∈ 1 . Then for some integer n, P can be reached by laying
off n copies of segment CD on r starting from A.
P By laying off one more copy
of CD, we reach O. That contradicts that O ∈ 2 .
Thus,
in either cases, we obtain a contradiction. Hence our assumption that
P
is
non-empty
is false. 2
2.5. AXIOM OF PARALLELISM
2.5
53
Axiom of Parallelism
If we were to stop with the axioms we now have, we could do a bit of geometry,
but we still could not do all of Euclidean geometry. We would have the so called
Neutral geometry, it is called so, because in doing this geometry we remain
neutral about the parallel axiom.
Hilbert’s Parallel Axiom: For every line l and every point P not on l, there is
at most one line m through P such that m is parallel to l and m pass through P .
Note that this axiom is weaker than the Euclidean parallel axiom. This axiom asserts only that at most one line through P is parallel to l, whereas the
Euclidean parallel axiom asserts in addition that at least one line through P
and parallel to l. This is omitted from Hilbert’s parallel axiom, since it can be
proved from the other axioms. It is therefore unnecessary to assume this as part
of an axiom.
The axiom of parallelism completes our list of 16 axioms for Euclidean geometry.
The Euclidean plane is a model of these axioms.
2.6
Exercises
[1] Given A ∗ B ∗ C and A ∗ C ∗ D. Then
i. Show that A, B, C, and D are distinct points.
ii. Show that A, B, C, and D are collinear.
[2] Given A ∗ B ∗ C.
i. If P is a fourth point collinear with A, B, and C. Then show that:
∼ A ∗ B ∗ P implies that ∼ A ∗ C ∗ P .
−−→ −→
ii. Show that BA ⊂ CA.
[3] Given a line l, a point P on l, and Q not on l. Then show thatevery point
−−→
of the ray P Q except P is on the same side of l as Q.
[4] In the following figure CB ∼
= BD ∼
= DA ∼
= AC. Prove that ∠CAD ∼
=
∠CBD.
54
CHAPTER 2. HILBERT’S AXIOM SYSTEM
[5] In the following figure, AB ∼
= DC.
= AE. Show that BE ∼
= AC, and AD ∼
[6] In the following figure AB ∼
= AC, A0 B ∼
= A0 C. Prove that 4AA0 B ∼
=
0
4AA C.
2.6. EXERCISES
55
[7] Prove that a line cannot be contained in the interior of a triangle. The
interior of a triangle is defined to be the intersection of the interiors of
all angles of the triangle.
[8] Given a triangle 4ABC.
i. If ray r emanating from an exterior point of 4ABC intersects AB
in a point between A and B, then r also intersects AC or BC. A
point is called an exterior of a triangle if and only if it is not in
the interior of the triangle, and not on the sides of the triangle.
ii. If a ray r emanates from an interior point of 4ABC, then it intersects
one of the sides of the triangle, and if it does not pass through a
vertex, it intersects only one side.
[9] Prove that an equiangular
2
triangle is equilateral.
[10] Given two triangles 4ADC and 4A0 D0 C 0 . And given A ∗ B ∗ C and
A0 ∗ B 0 ∗ C 0 . If AB ∼
= A0 B 0 , BC ∼
= B 0 C 0 , AD ∼
= A0 D0 , and BD ∼
= B 0 D0 ,
0 0
∼
then show that CD = C D .
−−→
−−→
−−→
[11] Given ∠ABC ∼
= ∠DEF and BG between BA and BC. Prove that there
−−→
−−→
−−→
is a unique ray EH between ED and EF such that ∠ABG ∼
= ∠DEH.
[12] If AB < CD then show that 2 · AB < 2 · CD.
2 Equiangular
triangle is a triangle, in which all angles congruent to one another
56
CHAPTER 2. HILBERT’S AXIOM SYSTEM
Chapter 3
Neutral Geometry
A geometry satisfying Hilbert’s axioms of incidence, betweenness, congruences,
and continuity, in which we neither affirm or deny Hilbert’s parallel axiom, is
called a neutral geometry.
The purpose of studying neutral geometry is to clarify the role of the parallel postulate by seeing which theorems in the geometry do not depend on it.
That is, which theorems follow from the other axioms alone without ever using
the parallel postulate in proofs.
Note that all propositions, theorems and corollaries, that we proved in the previous chapter hold in neutral geometry.
3.1
Alternate Interior angle Theorem and Exterior Angle Theorem
Definition 3.1.1 A line t is called a transversal to two lines l and m, if t
intersects l at point A, and t intersects m at point B and A 6= B.
Let t be a transversal to lines l and m, with t intersects l at A and m at B.
Choose points C and D on l such that C ∗ A ∗ D, and choose points E and F
on m such that E ∗ B ∗ F , and D, F are on the same side of the line t. Then
the following angles are called interior angles:
∠CAB, ∠DAB, ∠EBA, and ∠F BA.
The two pairs ( ∠CAB, ∠F BA) and (∠EBA, ∠DAB) are called pairs of alternate interior angles. See the following figure:
57
58
CHAPTER 3. NEUTRAL GEOMETRY
Theorem 3.1.1 (Alternate Interior Angle Theorem) If two lines cut by
a transversal have a pair of congruent alternate interior angles, then the two
lines are parallel.
Proof. Given t to be a transversal to lines l and m, with t intersects l at A
and m at B. Choose points C and D on l such that C ∗ A ∗ D, and choose
points E and F on m such that E ∗ B ∗ F , and D, F are on the same side of t.
Let ∠CAB ∼
= ∠F BA. We show that l is parallel to m. We use contradiction,
assume that l and m meet at point P . Say P is on the same side of t as D. By
C.1, choose C such that AC ∼
= BP . Then by SAS, 4P BA ∼
= 4CAB. Thus,
∠DAB ∼
∠ABC.
Since
∠EBA
is
supplement
of
∠F
BA
then
by proposition
=
2.3.6, ∠DAB ∼
∠ABE.
By
C.5,
∠EBA
=
∠CBA.
Then
C
lies
on m. Then
=
l and m have two points in common C and P , which contradicts proposition
2.1.1. Corollary 3.1.1 i. Two lines perpendicular to the same lines are parallel.
ii. For any line l and any point P not on l, there is a unique perpendicular
from P to l.
Proof. i. Let n and m be two distinct lines that are perpendicular to a line
l. Then the alternate interior angles are right angles, and hence congruent by
proposition 2.3.12. Then by alternate interior angle theorem n||m.
ii. Exercise. Corollary 3.1.2 If l is any line and P any point not on l, then there is at least
one line through P and parallel to l.
Proof. Given any line l and any point P not on l. By previous corollary ii,
there is a unique line, say m through P and perpendicular to l. Now apply
3.1. ALTERNATE INTERIOR ANGLE THEOREM AND EXTERIOR ANGLE THEOREM59
proposition 2.3.7 on P and the line m, then there is a line, say n, through P
and perpendicular to m. Thus, both l and n are perpendicular to m, then by
previous corollary, l||n. Note that, Hilbert’s parallel axiom and this corollary imply the Euclidean parallel axiom.
Definition 3.1.2 An angle supplementary to an angle of a triangle is called exterior angle of the triangle. The two angles of a triangle that are not adjacent
1
to an exterior angle of a triangle are called remote interior angles.
Theorem 3.1.2 (Exterior Angle Theorem) An exterior angle of a triangle
is greater than either remote interior angles.
Proof. Given any triangle 4ABC. Let D be a point such that B ∗ C ∗ D.
First, we show that ∠ACD > ∠BAC. If not, then by trichotomy on angles,
either ∠ACD ∼
∠BAC.
= ∠BAC or ∠ACD <
←→
←→
Case 1: If ∠ACD ∼
∠BAC. Then AC is a trasversal of the two lines CD and
=
←→
BA with a pair of congruent alternate interior angles.
←→
←→
Hence, by alternate interior angle theorem, the line CD and BA are parallel, which contradicts that the two lines meet at B.
−→
−−→
Case 2: If ∠ACD < ∠BAC. Then there is a ray, say AF between AB and
−→
−
→
AC such ∠F AC ∼
= ∠ACD. By Crossbar theorem, AF intersects BC at point,
say M between B and C.
1 two
angles are called adjacent if they have a common vertex and a common side.
60
CHAPTER 3. NEUTRAL GEOMETRY
←→
←→
←→
Then AC is a trasversal of the two lines CD and AF with a pair of congruent
alternate interior angles. Hence, by alternate interior angle theorem, the line
←→
←→
CD and AF are parallel, which contradicts that the two lines meet at M .
Second, we show that ∠ACD > ∠ABC. If not, then by trichotomy on angles, either ∠ACD ∼
= ∠ABC or ∠ACD < ∠ABC.
Case 1: If ∠ACD ∼
= ∠ABC. Then their supplementary angles; ∠1 and ∠2, are
congeruent by proposition 2.3.6. See the figure
←→
←→
←→
Then BD is a transversal of the two lines AC and AB with a pair of congruent alternate interior angles; ∠1 and ∠2. Hence, by alternate interior angle
←→
←→
theorem, the lines CA and BA are parallel, which contradicts that the two lines
meet at A.
Case 2: Exercise. 3.1. ALTERNATE INTERIOR ANGLE THEOREM AND EXTERIOR ANGLE THEOREM61
Proposition 3.1.1 (SAA) Given two triangles 4ABC and 4DEF with AC ∼
=
DF , ∠ABC ∼
= ∠DEF , and ∠ACB ∼
= ∠DF E. Then 4ABC ∼
= 4DEF .
Proof. Given two triangles 4ABC and 4DEF with AC ∼
= DF , ∠ABC ∼
=
∼
∠DEF , and ∠ACB = ∠DF E. Then by trichotomy on segments, either BC ∼
=
EF , BC > EF , or BC < EF .
If BC ∼
= EF , then by ASA, 4ABC ∼
= 4DEF .
If BC > EF , then there is a point, say M such that B ∗ M ∗ C such that
CM ∼
= EF . Then 4DEF ∼
= 4AM C. then ∠AM C ∼
= ∠DEF . Thus by C.5,
∼
∠AM C = ∠ABC, which contradicts Exterior angle theorem.
If BC < EF is an exercise. Proposition 3.1.2 (SSA,the angle is right) Given two triangles 4ABC,
and 4DEF , such that, ∠B and ∠E are right, DE ∼
= AB, and AC ∼
= DF .
∼
Then 4ABC = 4DEF .
Proof. Given two right triangles 4ABC and 4DEF with ∠B
right, DE ∼
= DF . By C.1, on the opposite ray of
= AB, and AC ∼
a unique point G such that EF ∼
= BG. Then by SAS, 4DEF ∼
=
the figure
and ∠E are
−−→
BC, there is
4ABG. See
In the isosceles triangle 4AGC, the angles of the base are congruent; ∠AGC ∼
=
∠ACG. Then by SAA, 4AGB ∼
= 4ACB. Thus, 4ACB ∼
= 4DF E. Definition 3.1.3 A midpoint of a segment AB is a point M such that A∗M ∗B
and AM ∼
= M B.
Proposition 3.1.3 Every segment has a unique midpoint.
Proof. Existence: Given any segment AB. By proposition 2.1.1, let C be any
←→
←→
point not on AB. Then by C.4, on the opposite side of AB from C, there is a
−−→
unique ray BX such that ∠BAC ∼
= ∠ABX. By C.1, there is a unique point D
−−→
←→
∼
on BX with AC = BD. Now, since D and C are in opposite sides of line AB,
←→
then CD intersects the line AB at a point say, M . We show that A ∗ M ∗ B.
←→
Since M ∈ AB, then either M = A, M = B, M ∗ A ∗ B, A ∗ M ∗ B, or A ∗ B ∗ M .
62
CHAPTER 3. NEUTRAL GEOMETRY
If M = A, see the figure.
Then considering 4ABD and the exterior angle ∠BAC. Then this angle is
greater that the remote interior angle, this is a contradiction to that ∠CAB ∼
=
∠ABD.
If M = B, is an exercise. See the figure
If M ∗ A ∗ B, then ∠CM B is an exterior angle to the triangle 4M BD. Thus by
exterior angle theorem ∠CM B > ∠M BD, and ∠CAB is exterior to 4CM A,
then ∠CAB > ∠CM B. But ∠CAB ∼
= ∠M BD. Then by ordering of angles
∠CM B < ∠M BD. Thus, we get a contradiction. See the following figure.
3.1. ALTERNATE INTERIOR ANGLE THEOREM AND EXTERIOR ANGLE THEOREM63
.
If A ∗ B ∗ M is an exercise. Use the following figure.
64
CHAPTER 3. NEUTRAL GEOMETRY
Thus, A ∗ M ∗ B. See the figure.
The vertical angles ∠CM A ∼
= ∠BM D. Then by AAS 4M CA ∼
= 4M DB.
Then AM ∼
= M B.
Uniqueness: Suppose there are two midpoints say M and E. Then A ∗ M ∗ B
and A ∗ E ∗ B. Then apply B.2 on M , E, and B, either E ∗ M ∗ B, M ∗ E ∗ B,
or M ∗ B ∗ E.
If M ∗ B ∗ E, with A ∗ E ∗ B, we get by proposition 2.2.2, A ∗ B ∗ M contradiction
to that A ∗ M ∗ B.
If E ∗ M ∗ B, then BM < EB ∼
= AE. From E ∗ M ∗ B and A ∗ E ∗ B, by
proposition 2.2.2 we get A ∗ E ∗ M , then AE < AM ∼
= BM , thus AE < BM ,
this is a contradiction.
If M ∗ E ∗ B, then AE ∼
= EB < M B ∼
= AM . From M ∗ E ∗ B and A ∗ M ∗ B,
we get A ∗ M ∗ E. Hence AM < AE. Thus we get a contradiction. −−→
Definition 3.1.4 i. A bisector of an angle ∠ABC is a ray BD lies between
−−→
−−→
rays BA and BC such that ∠ABD ∼
= ∠DBC.
ii. A bisector of a segment AB is a line that pass through the midpoint of
←→
AB and perpendicular to AB.
Proposition 3.1.4 i. Every angle has a unique bisector.
ii. Every segment has a unique bisector.
Proof. i. Given any angle ∠ABC. Choose A and C such that BC ∼
= AB. See
figure.
3.1. ALTERNATE INTERIOR ANGLE THEOREM AND EXTERIOR ANGLE THEOREM65
Let M be the midpoint of AC. Then by proposition 2.3.2 ∠BAC ∼
= ∠BCA
−−→
∼
∼
and hence by SAS 4ABM = 4CBM . Then ∠ABM = ∠M BC. Thus BM is
the bisector of the given angle.
ii. It follows from proposition 2.3.7 and proposition 3.1.3. Proposition 3.1.5 In any triangle the greater angle lies opposite the greater
side, and the greater side lies opposite the greater angle.
Proof. Given any triangle 4ABC. We show that AB > BC if and only if
∠C > ∠A.
(=⇒) Assume that AB > BC. Then there is a unique point D between A
and B such that BC ∼
= ∠BDC. By
= BD. Then by proposition 2.3.2 ∠BCD ∼
exterior angle theorem ∠BDC > ∠CAB. Since A ∗ D ∗ B then D interior point
of the angle ∠BCA. Thus, ∠BCA > ∠BCD ∼
= ∠BDC > ∠CAB. Hence by
C.5, ∠BCA > ∠CAB.
(=⇒) Assume that ∠C > ∠A. We will show that AB > BC. Use contradiction, assume that AB is not greater that BC. Then either AB ∼
= BC or
66
CHAPTER 3. NEUTRAL GEOMETRY
AB < BC.
If AB ∼
= BC. Then by proposition 2.3.2 ∠A ∼
= ∠C. Thus we get a contradiction.
If AB < BC. Then there is a unique point D between C and B such that
BA ∼
= ∠BDA. By exterior an= BD. Then by proposition 2.3.2 ∠BAD ∼
gle theorem ∠BDA > ∠ACB. Since C ∗ D ∗ B then D interior point of the
angle ∠BAC. Thus ∠BAC > ∠BAD ∼
= ∠BDA > ∠ACB. Hence by C.5,
∠BAC > ∠ACB. Thus we get a contradiction. Proposition 3.1.6 Given any two triangles 4ABC and 4EF G, with AB ∼
=
EF and BC ∼
= F G. Then ∠B < ∠F if and only if AC < EG.
−−→
−−→
Proof.(=⇒) Assume that ∠B < ∠F . Then there is a ray F X between F G and
−−→
−−→
F E such that ∠XF E ∼
= ∠CBA. By Crossbar theorem, the ray F X intersects
−−→
EG at a point, say H. By C.1, there is a unique point D on the ray F X such
that F D ∼
= BC. Then either H = D, F ∗ D ∗ H, or F ∗ H ∗ D.
If H = D, see the figure
Then by SAS, 4ABC ∼
= 4EF D. Then DE ∼
= AC and E ∗ D ∗ G, thus
EG > AC.
If F ∗ D ∗ H, Then by SAS, 4ABC ∼
= 4EF D. See the following figure.
3.2. MEASURE OF SEGMENTS AND ANGLES
67
Since 4DF G is an isosceles triangle, then ∠F GD ∼
= ∠F DG. Applying exterior angle theorem, we get ∠HDG > ∠F GD ∼
= ∠F DG, we get also, ∠F DG >
∠DGH. Since H is interior to the angle ∠GDE. Then ∠GDE > ∠HDG. Thus
by C.5 and ordering of angles, ∠EDG > ∠EGD. Then by previous proposition,
EG > ED ∼
= AC. Then EG > AC.
If F ∗ H ∗ D, see the following figure.
Then ∠F GD ∼
= ∠F DG. Since H is interior to the angle ∠DGF . Then,
∠DGF > ∠DGH. Since H is interior to the angle ∠EDG. Then ∠GDE >
∠GDH. Then by angle ordering, ∠GDE > ∠DGH. Then by previous proposition, EG > DE ∼
= AC. Then, EG > AC.
(⇐=) Assume that EG > AC. Use contradiction. Assume that ∠F is not
greater than ∠B. Then by trichotomy, either ∠F ∼
= ∠B or ∠F < ∠B.
If ∠F ∼
= ∠B. Then by SAS, 4ABC ∼
= 4EF G. Then AC ∼
= EG, we get a
contradiction.
If ∠F < ∠B, then by the first direction EG < AC, which is a contradiction. .
3.2
Measure of Segments and Angles
Up to this point we have not really needed the real numbers R. We have only
used R for models of geometry, but not in the theoretic development of geometry
68
CHAPTER 3. NEUTRAL GEOMETRY
itself. This is in the spirit of Euclid who did not assume the existence of such a
system of numbers.
However, to do geometry in a modern way, real numbers are useful. For example,
they measure segments and angles. They can also be used to measure areas and
volumes, but this is beyond the scope of this course.
The following theorem is a theorem of neutral geometry since its proof requires
the Archimedean principle.
Theorem 3.2.1 (Segment measure)
Let OI be a fixed segment (unit segment). There is a unique way to assign to
each segment AB a positive real number, denoted by AB ∈ R, called the length
of the segment AB, in such away that:
i. OI = 1.
ii. AB = BC if and only if AB ∼
= BC.
iii. A ∗ B ∗ C if and only if AC = AB + BC.
iv. AB < CD if and only if AB < CD.
v. ∀x ∈ R+ , there is a segment AB such that AB = x.
Proof. Let I1 be the midpoint of OI, I2 be the midpoint of OI1 . In general,
1
let Ik+1 be the midpoint of the segment OIk . Intuitively, OIk should be k .
2
Given any segment AB. For each integer i ≥ 0 let ni be the maximum natural
number such that ni .OIi is not greater than AB. The Archimedean principle
guarantees that such ni exists. Define the i-th approximation to the length of
AB as follows:
ni
mi := i
2
One can show that ni+1 = 2ni or ni+1 = 2ni + 1. In particular,
1
mi ≤ mi+1 ≤ mi+1 + i+1 . One can also show that mi is bounded below by
2
m0 and bounded above by m0 + 1. One can also show that, by Archimedean
principle, that ni > 0 for sufficiently large i. This means that the sequence
mi is monotonic and bounded. There is a property of the real numbers R that
guarantees that the limit exists. Define the length as
AB := lim mi .
n→∞
In the special case that AB = OI we see that ni = 2i , and so mi = 1 for all i.
This means that the length, that is the limit is 1.
For now, we skip the proof of the other parts. Corollary 3.2.1 (Triangle Inequality) Given any triangle 4ABC, then AC <
AB + BC.
3.2. MEASURE OF SEGMENTS AND ANGLES
69
Proof. By B.2 and C.1 let D be the point such that A ∗ B ∗ D and BD ∼
= BC.
Then 4CBD is an isosceles triangle, thus ∠DCB ∼
= ∠BDC. Since A ∗ B ∗ D,
then B is interior point of ∠ACD, thus ∠ACD > ∠BCD ∼
= ∠BDC. Hence, apply proposition 3.1.5 on the triangle 4CAD, AC < AD. Now, by the previous
theorem, AD = AB + BD. Then AD = AB + BC. Theorem 3.2.2 (Angle Measure) There is a unique way to assign to each
angle ∠A a positive real number x ∈ R ( called the degree measure of angle
∠A), in such a way that the followings hold:
i. (∠A)◦ is a real number such that 0 < (∠A)◦ ) < 180◦ .
ii. (∠A)◦ = 90◦ if and only if ∠A is a right angle.
iii. (∠A)◦ = (∠B)◦ if and only if ∠A ∼
= ∠B.
iv.
If D is an interior point to ∠ABC, then (∠ABC)◦ = (∠ABD)◦ +
(∠CBD)◦ .
v. For every real number x between 0 and 180, there exists an angle ∠A such
that (∠A)◦ = x◦ .
vi. If ∠B is supplementary angle to ∠A, then (∠A)◦ + (∠B)◦ = 180◦ .
vii. (∠A)◦ < (∠B)◦ if and only if ∠A < ∠B.
Definition 3.2.1 An angle ∠A is called an acute if (∠A)◦ < 90◦ .
An angle ∠A is called obtuse if (∠A)◦ > 90◦ .
Corollary 3.2.2 In any triangle, the sum of degree measures of any two angles
of the triangle is less than 180◦ .
Proof. Consider any triangle 4ABC. We show that (∠A)◦ + (∠B)◦ < 180◦ .
By B.2, let D be the point such that A ∗ B ∗ D. Then by Exterior angle theorem
70
CHAPTER 3. NEUTRAL GEOMETRY
∠CAB < ∠CBD. Since ∠CBD and ∠CBA are supplementary angles then by
previous theorem then
∠DBC)◦ + (∠CBA)◦ = 180◦
180◦ = ∠DBC)◦ + (∠CBA)◦ > (∠CAB)◦ + (∠CBA)◦ .
Theorem 3.2.3 (Sacchari-Legendre Theorem)
The sum of degree measures of the three angles of any triangle is less than or
equal to 180◦ .
Proof. We will use contradiction. Assume that, there is a triangle 4ABC with
angle sum 180◦ + ρ◦ . Where ρ is a positive real number.
←→
Let D be the midpoint of BC. By C.1, take E on AD so that A ∗ D ∗ E and
AD ∼
= 4CDE. See the following figure.
= DE. Then by SAS, 4BDA ∼
Then, the two triangles 4AEC and 4ABC have the same angle sum. Since
angle sum of 4 AEC = (∠DAC)◦ + (∠BCA)◦ + (∠DCE)◦ + (∠AEC)◦
= (∠DAC)◦ + (∠BCA)◦ + (∠ABC)◦ + (∠BAD)◦
= angle sum of 4 ABC
At least one of the angles ∠CAE or ∠BAE has at most half the number of
degrees of ∠BAC, since (∠BAE)◦ + (∠DAC)◦ = (∠BAC)◦ , and if the angles
∼ ∠DAC, then each is equal to 1 (∠BAC)◦ . If they are unequal, then
∠BAE =
2
1
one of them with degree less than (∠BAC)◦ .
2
∼ ∠BAD, then either
Since ∠CEA =
1
1
(∠CAE)◦ ≤ (∠BAC)◦ or (∠CEA)◦ ≤ (∠BAC)◦ .
2
2
1
If (∠CAE)◦ ≤ (∠BAC)◦ , then apply the same process to 4AEC, by taking
2
the midpoint of EC, name it G. Let F be the unique point on the opposite ray
−→
of GA such that F G ∼
= AG.
Thus, we obtain a triangle 4AF C with same angle sum 180◦ + ρ◦ , and in which
at least one of the angles ∠CAF , ∠CF A with degree measure does not exceed
3.2. MEASURE OF SEGMENTS AND ANGLES
71
1
(∠CAE)◦ . If that one is ∠CAF then
2
(∠CAF )◦ ≤
1
(∠CAE)◦
2
Hence,
1
(∠BAC)◦ .
22
If this process is applied n times, we obtain a triangle, say 4P QR with angle
sum 180◦ + ρ◦ and containing an angle, say ∠R such that
(∠CAF )◦ ≤
(∠R)◦ ≤
1
(∠BAC)◦
2n
Thus, right hand side of the inequality can be made arbitrary small by taking
n sufficiently large. In particular, it can be made less that ρ.
Thus, if this is done then
180◦ + ρ◦ = (∠P )◦ + (∠Q)◦ + (∠R)◦
< (∠P )◦ + (∠Q)◦ + ρ◦
Hence, 180◦ < (∠P )◦ + (∠Q)◦ . This contradicts that the sum of any two angles
of any triangle must be less than 180◦ . Corollary 3.2.3 The sum of degree measures of two remote interior angles in
a triangle is less than or equal to degree measure of their exterior angle.
Proof. Given any triangle 4ABC, by B.2, let D be a point such that A ∗ C ∗
D. Then (∠ABC)◦ + (∠BCA)◦ + (∠CAB)◦ ≤ 180◦ , and hence, (∠ABC)◦ +
(∠BAC)◦ ≤ 180◦ − (∠BCA)◦ = (∠BCD)◦ . Definition 3.2.2 A quadrilateral ABCD consists of four points, say A, B, C
and D, no three of which are collinear, and four segments AB, BC, CD, and
DA, such that the segments have either no point in common, or only have an
endpoint in common. Then the points A, B, C and D are called vertices of the
quadrilateral, and the segments are called the sides of the quadrilateral.
A diagonal of the quadrilateral is the segment connecting non-consecutive vertices. Thus, the quadrilateral ABCD has two diagonals, AC and BD.
A quadrilateral ABCD is called convex if it has a pair of opposite sides, say
←→
DC and AB such that D, C lie in the same side of the line AB, and A, B lie
←→
in the same side of the line CD.
Lemma 3.2.1 If ABCD is a convex quadrilateral, where A, B lie on the same
←→
←→
side of CD, and C, D lie on the same side of AB. Then also
←→
i. B, C lie on same side of AD.
←→
ii. A, D lie on same side of BC.
72
CHAPTER 3. NEUTRAL GEOMETRY
iii. B is interior to the angle ∠CDA, D is interior to the angle ∠CBA, A is
interior to the angle ∠BCD, and C is interior to the angle ∠DAB
Proof. Let ABCD is a convex quadrilateral, where A, B are on the same side
←→
←→
of CD, and C, D are on the same side of AB.
←→
i. Use contradiction, assume that, B, C are in opposite sides of AD, then BC
←→
intersects AD at a point, say E, then either E = A, E = D, E ∗ A ∗ D, A ∗ D ∗ E,
or A ∗ E ∗ D.
If E = A, then B, C, and A are collinear, this contradicts that no three of the
vertices are collinear.
If E = D, we get also a contradiction as in the previous case.
If A ∗ E ∗ D, then we get a contradiction to that the sides intersects only at
endpoints.
←→
If E ∗ D ∗ A, then consider the triangle 4ABE. Then the line CD intersects BE
←→
or AB. If CD intersects BE, we get three collinear vertices of the quadrilateral,
←→
which is a contradiction. If CD intersects AB, this contradicts the fact that
←→
A, B lie on same side of CD.
If E ∗ A ∗ D, we get also a contradiction as in the previous case.
ii. Exercise.
←→
iii. By hypothesis, AB is on the same side of the line DC, and from i. BC is
←→
on same side of DA. Then B is interior to the angle ∠CDA. Now, we will prove the following consequence of Sacchari-Legendre Theorem.
Theorem 3.2.4 The sum of degree measures of a convex quadrilateral is less
than or equal to 360◦ .
Proof. Consider the convex quadrilateral ABCD. See the figure. Now, the
diagonal AC divides the quadrilateral into two triangles; 4ABC and 4ADC.
3.2. MEASURE OF SEGMENTS AND ANGLES
73
By Sacchari-Legendre theorem, each one of the triangles with angle sum less
than or equal to 180◦ . That is
(∠ABC)◦ + (∠BAC)◦ + (∠ACB)◦ ≤ 180◦
(∠ADC)◦ + (∠DAC)◦ + (∠ACD)◦ ≤ 180◦
By previous proposition, A is interior point of ∠DCB, and C is interior point
of ∠DAB, then by angle addition we get:
(∠DAC)◦ + (∠BAC)◦ = (∠BAD)◦
(∠ACB)◦ + (∠ACD)◦ = (∠BCD)◦
Hence, we get the conclusion. Definition 3.2.3 A quadrilateral ABCD is called Sacchari-Quadrilateral
if ∠ABC and ∠BAD are right angles, and AD ∼
= BC. The side AB is called
the base. And the side DC is called summit. The angles ∠A and ∠B are
called base angles. The angles ∠C and ∠D are called summit angles.
Proposition 3.2.1 The diagonals of Sacchari quadrilateral are congruent.
Proof. Given a Sacchari quadrilateral ABCD, with ∠A and ∠B are right
angles, and BC ∼
= AD. The two triangles 4CBA and 4DAB are congruent
by SAS. Hence the diagonals AC and DB are congruent. .
Proposition 3.2.2 The summit angles of a Sacchari quadrilateral are congruent.
Proof. Given a Sacchari quadrilateral ABCD, with ∠A and ∠B are right
angles, and BC ∼
= AD. Consider the two triangles 4DCB and 4CDA, they
are congruent by previous proposition and SSS. Definition 3.2.4 We define the defect of a triangle 4ABC, to be 180◦ −
[(∠A)◦ + (∠B)◦ + (∠C)◦ ]. It will be denoted by δ(4ABC).
Theorem 3.2.5 Let 4ABC be any triangle and D any point between A and
B. Then δ(4ABC) = δ(4ADC) + δ(4DBC).
Proof. Let 4ABC be any triangle and D any point between A and B. Since
−−→
−→
−−→
CD lies between CA and CB. Then by previous theorem,
(∠ACD)◦ + (∠DCB)◦ = (∠ACB)◦ .
Since ∠CDA and ∠CDB are supplementary angles, then
(∠ADC)◦ + (∠CDB)◦ = 180◦ .
74
CHAPTER 3. NEUTRAL GEOMETRY
Then
δ(4ADC) + δ(4DBC) = 180◦ − [(∠ACD)◦ + (∠CAD)◦ + (∠ADC)◦ ] + 180◦
− [(∠CDB)◦ + (∠DCB)◦ + (∠CBD)◦ ]
= 360◦ − [(∠CAD)◦ + (∠DBC)◦
+ [(∠ACD)◦ + (∠DCB)◦ ] + [(∠CDB)◦ + (∠CDA)◦ ]]
= 360◦ − (∠CAD)◦ + (∠DBC)◦ + (∠ACB)◦ + 180◦
= 180◦ − δ(4ABC).
Corollary 3.2.4 Let 4ABC be any triangle and D any point between A and
B. Then angle sum of 4ABC is 180◦ if and only if the angle sums of both
4ACD and 4DCB is 180◦ .
Proof. Exercise. 3.3
Equivalence of Hilbert’s Parallel Axiom
First, we will prove the equivalence of Euclid’s fifth postulate and Hilbert’s
parallel axiom. Note that, we are not proving either or both of the axiom, we
are only proving that we can prove one if we first assume the other.
Theorem 3.3.1 Euclid’s fifth postulate is equivalent to Hilbert’s parallel axiom.
Proof. (=⇒) First, we assume that Euclid’s fifth postulate. Let l be any line
and P be any point not on l. By proposition 2.3.7, let t be the perpendicular
from P to l, which intersects l at point Q, and m be the perpendicular through
P to t. Then by theorem 3.1.1, m||l. Let n be another line through P . We
have to show that n must intersect l. Since n 6= m, n and t form an acute
angle. Let θ denote such an angle. Then we have two lines l and n cut by the
transversal t, and the the sum of two angles on one side of t is θ◦ + 90◦ < 180◦ .
Euclid’s fifth postulate implies that n meets l, and therefore n is not parallel to l.
(⇐=) Let us assume Hilbert’s parallel axiom. Let l and m denote a pair of
lines cut by a transversal t, where A is the common point of t and l, and point
B is the common point of m and t. By B.2, let C and D on l such that C ∗A∗D,
and E and F on m such that E ∗ B ∗ F , and C, E are on same side of t. The
assumption of Euclid’s fifth postulate is that
(∠ABE)◦ + (∠CAB)◦ < 180◦
(∠ABE)◦ < 180◦ − (∠CAB)◦
We have (∠CAB)◦ + (∠DAB)◦ = 180◦ , then
(∠ABE)◦ < 180◦ − (∠CAB)◦ = (∠DAB)◦
3.3. EQUIVALENCE OF HILBERT’S PARALLEL AXIOM
75
−→
−−→
−−→
Then by C.4, there is a unique ray AG between AD and AB, such that ∠BAG ∼
=
←→
←→
∠EBA. Then by alternate angle theorem AG||m. Since l 6= AG, Hilbert’s
parallel axiom implies that m intersects l.
To finish the proof, we must show that m intersects l on the same side of t as
C. If not, that is they meet at the point, say H, on the opposite side of t from
C. Then ∠EBA is an exterior angle of the triangle 4AHB. Thus by exterior
angle theorem, ∠EAB > ∠DAB, which is a contradiction. Proposition 3.3.1 Hilbert’s parallel axiom is equivalent to that, if a line intersects one of two parallel lines, then it also intersects the other.
Proof. (=⇒) Assume Hilbert’s parallel axiom. Let m, n be two parallel lines,
and the line l intersects m at a point say, P . Use contradiction, assume l does
not intersects n. Then l||n. Hence m and n are two lines through P that are
both parallel to l. This contradicts Hilbert’s parallel axiom.
(⇐=) Let l be any line and P be any point not on l. By proposition 2.3.7,
let t be the perpendicular from P to l, and m be the perpendicular through P
to t. Then by theorem 3.1.1, m||l. Let n be another line through P . Since n
intersects m, and l||m, then by assumption, n must intersects l. Proposition 3.3.2 Hilbert’s parallel axiom is equivalent to the converse of alternate interior angle theorem.
Proof. (=⇒) Assume Hilbert’s parallel axiom. Given two parallel lines m and
l, and t be a transversal to l and m, meets l at P , and meets m at Q. By
B.2, let A and B be on m such that A ∗ Q ∗ B. Let C on l on the same side
of t as A. We want to show that, ∠BQP ∼
= ∠CP Q. If not, without loss of
−−→
−−→
generality, let ∠BQP > ∠CP Q. Then there is a unique ray QX between QB
−−→
and QP such that, ∠XQP ∼
= ∠QP C. Then by alternate interior angle theo←→
rem, QX is another line parallel to l, which contradicts Hilbert’s parallel axiom.
(⇐=) Assume converse of alternate interior angle theorem. Let l be any line and
P be any point not on l. By proposition 2.3.7, let t be the perpendicular from
P to l and meets l at Q. Let m be the perpendicular through P to t. Then by
theorem 3.1.1, m||l. We use contradiction, let n be another line through P that
is parallel to l. Let A on n and B on l such that A, B are on the opposite sides
of t. Then our assumption implies that ∠BQP ∼
= ∠AP Q and hence ∠AP Q is
a right angle. Therefore line n is perpendicular to t, implying that m = n, by
the uniqueness of the perpendicular to t through P . Proposition 3.3.3 Hilbert’s parallel axiom is equivalent to that, if t is a transversal to l and m, l||m, and t is perpendicular to l, then t is perpendicular to m.
Proof. (⇐=) Assume that the second statement is true. Let l be any line and
P be any point not on l. By proposition 2.3.7, let t be the perpendicular from
P to l, and m be the perpendicular through P to t. Then by theorem 3.1.1,
m||l. We will use contradiction, let n be another line through P that is parallel
76
CHAPTER 3. NEUTRAL GEOMETRY
to l. Apply our assumption on n||l and t ⊥ l. Then t ⊥ n. By uniqueness of
the perpendicular, we have m = n.
(=⇒) Assume Hilbert’s parallel axiom. Let t is a transversal to l and m, l||m,
and t is perpendicular to l. By previous proposition converse of alternate interior angle theorem is true. Since m||l then the alternate interior angle formed
by l and m cut by the transversal t are congruent. Therefore, t ⊥ m. Proposition 3.3.4 Hilbert’s parallel axiom is equivalent to that, if k||l, m ⊥ k
and n ⊥ l then either m = n or m||n.
Proof. (=⇒) Assume Hilbert’s parallel axiom. Let k||l, m ⊥ k and n ⊥ l.
Since m intersects k, then by proposition 3.3.1 m must intersects l. By previous
proposition m ⊥ l. Thus both m and n are perpendicular to l. Then either
m = n or m 6= n.
If m = n. Then we are done.
If m 6= n. Then either m||n or not parallel.
If n and m are not parallel, then they have a common point, say P . This is a
contradiction to the uniqueness of perpendicular form a point P to the line l.
Thus, if m 6= n, then m||n.
(⇐=) Assume the second statement. Let l be any line and P be any point
not on l. By proposition 2.3.7, let t be the perpendicular from P to l, and m
be the perpendicular through P to t. Then by theorem 3.1.1, m||l. We will use
contradiction, let n be another line through P that is parallel to l. Let k be the
perpendicular through P to n. Then by our assumption either k = t or k||t.
Since P is a common point to k and t. Then k = t, Since t ⊥ m and t = k ⊥ n,
by uniqueness of perpendicular n = m. Definition 3.3.1 A quadrilateral whose four angles are right is called a rectangle.
Theorem 3.3.2 If a triangle with angle sum 180◦ exists, then every triangle
has angle sum 180◦ .
The proof of this theorem will be given in 5 steps:
i. If a triangle exists whose angle sum is 180◦ , then there is a right triangle
with angle sum 180◦ .
ii. From a right triangle with angle sum 180◦ , we can construct a rectangle.
iii. If a rectangle exists, then we can construct an arbitrarily large rectangle.
iv. If we have arbitrary large rectangle, then every right triangle has an angle
sum equal to 180◦ .
3.3. EQUIVALENCE OF HILBERT’S PARALLEL AXIOM
77
v. If every right triangle has angle sum 180◦ , then every triangle has angle
sum 180◦ .
Proof. i. Given a triangle 4ABC with angle sum 180◦ .
First, we construct a right triangle with angle sum 180◦ . If 4ABC is right,
then we are done. If not, Then at least two angles of this triangle are acute,
since the angle sum of two angles in a triangle must be less than 180◦ , assume
∠A and ∠B are acute. By proposition 2.3.7, drop a perpendicular from C to
←→
the line l := AB, which intersects l at D. Then either D ∗ A ∗ B, A ∗ B ∗ D, or
A ∗ D ∗ B.
If D ∗ A ∗ B, then by exterior angle theorem, ∠CAB is exterior to the triangle
4DCA, and hence ∠CAB > ∠CDA, which is a contradiction, since ∠CAB is
acute.
If A ∗ B ∗ D, we get also a contradiction. It is an exercise.
If A ∗ D ∗ B, then both 4ACD and 4DCB are right angles. By previous corollary both have angle sum 180◦ . Thus, we get a right triangle with angle sum
180◦ .
78
CHAPTER 3. NEUTRAL GEOMETRY
ii. From a right triangle with angle sum 180◦ , we construct a rectangle.
Let 4CDB be a right triangle at D and with angle sum 180◦ . See the following
figure.
−−→
←→
By C.4, there is a unique ray CX on the opposite side of BC from the point
−−→
D, such that ∠DBC ∼
= ∠XCB. By C.1, there is a unique point E on CX such
that CE ∼
= DB. Then by SAS, 4CDB ∼
= 4BEC. Hence, ∠CEB is right.
Also, since (∠DBC)◦ + (∠BCD)◦ = 90◦ by the hypothesis that the triangle has
angle sum 180◦ . We obtain by substitution and (∠ECB)◦ + (∠BCD)◦ = 90◦
and (∠DBC)◦ + (∠EBC)◦ = 90◦ .
Claim, B is an interior point of ∠ECD.
←→ ←→
By alternate interior angle theorem, DB||CE. Then D and B are on the same
←→
←→
side of CE, also C and E are on the same side of the line DB.
Similarly, we can show that C is an interior point of ∠EBD. Then by the
theorem on measure of angles, we conclude that (∠ECD)◦ = 90◦ = (∠EBD)◦ .
Thus, DBEC is a rectangle. iii. Given one rectangle, we want to construct arbitrarily large rectangle.
Given a rectangle BCDE. Construct a rectangle ACGH such that AC >
BC and GC > DC. This can be done using Archimedes’ axiom. We simply lay
off enough copies of the rectangle to get the desired result.
iv. Given arbitrary large rectangle, then all right triangles have defect zero.
To prove this. From previous step, we have arbitrary large rectangle, then any
right triangles can embedded in that rectangle. Given any right triangle 4RQS,
which is right at angle ∠Q, then it can be embedded in the rectangle ACGH,
3.3. EQUIVALENCE OF HILBERT’S PARALLEL AXIOM
79
see the following figure.
Since AC > QS, there is a point P , A ∗ P ∗ C, such that QS ∼
= P C. Also, there
is a point I with C ∗ I ∗ G such that CI ∼
= RQ. Then by SAS, 4QSR ∼
= 4CP I.
Since ACGH is a rectangle, then the right triangle 4AGC has defect zero.
Since A∗P ∗C, then the triangle 4GP C has also defect zero. And C ∗I ∗G, then
the triangle 4P CI has defect zero. The last triangle is congruent to 4QSR,
then it has defect zero, that is its angle sum is 180◦ .
At this point, we have proved that: If a triangle with defect zero exists, then
every right triangle has defect zero.
v. We prove now that, if all right triangles have defect zero, then every triangle with defect zero. Given any triangle 4KLM . If 4KLM is right, then
we are done. If not, Then at least two angles in this triangle are acute, since
the angle sum of two angles in a triangle must be less than 180◦ , assume ∠K
and ∠L are acute. By proposition 2.3.7, drop a perpendicular from M to the
←→
line l := KL, which intersects l at N . Then either N ∗ K ∗ L, K ∗ L ∗ N , or
K ∗ N ∗ L.
If N ∗ K ∗ L, then by exterior angle theorem, ∠M KL is exterior to the triangle
4N M K, and hence ∠M KL > ∠M N K, which is a contradiction, since ∠M KL
is acute.
If K ∗ L ∗ N , we get also a contradiction. It is an exercise.
If K ∗ N ∗ L, then both 4KM N and 4N M L are right angles. By assumption
both have angle sum 180◦ . Then by previous corollary, 4KLM have angle sum
180◦ 80
CHAPTER 3. NEUTRAL GEOMETRY
The contrapositive of the previous theorem, will give us the following corollary.
Corollary 3.3.1 If there is a triangle with positive defect, then all triangles
have positive defect.
Theorem 3.3.3 Hilbert’s parallel axiom is equivalent to that the angle sum of
every triangle is 180◦ .
Proof. (=⇒) Assume that Hilbert’s parallel axiom hold. Given any triangle
−−→
4ABC. Let AD be the unique ray such that ∠DAC ∼
= ∠ACB, by C.4. Then
←→
←→
by the alternate angle theorem, the line AD is parallel to the line BC. By B.2,
←→
let G be a point on AD such that G ∗ A ∗ D. By measure of angle theorem,
(∠GAB)◦ + (∠BAC)◦ + (∠CAD)◦ = 180◦ .
By the converse of alternate angle theorem, ∠GAB ∼
= ∠ABC. Hence, (∠ABC)◦ +
◦
◦
◦
(∠BAC) + (∠ACB) = 180 .
(⇐=) Assume that the angle sum of every triangle is 180◦ . Given any line
l and point P not on l. By proposition 2.3.7, let t be the perpendicular from P
to l, and meets l at point Q. Let m be the perpendicular through P to t. Then
by theorem 3.1.1, m||l. We will use contradiction, let n be another line through
−−→
−→
P . Then a ray of n lies between P Q and a ray of m emanating from P , say P A.
Let Y be a point on that ray of n. See the figure.
3.4. EXERCISES:
81
Drop a perpendicular from Y to on m at point, say X. Drop a perpendicu←
→
lar from Y on t at a point, say S. Since SY is parallel to m, then S, Y , and Q
are on the same side of m. The quadrilateral SY XP has three right angles.
Since every triangle has angle sum 180◦ , then each of the two triangles 4P XY
and 4P Y S has angle sum 180◦ . Then the quadrilateral SY XP has angle
sum 360◦ . Hence, the angle ∠XY S is also right angle. Thus, SY XP is a
rectangle. Then P S ∼
= XY , see by exercise 1.
By C.1 and exercise 12, Y can be chosen such that XY > P Q. Since P Q <
XY ∼
= P S, then P ∗ Q ∗ S.
Then Y and S are on the same side of l. Otherwise, if Y and S are on opposite
←
→
side of l, then Y S intersects l at some point, this contradicts the fact that SY
is parallel to l, since ∠P SY and ∠Q both are right and hence congruent, then
use alternate interior angle theorem.
Since P and S are in opposite sides of l, then P and Y are in opposite sides of
l. Thus, P Y intersects l, and hence n intersects l. 3.4
Exercises:
←→ ←→
←→ ←→
[1] A quadrilateral ABCD is called parallelogram if AB||CD and BC||AD.
i. Show that every parallelogram is convex.
ii. In Euclidean geometry:
a. Show that the opposite sides of the parallelogram are congruent.
b. Find a condition on a parallelogram that ensures the congruence
of its diagonals.
c. Show that the diagonals bisect each other.
[2] A rhombus is a parallelogram with all sides are congruent. Show that,
the diagonals of a rhombus are perpendicular.
[3] In a Sacchari quadrilateral.
i. Show that the segment joining the midpoints of the base and summit
is perpendicular to the base and the summit.
ii. Show that Sacchari quadrilateral is a parallelogram.
iii. Show that Sacchari quadrilateral is convex.
iv. Show that the summit angles of a Sacchari quadrilateral are either
right or acute.
[4] In Euclidean geometry, prove that, if two triangles are similar then their
corresponding sides are proportional.
82
CHAPTER 3. NEUTRAL GEOMETRY
[5] Let γ be a circle with centre O, and let A, and B be two points on the
circle γ. The segment AB is called a chord of γ.
←−→
i. Let M be the midpoint of AB, and O 6= M . Show that OM is perpendicular to AB.
ii. Show that, the perpendicular bisector of any chord CD of γ pass
through the centre of γ.
[6] In Euclidean Geometry:
i. Prove that an angle inscribed a semicircle is a right angle.
ii. Let γ be a circle with radius r and centre O. Then a line t is tangent
to γ at the point P if and if OP ⊥ t at P .
iii. If AB is a chord of a circle γ, and C is the intersection of the tangent
lines to γ at A and B. Then prove that 4ABC is an isosceles triangle.
iv. Given a circle γ and a point P outside γ, then show that there is two
tangent lines to γ through P .
←→ ←→
[7] Given A ∗ B ∗ C and DC ⊥ AC. Show that AD > BD > CD.
[8] In Euclidean geometry: Show that the bisectors of the angles of any triangle
all meet at a single point.
[9] In Euclidean geometry: Show that, any three distinct non-collinear points
determine a unique circle.
[10] In Euclidean geometry, prove that the shortest segment from a point P
to a line l is perpendicular from P to l.
[11] If l k m and A and B lie on opposite side of m from l. Show that, A and
B lie on the same side of m.
[12] Prove that: Given any segment AB and any side of a cute angle, then
there is a point Y on the given side of the angle such that if X is the foot of
the perpendicular from Y to the other side of the angle, then XY > AB.
This statement is known as Aristotele’s axiom.
Chapter 4
Hyperbolic Geometry
Hyperbolic geometry is often called Bolyai-Lobachevskiian geometry after two
of its discoverers Janos Bolyai and Nikolai Ivanovich Lobachevskii. Bolyai first
announced his discoveries in a 26 page appendix to a book by his father, the
Tentamen, in 1831. Another of the great mathematicians who seems to have
preceded Bolyai in his work is Karl Friedreich Gauss. He seems to have done
some work in the area dating from 1792, but never published it. The first
to publish a complete account of non-Euclidean geometry was Lobachevskii in
1829. It was first published in Russian and was not widely read. In 1840 he
published a treatise in German.
Hyperbolic geometry is, by definition, the geometry obtained by, assuming the axioms of neutral geometry and the negation of Hilbert’s parallel axiom.
The consistency of hyperbolic geometry was demonstrated by Beltrami, Klein,
and Poincare in the late 1800’s and early 1900’s. They created models of hyperbolic geometry inside Euclidean geometry, with strange definitions of points,
lines, circles, and angles. in each of their models, they showed that Euclid’s first
four postulates were true and that the hyperbolic postulate was true as well.
Since each model was created within Euclidean geometry, if hyperbolic geometry
had an internal contradictory statement, then that statement, when translated
into its Euclidean environment, would be an internal contradiction in Euclidean
geometry. Thus, if one believed that Euclidean geometry was consistent, then
hyperbolic geometry was equally as consistent.
4.1
Hyperbolic Axiom and some basic Consequences
In neutral geometry we have that for every line l and every point P not lying
on l, there exists at least one line through P that is parallel to l. Thus, in
Hyperbolic geometry, we assume the axioms of neutral geometry, and the nega83
84
CHAPTER 4. HYPERBOLIC GEOMETRY
tion of the Euclidean Parallel axiom, which will be called Hyperbolic Parallel
Axiom.
Hyperbolic parallel axiom: There is a line l and a point P not on l such
that at least two distinct lines parallel to l pass through P .
The following proposition is the first important consequence of hyperbolic axiom.
Proposition 4.1.1 In hyperbolic geometry, rectangles do not exists.
Proof. Use contradiction. Assume that there exists a rectangle. Then by the
proof of theorem 3.3.2, every triangle has angle sum 180◦ . Then by theorem
3.3.3, Hilbert’s parallel axiom hold, which is a contradiction. Theorem 4.1.1 (Universal Hyperbolic Theorem)
In hyperbolic geometry, for every line l and every point P not on l, there exists
at least two distinct lines through P and parallel to l.
Proof. Given any line l and any point P not on l. By proposition 2.3.7, let t
be the perpendicular from P to l at the point Q, and m be the perpendicular
through P to t. Then by alternate interior angle theorem, m||l. See the figure.
Let R be another point on l, by I.2. Let n be the perpendicular to l at R. Let
4.1. HYPERBOLIC AXIOM AND SOME BASIC CONSEQUENCES
85
S denote the foot of the perpendicular dropped from P to n at S. Then by
←
→
←
→
alternate interior angle theorem, P S||l. We claim that m 6= P S.
If not, S ∈ m. Then P SRQ is a rectangle, which contradicts that no rectangles
exists in hyperbolic geometry. Corollary 4.1.1 In hyperbolic geometry, for every line l and every point P not
on l, there exist infinitely many lines through P and parallel to l.
Proof. Note that the proof of the last theorem implies that to each R ∈ l,
R 6= Q, there corresponds a line through P that is parallel to l. Also, note that
different points on l give rise to different parallel lines. Theorem 4.1.2 In hyperbolic geometry, all triangles have angle sum less that
180◦ , that is, all triangles have a positive defect.
Proof. Use theorem 3.3.2 and proposition 4.1.1. As a consequence, we have:
Corollary 4.1.2 In hyperbolic geometry, all convex quadrilaterals have angle
sum less than 360◦ .
Proof. Exercise.
Definition 4.1.1 Two triangles are called similar if there is a one-to-one correspondence between vertices such that the corresponding angles are congruent.
Theorem 4.1.3 (AAA) In hyperbolic geometry, if two triangles are similar,
then they are congruent.
Proof. Given any two similar triangles 4ABC and 4DEF . If two corresponding sides are congruent then by ASA, the two triangles are congruent. If
no corresponding sides are congruent. Then consider the triples, (AB, BC, AC)
and (DE, EF, DE). One of the triples must contain at least two segments that
are larger than the two corresponding sides of the other triple. Say, AB > DE,
and AC > DF . Then there exists I with A ∗ I ∗ B such that AI ∼
= DE,
and there is a point J with A ∗ J ∗ C such that AJ ∼
= DF . Then by SAS,
4AIJ ∼
= 4DEF . Hence corresponding angles are congruent; ∠AIJ ∼
= ∠DEF ,
and ∠AJI ∼
= ∠DF E. By the hypothesis that 4ABC is similar to 4DEF ,
←→ ←
→
we have ∠AIJ ∼
= ∠ABC, and ∠AJI ∼
= ∠ACB. This implies that, BC||IJ by
alternate interior angle theorem and that vertical angles are congruent. See the
figure.
86
CHAPTER 4. HYPERBOLIC GEOMETRY
Thus the quadrilateral IJCB is convex. Also,
(∠B)◦ + (∠BIJ)◦ = 180◦ = (∠C)◦ + (∠CJI)◦ .
Then it follows, that IJCB has angle sum 360◦ . This contradicts the previous
corollary. Note that, in hyperbolic geometry it is impossible to magnify or shrink a triangle
without distortion. In hyperbolic geometry a segment can be determined with
the aid of an angle. For example, an angle in an equilateral triangle determines
the length of the side uniquely.
4.2
Perpendicular Lines in Hyperbolic Geometry
Definition 4.2.1 i. If l is any line and P ∈
/ l, drop a perpendicular from P to
l with foot P 0 , then the distance from P to l, denoted by d(P, l) is defined to be
the length of the segment P P 0 . That is d(P, l) := P P 0 . By theorem on measure
of segments, 0 < P P 0 , and hence d(P, l) > 0
If P ∈ l, then d(P, l) := 0
ii. Given any two lines l and m, and let P and Q lie on l. We say that P and
Q are equidistant from m, if d(P, m) = d(Q, m), that is, if the perpendiculars
from P to m at the point P 0 , and from Q to m at Q0 , such that P P 0 ∼
= QQ0 .
In Euclidean geometry, if two lines l and m are parallel, then all points on l are
equidistant from m. In other words, parallel lines are a fixed distance apart.
4.2. PERPENDICULAR LINES IN HYPERBOLIC GEOMETRY
87
In hyperbolic geometry, if l and m are any two lines, then there are at most two
points on l that are equidistant from m.
Theorem 4.2.1 In hyperbolic geometry, given any two parallel lines l and m,
then any set of points on l equidistant from m has at most two points on it.
Proof. Given any two parallel lines l and m, we will use contradiction. Assume that there are three points P , Q and R on l that are equidistant from m.
Drop perpendiculars from P , Q, and R to m with P 0 , Q0 , and R0 be the feet of
perpendiculars dropped from P , Q, and R to m respectively. See the figure.
Then the quadrilaterals P 0 Q0 QP , Q0 R0 RQ, and P 0 R0 RP are Sacchari
quadrilaterals (the base angles are right angles and the sides are congruent.
Thus ∠P 0 P Q ∼
= ∠R0 RQ , ∠P 0 P Q ∼
= ∠R0 RP . By C.5,
= ∠Q0 QP , ∠Q0 QR ∼
0
it follows that the supplementary angles ∠Q QP and ∠Q0 QR are congruent,
hence, they are right angles. Therefore, these Sacchari quadrilateral are rectangles. But rectangles do not exist in hyperbolic geometry. Thus P, Q, and R
cannot be equidistant. The theorem states that at most two points at a time on l can be equidistant from m. It allows the possibility that there are pairs of points (P, Q),
(R, S), (T, U ), ... on l such that each pair is equidistant from m.
However, the last theorem allows another possibility, that there is no pair of
points on l equidistant from m.
Lemma 4.2.1 In neutral geometry.
i. The segment joining the midpoints of the base and the summit of a Sacchari
quadrilateral is perpendicular to both the base and the summit.
ii. The segment joining the midpoints of the base and the summit of a Sacchari
quadrilateral is shorter than or equal the sides of the Sacchari quadrilateral.
88
CHAPTER 4. HYPERBOLIC GEOMETRY
Proof. Consider the Sacchari quadrilateral ABCD. See the following figure.
∼ BC. We have
Where the base right angles at the point A and B and DA =
shown before that, ∠D ∼
= ∠C. Let M 0 be the midpoint of the base, and M be
the midpoint of the summit.
←−−→
i. We want to show that M M 0 perpendicular to both the summit and the base.
First, by SAS the triangle 4DAM ∼
= 4CBM , and then by SSS, 4AM 0 M ∼
=
4BM 0 M . Then the supplementary angles ∠AM 0 M and ∠BM 0 M are
congruent, and hence right.
Now, since ∠AM M 0 ∼
= ∠BM M 0 , and ∠CM B ∼
= ∠DM A, then by angle
0
addition, ∠M 0 M D ∼
∠M
M
C,
and
since
they
are
supplementary angles,
=
then they are right.
∼ DA, use contradiction. Assume that
ii. To show that M M 0 < DA or M M 0 =
M M 0 > DA, then there is a point G such that M 0 G ∼
= DA. The angle
∠DGM 0 is exterior to the triangle 4DM G, and hence is obtuse. But
DGM 0 A is Sacchari quadrilateral, thus ∠GDA is also obtuse. Then the
angle sum of the quadrilateral is greater than 360◦ , which is a contradiction. Theorem 4.2.2 In hyperbolic geometry, if l and m are parallel lines, for which
there exists a pair of points P and Q on l equidistant from m. Then:
i. l and m have a common perpendicular.
ii. This common perpendicular is unique.
iii. The common perpendicular segment is the shortest segment between l and
m.
Proof. Given any two parallel lines l and m. Assume that the points P and
Q on l that are equidistant from m. Drop perpendiculars from P and Q to
4.3. CLASSIFICATION OF PARALLEL LINES
89
m with P 0 and Q0 be the feet of perpendiculars dropped from P and Q to m
respectively. Then P P 0 ∼
= QQ0 .
0
i. Let M and M be the midpoints of l and m respectively. Then by the lemma
←−−→0
M M is the common perpendicular.
←→
ii. Let LL0 be another common perpendicular. Then M M 0 L0 L is a rectangle, thus we get a contradiction.
iii. Let K be any point on l distinct from M , and S be any point on m.
We show that KS > M M 0 . Drop a perpendicular from K to m, say at the
point K 0 . Then either K 0 6= S or K 0 = S. If K 0 = S, then by previous lemma
KS < M M 0 . If K 0 6= S, then KK 0 < KS by proposition 3.1.5. By previous
lemma, KK 0 > M M 0 . Then KS < KK 0 < M M 0 . Theorem 4.2.3 In hyperbolic geometry, if l and m have a common perpendicular M M 0 , where M is on line l, and M 0 on the line m. Then:
i. l and m are parallel.
ii. If P and Q are any two points on l such that M is the midpoint of the
segment P Q, then P and Q are equidistant from the line m.
Proof. Exercise.
4.3
Classification of Parallel Lines
In this section we will see that, in hyperbolic geometry, there are two types
of parallel lines to any given line l, some parallel lines to l have a common
perpendicular with l, and some that have no such common perpendicular.
Theorem 4.3.1 In hyperbolic geometry, given any line l, and any point P ∈
/ l,
let Q be the foot of perpendicular dropped from P to l. Then there are two unique
−→
−→
←→
non-opposite rays P R and P L on opposite sides of the line P Q such that:
−→
−→
i. The rays P L and P R do not intersects l.
−−→
−−→
−→
ii. A ray P X intersects the line l if and only if P X is between and P L and
−→
P R.
iii. ∠QP R ∼
= ∠QP L.
iv. (∠QP R)◦ < 90◦ .
Proof. Given any line l, and any point P ∈
/ l, let Q be the foot of perpendicular
←→
dropped from P to l. Let m be the perpendicular at P on P Q.
−→
First, we prove the existence of the ray P L. Let S be a point on line m to the
90
CHAPTER 4. HYPERBOLIC GEOMETRY
←
→
left of P . Consider the line SQ. See the figure
P
−→
Let
1 := {T ∈ SQ; P T intersects l} ∪ { all points on the opposite ray of
−→
QS}.P
P
P
P
Let 2 be the complement of 1 . Then S ∈ 2 and Q ∈ 1 , thus the subsets
←
→
←
→
of SQ are non-empty,
and their union is all P
points on SQ. Now, we show that
P
P
no point in 1 lies between two points in 2 and vice versa. Let T ∈ 1 .
−→
Then P T intersects l at a point say H. Then any point J on T Q also lies in
P
−→
−−→
−−→
1 . Since T ∗ J ∗ Q, then P J lies between P H and P Q, and hence by Crossbar
−→
theorem,
HQ, and hence intersects l.
P P J intersects
P
Thus, 1 and 2 form a Dedekind cut. By Dedekind’s axiom, there is a unique
←
→
←
→
point,Psay L, on SQPsuch that for P1 and P2 on SQ, P1 ∗ L ∗ P2 if and only if
P1 ∈ 1 and P2 ∈ 2 .
P
P
−→
−→
By definition of 1 and 2 , rays below P L all meet l, and rays above P L do
not meet l.
−→
i. Now, we prove that the ray P L does not intersects l. If not, that is if
−→
P L intersects l at a point, say U . Let V be a point such that Q ∗ U ∗ V . Then
←
→
U and V are on the same side of SQ and the points P and U are in opposite
←
→
←
→
−−→
sides of SQ. Thus P and V are in opposite sides
P of SQ. Thus P V intersects
SQ in a point Y . We have Y ∗ L ∗ Q, then Y ∈ 2 , thus we get a contradiction.
−→
In a similar way we obtain the ray P R.
−−→
−−→
ii. Thus, a ray P X intersects the line l if and only if P X is between and
−→
−→
P L and P R.
4.3. CLASSIFICATION OF PARALLEL LINES
91
iii. Now, we prove that ∠QP R ∼
= ∠QP L. If not, then the two angles are
not congruent, say, ∠QP R > ∠QP L. See the following figure.
−−→
−→
−−→
Then, there is a ray P N between P R and P Q such that ∠N P Q ∼
= ∠LP Q. Then
−−→
−→
P N intersects l at a point, say M . Choose F on P L such that P F ∼
= P M . Then
by SAS, 4P QF ∼
4P
QM
.
Then,
∠P
QF
is
right,
which
is
a
contradiction
to
=
the uniqueness of C.4, unless F lies on the line l, which is also a contradiction,
−→
since P L does not intersect l by i.
−→
−→
iv. Choose I on P L and choose K on P R, such that P K ∼
= P I, the triangle 4P IK is an isosceles triangle, thus ∠P IK ∼
= ∠P KI. See the following
figure.
92
CHAPTER 4. HYPERBOLIC GEOMETRY
In hyperbolic geometry, the angle sum of this triangle is less than 180◦ . Let
α := (∠P IK)◦ = (∠P KI)◦
β := (∠IP Q)◦ = (∠KP Q)◦
Then
180◦ > 2α + 2β
90◦ > α + β > β
Hence (∠LP Q)◦ < 90◦ . Notice that, the previous theorem is proved in neutral geometry, except of the
last part, that (∠QP R)◦ < 90◦ , we used the assumption that the geometry is
hyperbolic.
−→
−→
The ray P R is called right limiting parallel ray, and P L is called left limiting parallel ray.
←
→
From the definition of the left and right limiting parallel rays, the lines P L and
←→
−→
P R are parallel to l. For if opposite ray of P L intersects l say at J, then 4P QJ
has angle sum greater that 180◦ , since the angle ∠JP Q is obtuse because it is
supplementary to acute angle ∠LP Q. And the angle ∠JP Q is right. Then, the
angle sum of the triangle 4P QJ is greater than 180◦ , which is a contradiction.
−→
In similar way, we show that, the opposite ray of P R does not intersect l. Hence
←
→
←→
the lines P L and P R are parallel to l.
In Euclidean geometry, there is a unique line through P and parallel to l,
←→ ←
→
thus m = P R = P L. Hence, the angles ∠QP L and ∠QP R are right angles.
Now, the angles ∠QP R, ∠QP L are called the angles of parallelism of l
at P . The following proposition shows that congruent segments have congruent
angle of parallelism. Which satisfies that, the degree measure of the angles of
parallelism ∠QP R, ∠QP L will be denoted by Π(P Q)◦ .
A natural unit segment OI in hyperbolic geometry is any segment OI
such that Π(OI)◦ = 45◦ .
Proposition 4.3.1 Let l and l0 be two distinct lines, P ∈
/ l and P 0 ∈
/ l0 . Let
←
−
→
←
→
Q ∈ l and Q0 ∈ l0 be such that P Q ⊥ l, P 0 Q0 ⊥ l0 . If P Q ∼
= P 0 Q0 then the angle
of parallelism of l at P is congruent to the angle of parallelism of l0 at P 0 .
Proof. Let l and l0 be two distinct lines, P ∈
/ l and P 0 ∈
/ l0 . Let Q ∈ l and
←−
→
←→
−→
0 0
0 0
0
Q ∈ be such that P Q ⊥ l, P Q ⊥ l . Let P L be the limiting parallel ray of l
−−→
at P , and P 0 L0 be the limiting parallel ray of l0 at P 0 . See the figures.
4.3. CLASSIFICATION OF PARALLEL LINES
93
By C.4 and the theorem on measure of angles, for every real number x ∈
−−→
←→
[0, 180], there is a ray P X on one side of P Q such that x = (∠XP Q)◦ . Define
−−−→
−−→
the set S := {x ∈ [0, 180]; P X intersects l}. Also, there is a ray P 0 X 0 on one
←−
→
−−−→
side of P 0 Q0 such that x = (∠X 0 P 0 Q0 )◦ . Define the set S 0 := {x ∈ [0, 180]; P 0 X 0
intersects l0 }.
−−→
If x ∈ S, let Y denote the point on l where P Y intersects l, and ∠QP Y = x.
Consider the point Y 0 on l0 such that QY = Q0 Y 0 . Then by SAS, 4P QY ∼
=
4P 0 Q0 Y 0 . Then, (∠Y 0 P 0 Q0 )◦ = x, which implies that x ∈ S 0 . That is S ⊆ S 0 .
Similarly, we can show that S 0 ⊆ S. Thus S = S 0 . And hence they have the
same least upper bound, which is the measure of the angle of parallelism. −−→
Proposition 4.3.2 Let P Y be a limiting parallel ray to a line l at the point P .
−→
If P ∗ A ∗ Y , then AY is a limiting parallel ray to l at the point A.
−−→
Proof. Let P Y be a limiting parallel ray to a line l at the point P . Assume
that P ∗ A ∗ Y . See the following figure.
Let A0 and P 0 be the feet of perpendiculars from A and P to l respectively.
−−→
−−→
−→
−−→
It is enough to show that, if AD lies between the rays AY and AA0 , then AD
intersects l.
94
CHAPTER 4. HYPERBOLIC GEOMETRY
−−→
−−→
−→
If AD lies between the rays AY and AA0 , then D is interior point to the angle
←→
∠A0 AY , since also A0 and P 0 are on same side of P Y , then D is an interior point
−−→
−
−
→
−−→
−−→
to the angle ∠P 0 P Y . Thus, P D lies between the rays P Y and P P 0 . Since P Y
−−→
is a limiting parallel ray to the line l at the point P , then P D intersects l. Let
H be the point of intersection. Since A is exterior to the triangle 4HP P 0 and
−−→
the ray AD intersects the side P H of the triangle 4HP P 0 , and then it must
intersects P 0 H or P P 0 .
−−→
For all Q ∈ AD such that A ∗ D ∗ Q, Q is interior to the angle ∠Y AA0 , then Q
←−→
←−→
and Y are in opposite sides of AA0 . But P and Y are in opposite sides of AA0 .
←−→ ←−→
←−→
←−→
Since P P 0 ||AA0 , then all points of P P 0 are in opposite sides of AA0 from the
−−→
←→
point Q. Then AD does not intersect P P 0 . Thus, AD must intersect the side
−
−
→
P 0 H. Therefore, AD intersects l. The following proposition shows that the line that contains a limiting parallel ray to a line l asymptotically approaches l in the direction that contains the
limiting parallel ray and diverges from l in the other direction.
−→
Proposition 4.3.3 Let P L be a limiting parallel ray to a line l through a point
P and let Q be the foot of the perpendicular on l.
−→
i. If A ∈ P L and A0 the foot of the perpendicular from A to l, then AA0 < P Q.
−→
ii. If B in the opposite ray of P L and B 0 the foot of the perpendicular from B
to l, then BB 0 > P Q.
−→
Proof. Let P L be a limiting parallel ray to a line l through a point P and let
Q be the foot of the perpendicular on l.
−→
i. Let A ∈ P L, A 6= P , and A0 the foot of the perpendicular from A to l. See
the following figure.
Let Y be a point such that P ∗ A ∗ Y . Then by previous proposition,
4.3. CLASSIFICATION OF PARALLEL LINES
95
−→
AY is a limiting parallel ray to l at the point A. Then the angle of parallelism ∠A0 AY is acute. Thus, ∠A0 AP is obtuse. Then ∠QP A < ∠A0 AP .
To show that AA0 < P Q, use contradiction, that is either AA0 ∼
= P Q or
AA0 > P Q.
If AA0 ∼
= P Q, then since AP QA0 is Sacchari quadrilateral and hence
∠QP A ∼
= ∠P AA0 . This is a contradiction, since ∠QP A is an angle of
parallelism and hence, it is acute, and the angle ∠P AA0 is supplementary
of an angle of parallelism and hence must be obtuse.
If AA0 > P Q, then there is a unique point B such that A ∗ B ∗ A0 with
BA0 ∼
= P Q. Then, ∠AP Q > ∠BP Q, and the exterior angle ∠A0 BP >
0
∠A AP . Then ∠AP Q > ∠A0 AP which is a contradiction.
−→
ii. If B is in the opposite ray of P L and B 0 the foot of the perpendicular from
B to l. See the figure.
Use contradiction, that is either BB 0 ∼
= P Q or BB 0 < P Q.
0 ∼
0
∼
If BB = P Q, then ∠QP B = ∠B BP , which is a contradiction, since the
first must be obtuse and the second must be acute.
If BB 0 < P Q is an exercise. Lemma 4.3.1 If a line m contains a limiting parallel to a line l, then l also
contains a limiting parallel ray to m.
Proof. Let m be a line that contains a limiting parallel ray to a line l. Let P
be a point on m, drop a perpendicular from P to l, with foot Q. Drop a perpendicular from Q to m at the point T . Assume that m contains the limiting
−→
−→
parallel ray P S. Then either T ∈ P S, or T belongs to the opposite ray.
If T belongs to the opposite ray, since measure of angle of parallelism ∠SP Q is
acute angle, then (∠T P Q)◦ > 90◦ . See the following figure.
96
CHAPTER 4. HYPERBOLIC GEOMETRY
Then 4P T Q has angle sum greater than 180◦ . Which is a contradiction. Thus,
−→
T ∈ P S.
←→
Now, let R ∈ l be on the same side of P Q as T . See the following figure.
−−→
Claim: QR is a limiting parallel ray to m.
−−→
To prove the claim, it is enough to show that any ray, say QW , that is between
−→
−−→
QT and QR, intersects m.
←−→
−−→
Drop a perpendicular from P to the line QW with foot A. Then A ∈ QW ,
otherwise 4P QA has angle sum greater than 180◦ .
In the triangle 4P QA, the side opposite to the angle ∠A is greater than the
side opposite to the angle ∠Q. That is, P Q > P A.
Then there exists a unique point, say B, such that P ∗B ∗Q such that P B ∼
= P A.
←→
And let t be the line through B that is perpendicular to P Q. Consider the
−−→
←→
ray P D on the same side of P Q as A, such that ∠BP D ∼
= ∠AP T . Since
4.3. CLASSIFICATION OF PARALLEL LINES
97
−−→
∠BP D < ∠BP T , we conclude that P D intersects l at a point, say C.
−−→
Then either P D intersects t or does not intersect t.
−−→
←→
If P D does not intersect t, then all points of P D lie on the same side of t as P .
Since t k l, then all points of l are on the same side of t as Q. But, Q and P are
←→
in opposite sides of t, and since C and P both belong to P D, then C, P are on
same side of t, then C and Q are in opposite sides of t, which is a contradiction.
−−→
Then P D intersects t at a point, say E.
−→
Now, let F ∈ P T be such that P F ∼
= P E. Since P E > BP and P B ∼
= PT,
then P F > P T , and hence, P ∗ T ∗ F .
Now, by SAS 4P BE ∼
= 4P AF . Then ∠P AF ∼
= ∠P BE. Then ∠P AF is a
−−→ −→
−−→
right angle, that is AW = AF . Then QW intersects m at F . Lemma 4.3.2 Let l and m be two distinct lines that both contains a limiting
parallel ray to another line n and both converges to n in the same direction, then
l contains a limiting parallel ray to m in the same direction.
Proof. Let l and m be two distinct lines that both contains a limiting parallel
ray to another line n and both converges to n in the same direction.
Case 1: Assume that l and m are in opposite sides of n. Let P be a point on
m. Drop a perpendicular from P to l with foot Q on l. Then n intersects P Q
at a point, say U . See the following figure.
We show that l k m, if not, let A be the point of intersection. Consider the
triangle 4P QA, then by Pasch’s theorem n must intersects either QA, or P A
98
CHAPTER 4. HYPERBOLIC GEOMETRY
which is a contradiction. Thus l k m.
←→
Let R be on m on the side of P Q as the limiting parallel ray on m to n. Consider
−→
−→
−−→
−→
P T that is between P R and P Q. Enough to show that P T intersects l.
∠1 is obtuse since its supplementary angle ∠2 is acute. Drop a perpendicular
←→
from P to n, say at G.If G is on the same side of P Q as R, then ∠1 must be
←→
acute, which is a contradiction. Then G is in opposite side of P Q from R. The
−→
ray P T intersects n at a point say X. Let Y be such that P ∗ X ∗ Y , and V such
−−→
−−→
that U ∗ X ∗ V . Since XV is a limiting parallel ray to l, then XY intersects l,
−→
then P T intersects l.
Case 2: Assume that l and m are on the same side of n. Without loss of
generality assume that l and n are in opposite sides of m. Let D be a point
on l. Let t be the limiting parallel line through D to the line m in the same
direction of parallelism. See the following figure.
Now, both t and n are limiting parallel lines to m and in opposite sides of
m. Then by case 1, t and n are limiting parallel lines. By assumption l is
limiting parallel ray to n, hence at D, n has two parallel rays t and l, which is a
contradiction unless t = l. Thus, t = l, since given one direction of parallelism,
there is only one line through D that contains a limiting parallel ray to n. In the next theorem, we show that, if a line m that is parallel to l and does not
contain a limiting parallel ray to l, then m and l have a common perpendicu-
4.3. CLASSIFICATION OF PARALLEL LINES
99
lar. Note that in this case, m diverges from l from both sides to the common
perpendicular.
Theorem 4.3.2 Let the line m be parallel to l. Then m does not contain a
limiting parallel ray to l in either direction, if and only if m and l have a
common perpendicular.
Proof. Let the line m be parallel to l.
(=⇒) Let m does not contain a limiting parallel ray to l in either direction. We
need to find two points on l that are equidistant from m. If we get such points,
then the common perpendicular is the perpendicular bisector of the segment
with these points as endpoints.
Given any two points on l, say A and B. Let A0 and B 0 be the feet of the
perpendiculars from A and B respectively to the line m. If AA0 ∼
= BB 0 , then
we are done.
If not, then we may assume that AA0 > BB 0 , since one must be greater than
the other. Then there is a point E with A ∗ E ∗ A0 such that A0 E ∼
= BB 0 . See
the following figure.
←−→
−−→
Then there is a unique ray EF on the same side of AA0 as B, such that
∠F EA0 ∼
= ∠B 0 BG, where G, with A ∗ B ∗ G.
−→ −−→
Claim: AG ∩ EF 6= Φ.
−→
−−→
If this claim is true, then AG and EF intersects at a point, say H. Let K
−−→
be the unique point on BG such that EH ∼
= BK.
←−→0
←−→0
Drop perpendiculars HH and KK to the line m. Then by SAS, 4EHA0 ∼
=
4BKB 0 . Then ∠EA0 H ∼
= ∠BB 0 K 0 , and ∠KB 0 H 0 ∼
= ∠HA0 B 0 by angle subtraction.
By AAS, 4KB 0 K 0 ∼
= 4HA0 H 0 . Thus, HH 0 ∼
= KK 0 .
Hence, the points H and K are two points on l that are equidistant from m.
Therefore, if M is the midpoint of HK and M 0 is the midpoint of H 0 K 0 , then
M M 0 is the desired common perpendicular.
100
CHAPTER 4. HYPERBOLIC GEOMETRY
−→ T −−→
Now, we prove the claim that AG EF 6= Φ :
−−−→
−−→
Let A0 W be the limiting parallel ray to EF ,
−−0→
−→
A N be the limiting parallel ray to AG, and
−−0→
−−→
B P be the limiting parallel ray to BG.
Let L be a point on m, such that A0 ∗B 0 ∗L. Since m does not contain a limiting
−−→ −−→
−−→ −−→
parallel ray to l, then B 0 L 6= B 0 P , and A0 L 6= A0 N .
−−−→
−−→
−−→
Since EA0 ∼
= BB 0 , A0 W be the limiting parallel ray to EF , and B 0 P be the lim−−→
iting parallel ray to BG then ∠W A0 E ∼
= ∠P B 0 B. Then by angle subtraction,
0 ∼
0
∠W A L = ∠P B L.
Hence by alternate interior angle theorem and that vertical angle are congruent,
−−0→
−−−→
B P is parallel to A0 W .
−−0→
−−→
−→
And A N be the limiting parallel ray to AG implies that A0 N be the limiting
−−→
−−→
−−→
parallel ray to BG. And since BG be the limiting parallel ray to B 0 P . Then, by
−−0→
−−0→
Lemma1 B P be the limiting parallel ray to A N . The last statement implies
−−−→ −−→
that ∠W A0 L > ∠N A0 L since also, A0 W ||B 0 P .
−−→
−−−→
−−→
Then A0 N lies between A0 W and A0 L, and ∠W A0 L ∼
= ∠P B 0 L. Then ∠N A0 L <
−
−
−
→
−
−
→
−
−
→
−−→
∠P B 0 L. Then A0 W lies between A0 A and A0 N . Now, since A0 N be the limiting
−−−→
−→
−→
parallel ray to AG, then A0 W intersects AG at a point, say J.
−−−→
←
→
Now, J and A0 are on the same side of EF , since J ∈ A0 W , which is a limiting
−−→
parallel ray to EF .
←→
←→
Then J and A are in opposite side of EF . Thus, AJ intersects EF at a point,
←−→0
−−→
say H. But H ∈ EF since H lies on the same side of AA as J does. So we
←→
←→
conclude our claim that, EF and AG meet at the point H.
(⇐=) We use contradiction, we assume that l contains a limiting parallel ray to
m, and we show that no common perpendicular exists.
−−→
Let P ∈ l such that P Y is a limiting parallel ray to m. Let M ∈ l such that
P ∗ M ∗ Y and M M 0 is a common perpendicular to l and m. Where M 0 the
foot of perpendicular from P to m. Let P 0 and Y 0 be the feet of perpendiculars
from P and Y to m respectively.
−−→
Now, M Y is a limiting parallel ray to m at the point M . Then by proposition 4.3.2, M M 0 > Y Y 0 . Which contradicts, that the common perpendicular is
shorter than any other segment between two parallel lines. .
Corollary 4.3.1 Let m be a line that is parallel to a line l. Then m contains
a limiting parallel ray to l in either direction, if and only if m and l have no
common perpendicular.
Proof. This corollary is the contrapositive of the previous theorem. .
Here is a summary of the main ideas in this section: Given a point P not
on a line l, there exists exactly two limiting parallel rays to l through P , one in
each direction. There are infinitely many lines through P that do not enter the
4.4. POINCARE DISK MODEL
101
region between the limiting parallel rays and l. Each such line is divergently
parallel to l and admits a unique common perpendicular with l. Only one of
these lines with common perpendicular will go through P , but for all the rest
the common perpendicular will pass through other points.
4.4
Poincare Disk Model
There are three traditional models for hyperbolic geometry. They are known as
the Klein model, the Poincare Disk model, and the Poincare Half-Plane model.
We will discuss only the Poincare Disk model. There are isomorphisms between
these models, it is just that some properties are easier to see in one model than
the other. The Poincare models tend to give us the opportunity to do computations more easily than the Klein model though the Klein model is somewhat
easier to describe. In order to give a model for hyperbolic geometry, we need
to decide on a set of points, then determine what lines are and how to measure
distance and angles.
For Poincare Disk Model we take the set of all points that lie inside the unit
circle, name it γ. Note that points on the circle itself are not in the hyperbolic
plane. However they do play an important part in determining our model. Euclidean points on the circle itself are called ideal points, or points at infinity.
The five undefined terms are interpreted as follows:
Points in the hyperbolic plane are interpreted to be the set of all interior points
of γ.
Definition 4.4.1 Two circles in Euclidean geometry are called orthogonal to
one another, if at the points of intersection their radii are perpendicular.
Lines in the hyperbolic plane are of two types:
1. The open diameters, that is diameters without endpoints.
2. The open arcs, that are portions inside γ of circles orthogonal to γ. These
lines are called Poincare lines.
Lies on is interpreted as in Euclidean sense.
Betweenness:
On an open diameter, betweenness is interpreted as in Euclidean sense.
−−→
−→
−−→
On Poincare line: A ∗ B ∗ C if and only if OB, lies between OA and OC in
Euclidean sense. See the following figure.
102
CHAPTER 4. HYPERBOLIC GEOMETRY
The congruence of segments in the Poincare model is complicated. If we
used the Euclidean length of segments, then all lines in the Poincare model
would be of finite length which contradicts B.2, that guarantees that lines are
infinite. So we have to define the distance in Poincare model. But first, we have
to show that, any two distinct points in the Poincare model determine a unique
line. That is, I1 is true statement in this model.
Proposition 4.4.1 Let A and B be any two points inside γ, where γ is a unit
circle with centre O, then there is a unique line through A and B in the model.
Proof. First, if A, B and the centre O are collinear in the Euclidean sense,
←→
then there is a line AB in the Euclidean sense. Thus we get an open diameter,
←→
which is a portion of the line AB, that contains A and B.
Second, if A, B and O are not collinear in the Euclidean sense. We want
to construct a circle δ that contains A and B and is orthogonal to γ.
−→
←→
Consider the ray OA. Let m be the perpendicular to the line OA at A. Then
m intersects γ in two points, say P and R. Let t be the line through P that
is tangent to γ. Then P O ⊥ t. In the triangle 4P AO, ∠A is right then the
other two angles are acute. That is ∠P OA is acute. Since we are in Euclidean
geometry, and angle sum of the alternate interior angles ∠P OA and ∠P is less
−→
than 180◦ , then by Euclid’s fifth postulate, t and OA intersects at a point, say
A0 . A0 is usually called the pole of A. See the figure.
4.4. POINCARE DISK MODEL
103
Since 4P AO is similar to the triangle 4A0 P O, then
OA0
OP
=
.
OP
OA
That is
2
(OA0 )(OA) = OP = 1.
The points A, B and A0 determine a unique circle, called δ. This circle is orthogonal to γ. To show this, first, we need to find the centre of δ, called Q.
Let G be the intersection of δ with m. Since ∠A0 AG is right then, A0 G is a
diameter in δ. Thus, the midpoint of A0 G is the centre Q of δ
Now, we have to show that, the radii of γ and δ are perpendicular.
Let l and n be the tangents from the point O to the circle δ at the points, say
I and J. We claim that I and J lie on γ.
Since ∠OJQ and ∠OIQ are right angles, then by SSA, the triangles 4OIQ
and 4OJQ are congruent. Then, ∠JQO ∼
= ∠IQO. Since 4IQJ is an isosceles
−−→
∼
triangle, then ∠QIJ = ∠QJI. Then the angle bisector QO is perpendicular
to the base IJ at the point, say K. Then, O is the pole of K, where K is
considered as a point inside δ. Then OQ OK = 1. Now, 4OIQ is similar to
4OKI. Then
OI
OQ
=
.
OK
OI
104
CHAPTER 4. HYPERBOLIC GEOMETRY
2
Thus, OI = OQ OK = 1. Then, OI = 1.
Hence, I lies on γ. Since the tangent and the radius are perpendicular, then the
tangent l is perpendicular to IQ the radius of δ. From this proposition, we conclude that incidence axiom 1 is true in this model.
Also, incidence axiom 2 and 3 are trivially true in this model, which is embedded in Euclidean plane.
Note that the boundary of Poincare model should not be reachable, so we need
to define distance such that the distance goes to infinity as we approach the
boundary of Poicare disk.
Definition 4.4.2 Let A and B be any two points in the Poincare disk, and let
P and Q be the endpoints of the line containing A and B. Then the Poincare
distance dP (A, B) between A and B is defined by
AP BQ dP (A, B) := ln
.
AQ BP where Q ∗ A ∗ B ∗ P , and AP , BP , AQ, and BQ are the Euclidean length.
We then define congruence of segments in Poincare model by
AB ∼
= CD ⇐⇒ dP (A, B) = dP (C, D).
Now, suppose we fix a point A on the Poincare line l from P to Q on γ, and
let B the point on l, that moves continuously from A to P , where Q ∗ A ∗ B ∗ P ,
4.4. POINCARE DISK MODEL
105
as in the following figure
AP BQ
AP
will increase continuously from 1 to ∞, since
is constant,
AQ BP
AQ
BP approaches zero, and BQ approaches P Q. Hence, dP (A, B) will increase
from 0 to ∞ continuously. If we fix B and let A moves continuously from P
to Q, we get the same result. It follows immediately that for any Poincare ray
−−→
−−→
CD, there is a unique point E on CD such that dP (C, E) = dP (A, B). This
verifies C1.
Then
Note that, this length does not depend on the order in which we write A and B.
That is dP (A, B) = dP (B, A). We will prove this in the next proposition part
i. With this interpretation of congruent of segments, axiom C2 is immediately
verified.
Next we verify C3. This will follows from the additivity of of the Poincare
length, which is proved in the next proposition part iv.
Proposition 4.4.2 The Poincare distance satisfies the defining properties of
distance function. That is:
i dP (A, B) = dP (B, A)
ii dP (A, B) ≥ 0
iii dP (A, A) = 0
iv If A ∗ B ∗ C on a Poincare line with endpoints P and Q, then
dP (A, B) + dP (B, C) = dP (A, C).
Proof.
106
CHAPTER 4. HYPERBOLIC GEOMETRY
[i] We will use the following property of logarithmic function:
| ln
1
|=| ln x |
x
Let A and B be any two points in the Poincare disk, and let P and Q be
the endpoints of the line containing A and B.
AP BQ dP (A, B) = ln
AQ BP BP AQ = ln
BQ AP = dP (B, A).
[ii] dP (A, B) ≥ 0, since for any real number x, | (ln x) |≥ 0.
AP AQ [iii] dP (A, A) := ln
.
AQ AP Thus, dP (A, A) = ln1 = 0.
[iv] Let A ∗ B ∗ C be on a Poincare line with endpoints P and Q, and Q ∗ A ∗ B
See the figure.
AP BQ
·
> 1.
AQ BP
BP CQ
Similarly, BP > CP and CQ > BQ, then
·
> 1.
CP BQ
Since AP > BP and BQ > AQ, then
4.4. POINCARE DISK MODEL
107
Then,
AP BQ BP CQ ln
+
dP (A, B) + dP (B, C) = ln
AQ BP BQ CP AP BQ
BP CQ
= ln
+ ln
AQ BP
BQ CP
AP BQ
BP CQ
= ln
AQ BP
BQ CP
= dP (A, C).
Congruence of angles in Poincare model:
If the two sides of the angle are arts of open diameter, and the vertex is the
centre of γ. Then the angle is measured as in euclidean sense. See the
following figure.
If one side is an open diameter and the other side is a part of Poincae line,
then the angle is considered to be the angle in Euclidean sense between
the open diameter and the tangent of the Poincare line at the vertex of
the angle. See the following figure.
108
CHAPTER 4. HYPERBOLIC GEOMETRY
If the two sides of the angle are parts of two Poincare lines, then the angle is
the angle in Euclidean sense between the tangents of the Poincare lines.
See the following figure.
Now, axiom C5 is trivially verified. To verify axiom C4 (laying off a congruent copy of a given angle, say ∠H at some ray with initial point A). We
consider the following cases:
4.4. POINCARE DISK MODEL
109
If A is the centre O of γ, the angle is formed by diameters and the laying off
is accomplished in the Euclidean way.
If A is not the centre of γ and pass through the line l, which is either an open
diameter or a tangent to some Poincare line that pass through A. If l is
an open diameter. Then we have to find a unique circle through A that
is orthogonal to γ and tangent to a given Euclidean line l that passes
through A, such that the two lines l and k determine the angle measure
of ∠H.
Let δ be a circle through A. Now, δ meets γ orthogonally if and only if δ
passes through the pole A0 of A with respect to γ.
The centre C of δ must lie on the perpendicular bisector of chord AA0 at
M , call this bisector m. If k is the tangent to δ at A, then C must also
lie on the perpendicular n to k at A. So δ must be the circle whose centre
is the intersection C of m and n and whose radius is CA.
See the following figure.
If A is not the centre of γ and pass through the line l, which is a tangent to
some Poincare line that pass through A. Then we have to find a unique
circle through A that is orthogonal to γ and tangent to a given Euclidean
110
CHAPTER 4. HYPERBOLIC GEOMETRY
line k that passes through A, such that the two lines l and k determine
the angle measure of ∠H.
Let δ be a circle through A. Then δ meets γ orthogonally if and only if δ
passes through the pole A0 of A with respect to γ.
The centre C of δ must lie on the perpendicular bisector of chord AA0 at
M , call this bisector m. If k is the tangent to δ at A, then C must also
lie on the perpendicular n to k at A. So δ must be the circle whose centre
is the intersection C of m and n and whose radius is CA.
See the following figure.
To show that C6(SAS) is true in this model, we should study the inversion of
circles. So, we leave it without justification.
Betweeness axioms are all trivially true in this model as it is embedded in
Euclidean plane.
Having interpreted all the undefined terms of hyperbolic geometry in the Poincare
model, we get interpretations of all the defined terms. For example, two Poincare
lines are parallel if and only if they have no point in common. The limiting
parallel rays in the Poincare model are illustrated in the following figures:
4.4. POINCARE DISK MODEL
111
In the above figure, we have chosen l to be an open diameter, with endpoints
P and Q, and A any point inside γ not on l. Then the limiting parallel rays
through A and parallel to l are the rays emanating from A and are arcs of circles
orthogonal to γ through P and Q.
The above figure illustrates two parallel Poincare lines with a common perpendicular. The figure shows that how the line m diverges from l on either side
of the common perpendicular AB.
Lemma 4.4.1 Let A be a point inside γ, and d := dP (O, A). Then
OA =
ed − 1
ed + 1
112
CHAPTER 4. HYPERBOLIC GEOMETRY
Proof. Let P and Q be the end of the diameter containing A such that Q ∗ A ∗
O ∗ P . Then
AP OQ AP = ln
.
d = dP (A, O) := ln
OP AQ AQ Then
ed =
AP
1 + AO
=
AQ
1 − AO
2OA
1 − OA
2
ed + 1 =
.
1 − OA
ed − 1 =
Thus, ed − 1 = OA ed + 1 Theorem 4.4.1 (Boylai Lobachevsky
Formula)
Q
The angle of parallelism θ(d) := (AB)◦ in the Poincare Disk model of Hyperbolic geometry satisfies
θ(d)
tan
= e−d
2
where d := dP (A, B)
Proof. Since θ(d) depends only on the distance we are free to choose any line
and any point. Let l be a diameter and let A be a point inside γ on a diameter
perpendicular to l. Let d := dP (A, O). We will find θ(d).
The limiting parallel ray through A to l will be an arc of the circle, say δ, that
is tangent to l at C on the circle γ. See the following figure.
4.5. EXERCISES
113
The tangents to δ at A and C meet at a point R on l such that O ∗ R ∗ C
, see the figure. Then 4ACR is an isosceles triangle. Hence ∠ACR ∼
= ∠CAR.
Let β := (∠CAR)◦ . Then
(∠CRA)◦ = π − 2β
π
− 2β. Then
2
π
tan( ) − tanβ
π
θ(d)
1 − tanβ
4
) = tan( − β) =
tan(
=
π
2
4
1 + tanβ
1 + tan( )tanβ
4
Thus, (∠ARO) = 2β. Hence θ(d) =
By previous lemma,
ed − 1
OA
=
= tan β. Then
d
e +1
1
tan
4.5
θ(d)
2
ed − 1
ed + 1 = e−d .
=
ed − 1
1+ d
e +1
1−
Exercises
[1.] In hyperbolic geometry: Prove that the summit is always greater than the
base in a Sacchari quadrilateral.
[2.] In hyperbolic geometry: Prove that two Sacchari quadrilaterals with congruent bases and congruent summit angles must be congruent quadrilaterals.
[3.] In hyperbolic geometry: Given two lines l and m with a common perpendicular M M 0 . If A and B are two points on l such that A ∗ M ∗ B, where
M is not the midpoint of the segment AB. Then show that A and B are
not equidistant from m.
[4.] In hyperbolic geometry: Given two lines l and m with a common perpendicular M M 0 . If A and B are two points on l such that M ∗ A ∗ B. Then
prove that d(A, m) < d(B, m).
[5.] In Euclidean geometry, all triangle have the same defect which is zero. In
hyperbolic geometry, show that not all triangles have the same defect.
[6.] In neutral geometry, we have proved the alternate interior angle theorem. In hyperbolic geometry, the alternate interior angle theorem can
be strengthened, so that the two lines are parallel and have a common
perpendicular.
[7.] Given any triangle 4ABC. Let I, J, and K be the midpoints of BC,
AC, and AB respectively. Drop perpendiculars from A, B, and C to the
←
→
line IJ at the points D, E, and F respectively.
114
CHAPTER 4. HYPERBOLIC GEOMETRY
i. Show that AD ∼
= CF ∼
= BE. That is, the quadrilateral EDAB
is a Sacchari quadrilateral. It is called the Sacchari quadrilateral
associated to the triangle 4ABC.
←→
ii. Show that the perpendicular to the line AB at K is also perpendicular
←
→
to the line IJ .
1
iii. Show that IJ = ED.
2
1
iv. In hyperbolic geometry, show that IJ < AB.
2
v. In hyperbolic geometry, show that AB > ED.
vi. Show that δ(4ABC) = δ(4EDAB).
vii. Suppose that the triangle 4ABC has right angle at C. In hyperbolic
geometry, show that Pythagorean theorem does not hold.
[8.] Let 4ABC be any triangle, and L, M , and N be the midpoints of BC,
AB, and AC respectively. In hyperbolic geometry: Prove that 4AM N is
not similar to 4ABC.
Chapter 5
Axiom System of Euclidean
3-space and Spherical
Geometry
Spherical geometry can be said to be the first non-Euclidean geometry. For at
least 2000 years humans have known that the Earth is almost a sphere and that
the shortest distance between two points on the Earth is along a great circle;
which is the intersection of the sphere with a plane through the centre of the
sphere.
In the first section , we shall introduce the Hilbert’s axiom system of 3-dimensional
Euclidean space. In the second section we shall see the spherical geometry as
the geometry of the sphere in the 3-dimensional Euclidean space.
In the third section, we shall introduce an axiom system for spherical geometry,
that is in the spirit of Hilbert’s axiom system of Euclidean geometry, which is
done by Borsuk.
5.1
Hilbert’s Axiom System of 3-dimensional Euclidean Space
We begin by the undefined terms that are basis for defining all other terms in
3-space Euclidean geometry. They are:
point
line
plane
incidence with (lie on or pass through)
between
congruent.
Hilbert’s axioms are divided into five groups: incidence axioms, betweenness
axioms, congruence axioms, continuity axioms, and parallelism axiom.
115
116CHAPTER 5. AXIOM SYSTEM OF EUCLIDEAN 3-SPACE AND SPHERICAL GEOMETRY
Note that, all planes will be assumed to be Euclidean planes, that is all axioms
of Eucildean plane geometry in chapter 2 and consequently all its theorems in
chapter 2 and chapter 3 hold for every plane.
The axioms of incidence are:
I.1 Given two distinct points P and Q there is a unique line incident with P
←→
and Q. This line will be denoted by P Q.
I.2 For every line l there exist at least two distinct points incident with l.
I.3 There exist three distinct points with the property that no line is incident
with all three of them.
I.4 Given three distinct points, that are not collinear, there is a unique plane
incident with them.
I.5 Given any plane Γ, then there exists at least one point in Γ.
I.6 If two distinct points of a line l are in one plane Γ, then all points of the
line l are in Γ.
I.7 There exists 4 distinct points with the property that no plane contains all
of them.
Definition 5.1.1 Two planes are called parallel if they have no point in
common.
I.8 the intersection of two not parallel planes contains at least two distinct
points.
Definition 5.1.2 A line l is said to be parallel to a plane if they do not have
any common point.
A set of points are called coplanar if there exists a plane that contains all of
them.
Two lines are called coplanar if there exits a plane that contains both of them.
Two not coplanar lines are called skew-lines
Proposition 5.1.1 Let l be a line and Γ be a plane. Then only one of the
following holds:
i. l is parallel to Γ.
ii. l intersects Γ in exactly one point.
iii. l lies in Γ.
Proposition 5.1.2 Let Γ1 and Γ2 be two planes. Then only one of the following
hold:
i. Γ1 and Γ2 are parallel.
5.1. HILBERT’S AXIOM SYSTEM OF 3-DIMENSIONAL EUCLIDEAN SPACE117
ii. The intersection of Γ1 and Γ2 is a line.
iii. Γ1 = Γ2 .
Note that two skew-lines do not have a common point. Therefore, two distinct
lines in space are said to be parallel if they are coplanar and have no point in
common.
Proposition 5.1.3 i. If two distinct lines have a common point, then there
exists only one plane containing them.
ii. If l and m are two distinct lines in space, then only one of the following
occurs:
1. l and m are skew-lines.
2. l and m are concurrent.
3. l and m are parallel.
The axioms also imply the following statements.
Proposition 5.1.4 i. if A and B are two distinct points, then there exists at
least two distinct planes containing them.
ii. For every line there exists at least two distinct planes containing it.
iii. There exists at leas three distinct planes that do not have a common line.
iv. There exists two lines with the property that no plane contains both of them.
v. For every line there exists at least one plane not containing it.
vi. For every plane there exists at least one line not contained in it.
Axioms on Betweenness:
B.1 If A ∗ B ∗ C then A, B, and C are three distinct points all lying on the
same line, and C ∗ B ∗ A.
B.2 Given any two distinct points B and D, there exists points A, C and
←→
E lying on BD such that A ∗ B ∗ D, B ∗ C ∗ D and B ∗ D ∗ E.
B.3 If A, B, and C are three distinct points lying on he same line, then one
and only one of the points is between the other two.
Betweenness axiom 4 will be modified as follows:
B.4 Let A, B, and C be three non-collinear points, and let l be a line in the
plane that contains the three points, and not passing through any of them, 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.
118CHAPTER 5. AXIOM SYSTEM OF EUCLIDEAN 3-SPACE AND SPHERICAL GEOMETRY
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 line l.
Definition 5.1.3 Let Γ be a plane and A and B are two distinct points not in
Γ. We say that A and B are on the same side of Γ if A and B are on the
←→
same side of every line lying in Γ that is coplanar with AB. That is A and B
are on the same side of γ if and only if the segment AB does not intersect Γ.
Let Γ be any plane and A any point not in Γ. The set of all points that are on
the same side of Γ as A is called half-space bounded by Γ containing A. We
denote it by HSA,Γ .
Theorem 5.1.1 Space separation property Every plane determines exactly
two disjoint half-spaces.
Note that, all lines and points are considered to lie on the same plane unless
otherwise stated.
Now we state the axioms of congruence:
C.1 If A and B are two distinct points and if C is any point, then for each given
ray r emanating from C there is a unique point D on r such AB ∼
= CD.
C.2 If AB ∼
= CD and AB ∼
= EF then CD ∼
= EF . Moreover, every segment is
congruent to itself.
C.3 (Segment Addition)If A ∗ B ∗ C, E ∗ F ∗ G, AB ∼
= EF , and BC ∼
= F G,
then AC ∼
= EG.
Congruence axiom 4 is modified as follows:
−−→
C.4 Given ∠BAC in a plane Γ1 and any ray EF in a plane Γ2 , then on a
←→
−−→
given side of the line EF in the plane Γ2 , there is a unique ray EG such
that ∠BAC ∼
= ∠F EG.
C.5 If ∠A ∼
= ∠B and ∠A ∼
= ∠C, then ∠B ∼
= ∠C. Moreover, every angle is
congruent to itself.
C.6 (SAS) If two sides and the included angle of one triangle are congruent
respectively to two sides and the included angle of another triangle, then
the two triangles are congruent.
The axiom of continuity is
Dedekind
Suppose that the set of all points on a line lP
is the disjoint
P Axiom:
P
union 1 ∪ 2Pof two non-empty subsets, such that no points of 1 is between
two points of 2 and vice versa. Then P
there is a unique
P point O lying on l
such that P1 ∗ O ∗ P2 if and only if P1 ∈ 1 and P2 ∈ 2 . That is one of the
subsets is equal to a ray of l with vertex O and the other subset is equal to the
complement.
Finally, the axiom of parallelism is
Hilbert’s Parallel Axiom: For every line l and every point P not on l, there is
at most one line m through P such that m is parallel to l and m pass through P .
5.2. SPHERICAL GEOMETRY: A 3-DIMENSIONAL EUCLIDEAN SPACE VIEW119
5.2
Spherical geometry: A 3-dimensional Euclidean Space View
In this section, spherical geometry is the study of the surface of a sphere. Where
the sphere is the set of all points in Euclidean 3-space, that have distance R
from a particular point O called the centre of the sphere, and R is called the
radius of the sphere.
Let S be a unit sphere in Euclidean 3-space with centre the origin.
Definition 5.2.1 A great circle on the sphere S is the set of all points in the
intersection of S and a plane passes through the origin.
Theorem 5.2.1 The shortest path on the sphere S between two points A and
B on the sphere, lies a long a great circle connecting the two points.
Proof.
Definition 5.2.2 An opposite point of A on the sphere S, is the point A
on S which lies on a diameter of S which contains A.
0
Now, suppose we have two points A, and B on a great circle l. These are two
circular arcs that connect A and B, the shortest of these arcs must be the shortd If B = A0 , then each of two
est path from A to B, we will denote it by AB.
arcs has the same length.
In fact, if B is the opposite point of A, then there are infinitely many paths
with shortest distance between A and its opposite B. For, there are infinitely
many planes that can pass through A, B, and centre O, since the three points
are collinear.
If B is not opposite to A, then A, B, and O are not collinear, and hence determine a unique plane, thus there is a unique great circle through A and B.
Thus, there is a unique shortest path between A and B.
Theorem 5.2.2 Any two great circles have exactly two opposite points in common.
Proof. Let Γ1 be the plane that determine the great circle l1 , and Γ2 be the
plane that determine the great circle l2 . That is,
l1 = Γ1 ∩ S.
l2 = Γ2 ∩ S.
Then
l1 ∩ l2 = (Γ1 ∩ S) ∩ (Γ2 ∩ S) = Γ1 ∩ Γ2 ∩ S.
From Euclidean 3-space geometry, any two not parallel planes intersect in a line.
The line of intersection contains the centre O. Hence, the line of intersection
crosses the sphere exactly twice. Thus, l1 ∩ l2 consists of exactly two points,
which are opposite points. 120CHAPTER 5. AXIOM SYSTEM OF EUCLIDEAN 3-SPACE AND SPHERICAL GEOMETRY
Definition 5.2.3 The angles between two great circles is defined to be the angle
between tangent lines to the great circles.
In spherical geometry it is possible to define a two-sided polygon, because any
two distinct great circles intersect at two antipodal points and divide the sphere
into four regions, each of which is called a lune(or biangle), since it has two
sides and two vertices and two angles.
The two angles of the lune are congruent, since the two angles are the two angles
between the planes from which the two great circles are created.
Proposition 5.2.1 A lune whose angle is θ radians, then it has area 2θR2 .
Proof. Let N be a point on S, divide the sphere into n equal lunes, then the
2π
. And the area of each lune is the area of S divided
angle of each lune will be
n
by n. That is
4πR2
2π
= 2( )R2 .
n
n
That is, if the lune has angle θ then the area of the lune is 2θr2 . Definition 5.2.4 Let A, B, and C be distinct points on S, such that no great
circles contains all of them. The spherical triangle 4S ABC is defined to be the
d ∪ BC
d ∪ AC.
d
arcs AB
0
0
0
Definition 5.2.5 If 4S ABC is a spherical triangle, the triangle 4S A B C ,
0
0
0
where A , B , and C are the opposites of A, B, and C respectively, is called
the opposite triangle of 4S ABC.
Lemma 5.2.1 If 4S ABC is any spherical triangle, then
0
0
0
4S ABC ∼
= 4S A B C
Theorem 5.2.3 (Girard’s Theorem)
If 4S ABC is any spherical triangle with angles α, β, and gamma given in
radians, then
area(4S ABC) = R2 (α + β + γ − π)
Corollary 5.2.1 If 4S ABC is any spherical triangle with angles α, β, and
gamma given in radians, then
α+β+γ =π+
area(4S ABC)
R2
Theorem 5.2.4 (Pythagorean Theorem on the Sphere)
Let 4S ABC be a spherical right triangle with right angle at the vertex C. Let
the arc length of the sides opposite to A, B, and C be a, b, and c respectively.
Then
c
a
b
cos( ) = cos( )cos( ).
R
R
R
5.3. SPHERICAL GEOMETRY; AXIOMATIC APPROACH
5.3
121
Spherical Geometry; Axiomatic Approach
In this section, we will introduce an axiom system for spherical geometry that is
in the spirit of Hilbert’s axiom system for Euclidean geometry. This axiom system is given in [1] Rational Geometry for Halstad. We begin by the undefined
terms that are basis for defining all other terms in spherical geometry. They
are:
point
line
incidence with (lie on or pass through)
between
congruent.
The set of axioms are divided into five groups: incidence axioms, betweenness
axioms, congruence axioms, continuity axioms, and parallelism axiom.
Incidence Axioms
0
IS.1 For every point P , there exists always one other point P which with P
0
does not determine a line. This second point P is called the opposite of
P.
IS.2 Any line incident with a point is also incident with its opposite.
IS.3 Any two points not each the others opposite determine a unique line.
IS.4 There are at least three points not incident with the same line.
Now we will prove some consequences of the incidence axioms.
0
0
Proposition 5.3.1 If P is opposite of point P , then P is also opposite of P .
0
Proof. We use contradiction. Assume P is opposite of point P and P is not
0
opposite of point P . Then by IS.3, P and P 0 determine a unique line l. By IS.4,
0
there exists a point A not on l and A is not opposite point of P , then A 6= P .
By IS.3, A and P determine a unique line, say m. By IS.2, m is incident with
0
0
P . Thus, P and P does not determine a unique line. Proposition 5.3.2 Two distinct lines cannot have three points in common.
Proof. Let l and m be two distinct lines. We use contradiction. Assume that
A, B, and C be three distinct points that lie on both m and l. At least two of
them not opposite points, say A, and B are not-opposite points. Then by IS.3,
they determine a unique line, hence l = m, which is a contradiction. Betweenness Axioms
BS.1 If A ∗ B ∗ C, then A, B, and C are distinct points and lie on one line,
and C ∗ B ∗ A.
BS.2 No point is between two opposites.
122CHAPTER 5. AXIOM SYSTEM OF EUCLIDEAN 3-SPACE AND SPHERICAL GEOMETRY
BS.3 Given any two non-opposite points A, and B, there exits points C, D,
and E, such that, A ∗ C ∗ B, A ∗ B ∗ D, and E ∗ A ∗ B.
BS.4 Given any three points, no more than one can be between the other two.
BS.5 If A ∗ B ∗ C and A ∗ C ∗ D then A ∗ B ∗ D.
BS.6 Between no two points are there two opposites.
Proposition 5.3.3 No point is between its opposites and any third point.
0
Proof.Let A be opposite point of A. Let B be any third point. Use
0
contradiction. Assume that B ∗ A ∗ A.
Now, since A , B are not opposites, then by BS.3 there is a point C such
0
that B ∗ A ∗ C. By BS.5 B ∗ A ∗ C. Then both A and its opposite lie
between B and C, which is a contradiction to BS.6 .
Definition 5.3.1 Let A and B be two non-opposite points on a line l. A
segment AB is defined to be the set of all points C such that A ∗ C ∗ B
together with A and B. A and B are called the endpoints of the segment
AB. The segment AB will be denoted by AB.
BS.7 (Pasch) Let A, B, and C be any three distinct points, that no line is
incident with all of them, and no two of them are opposite points. Let l
be a line that does not incident to any of them. IF l intersects AB then
it must intersects either BC or AC but not both.
Proposition 5.3.4 Every line l separates the points not on l into two regions.
Each of which is called a side of l. Let A and B be two points not on l. Then they
are said to be in opposite sides of l if and only if the segment AB intersects
l. They are said to be on same side of l if and only if AB does not intersect l.
Proof.
0
Proposition 5.3.5 The points of any line l other than two opposites P and P
0
are separated into two classes such that P or P is between any point of the one
0
and any non-opposite point of the other, but neither P nor P is between two of
the same class.
Definition 5.3.2 The parts of a line l determined by a point A and its opposite
−→
are called rays of l. If C is on l and not opposite of A, then the ray AC consists
0
of point A, point C, and all points D such that either A ∗ D ∗ C or C ∗ D ∗ A .
−→
We say AC is emanating from A. Two distinct rays part of the same line, and
emanating of the same point are called opposite rays.
Proposition 5.3.6 Two opposites cannot be on the same ray.
Proposition 5.3.7 Every line has exactly two points in common with any other
line
5.3. SPHERICAL GEOMETRY; AXIOMATIC APPROACH
123
Axioms of Congruence
CS.1 If A and B are two non-opposite distinct points and if C is any point,
then for each given ray r emanating from C there is a unique point D on
r such AB ∼
= CD.
CS.2 If AB ∼
= CD and AB ∼
= EF then CD ∼
= EF . Moreover, every segment
is congruent to itself.
CS.3 (Segment Addition) If A∗B ∗C, E ∗F ∗G, AB ∼
= EF , and BC ∼
= F G,
∼
then AC = EG.
−−→
−→
Definition 5.3.3 Let AB and AC be two distinct rays, that are parts of
−−→
−→
two distinct lines, then the angle ∠BAC consists of AB and AC. And A
−−→
−→
is called the vertex of ∠BAC. The rays AB and AC are called the sides
of the angle ∠BAC.
−−→
←→
CS.4 Given ∠BAC and any ray EF , then on a given side of the line EF
−−→
there is a unique ray EG such that ∠BAC ∼
= ∠F EG.
CS.5 If ∠A ∼
= ∠B and ∠A ∼
= ∠C, then ∠B ∼
= ∠C. Moreover, every angle is
congruent to itself.
We conclude from this axiom that: If ∠A ∼
= ∠B then ∠B ∼
= ∠A.
Definition 5.3.4 A spherical n-polygon consists of n segments A1 A2 ∪
A2 A3 ∪ A3 A4 ∪ · · · ∪ An A1 , such that any two of them meet in at most
one endpoint. The segments are called the sides of the polygon, and the
endpoints of the segments are called vertices of the polygon.
A spherical 3-polygon is called a triangle.
Two triangles are said to be congruent if there is a one to one correspondence between their vertices so that the corresponding sides are congruent
and corresponding angles are congruent.
CS.6 (SAS) If two sides and the included angle of one triangle are congruent
respectively to two sides and the included angle of another triangle, then
the two triangles are congruent.
Corollary 5.3.1 Given 4ABC and segment DE ∼
= AB, there is a unique point
←→
F on a given side of line DE such that 4ABC ∼
= 4DEF .
Proof. Given any 4ABC and segment DE ∼
= AB. Then by CS.4 there is
−−→
←→
a unique ray DF on a given side of the line DE such that ∠BAC ∼
= ∠F DE.
−−→
Then by CS.1 on the ray DF choose F the unique point such that AC ∼
= DF .
Hence by SAS, 4ABC ∼
4DEF
.
=
Proposition 5.3.8 (ASA) Given 4ABC and 4DEF with AB ∼
= DE, ∠CAB ∼
=
∼
∠F DE, and ∠CBA ∼
∠F
ED,
then
4ABC
4DEF
.
=
=
124CHAPTER 5. AXIOM SYSTEM OF EUCLIDEAN 3-SPACE AND SPHERICAL GEOMETRY
Proof. Given 4ABC and 4DEF with AB ∼
= DE, ∠CAB ∼
= ∠F DE, and
−→
∠CBA ∼
= ∠F ED. By CS.1 there is a unique point G on ray AC, such that
AG ∼
= DF . Then by SAS, 4ABG ∼
= 4DEF , and hence ∠ABG ∼
= ∠DEF .
But by assumption, ∠DEF ∼
= ∠ABC. Then by uniqueness of CS.4, we have
−−→ −−→
←→ ←→
BG = BC. We claim that G = C. If not, that is G 6= C then G ∈ AC ∩ BC,
←→ ←→
then by proposition ???, AC = BC. But this contradicts that A, B, and C are
non-collinear. Definition 5.3.5 By AB < CD or CD > AB we mean that there exists a
point E between C and D such that AB ∼
= CE.
Definition 5.3.6 A triangle is called isosceles if two of its sides are congruent.
Proposition 5.3.9 Given 4ABC, then:
i. If 4ABC is an isosceles triangle then ∠B ∼
= ∠C.
ii. If ∠B ∼
= ∠C then the triangle is isosceles.
Proof. i. Given a triangle 4ABC in which AB ∼
= AC. Consider the triangles
4ABC and 4ACB. Then by CS.5, ∠A ∼
= ∠A, by hypothesis AB ∼
= AC, and
by CS.2, we have AC ∼
= AB, then by SAS, 4ABC ∼
= 4ACB. Hence, ∠B ∼
= ∠C.
−−→
Definition 5.3.7 By ∠BAC < ∠DEF , we mean that, there is a ray EG be−−→
−−→
tween EF and ED such that ∠BAC ∼
= ∠GEF .
Definition 5.3.8 Two angles are called supplementary angles, if they have
the same vertex, have a common side, and the other sides are opposite rays.
An angle is called right angle, if it is congruent to its supplementary angle.
Two angles are called vertical angles if they have the same vertex and the
sides are opposite rays.
Two lines meet at a point, say P are called perpendicular, if the angles consists of P and one side is a ray in the first line emanating from P , and the other
side is a ray in the second line emanating from P , are right angles.
Proposition 5.3.10 i. Supplements of congruent angles are congruent.
ii.Vertical angles are congruent to each other.
iii An angle congruent to a right angle is right.
Proof. i. Given ∠CAD ∼
= ∠GF H. We show that ∠BAD ∼
= ∠EF H.
By CS.1, choose G such that F G ∼
= AC, E such that EF ∼
= AB, and H such
∼
that F H ∼
AD.
By
SAS,
4DAC
4HF
G.
By
segment
addition
BC ∼
=
=
= EG.
∼
∼
Then, since ∠DCA ∼
∠HGF
and
DC
HG,
so
by
SAS,
4DBC
4HEG.
=
=
=
5.3. SPHERICAL GEOMETRY; AXIOMATIC APPROACH
125
By congruence of triangles, HE ∼
= DB and ∠DBC ∼
= ∠HEG. Then by SAS,
∼
4DBA = 4HEF . Finally, by congruence of triangles ∠DAB ∼
= ∠HF E.
ii. Consider the following figure. ∠1 is supplementary to ∠2 and also supplementary to ∠4. By CS.5 ∠1 ∼
= ∠4.
= ∠1, then by part i ∠2 ∼
iii. Let ∠1 be a right angle and ∠2 is any angle congruent to ∠1.
Since ∠1 is right angle, then it is congruent to its supplementary angle ∠3.
Then by i, the supplementary angle of ∠2, which is ∠4, is congruent to angle
∠3. That is, ∠4 ∼
= ∠3 ∼
= ∠1 ∼
= ∠2. By CS.5, ∠4 ∼
= ∠2. Hence, ∠2 is right. Proposition 5.3.11 For every line l and every point P , there exists a line
through P that is perpendicular to l.
Proof. Given any line l, and P any point. Then, either P ∈ l or P ∈
/ l.
Case 1: let P ∈
/ l. By IS.2, there are two points, say A and B on l. By CS.4, on
−→
the opposite side of l from P , there exists a ray AG such that ∠GAB ∼
= ∠P AB.
−
→
By C.1, there is a point P 0 on AG such that AP 0 ∼
= AP . Now P P 0 intersects l
at a point Q. Either Q = A or Q 6= A.
If Q = A, since ∠P QB ∼
= ∠P 0 QB and they are supplementary angles, then
←−→
they are right angles. Hence, P P 0 is perpendicular to l.
If Q 6= A, then by SAS, 4P QA ∼
= ∠P 0 QA. Thus
= 4P 0 QA. Hence, ∠P QA ∼
←−→0
P P is perpendicular to l.
Case 2: Let P ∈ l. By proposition there is a point not on l, say H. by case
1, we can drop a perpendicular from H to l, call it m. Let l and m intersects
at a point G. Then either P = G or P 6= G.
If P = G, then we are done.
If P 6= G.
By CS.4, consider angle ∠HGP and any ray part of l emanating from P
and a given side of l, there is a unique ray emanating from P , which produce
an angle congruent to the right angle ∠HGP . −−→
−−→
−→
Proposition 5.3.12 (Angle Addition) Given AD lies between AB and AC
−−→
−−→
−−→
and ray EG lies between EF and EH, ∠CAD ∼
= ∠GEH, and ∠DAB ∼
= ∠GEF .
∼
Then ∠CAB = ∠F EH.
−−→
−−→
−→
−−→
−−→
Proof. Given AD lies between AB and AC and ray EG lies between EF and
−−→
EH, ∠CAD ∼
= ∠GEH, and ∠DAB ∼
= ∠GEF . See the figure
126CHAPTER 5. AXIOM SYSTEM OF EUCLIDEAN 3-SPACE AND SPHERICAL GEOMETRY
−−→
Then by Crossbar theorem, AD intersects BC at a point, say M , such that
B ∗ M ∗ C. By C.1, choose F , G and H, such that AB ∼
= EF , AC ∼
= EH, and
∼
AM = EG.
Then, by SAS, 4M AC ∼
=
= 4F EG. Then ∠HGE ∼
= 4GEH, and 4BAM ∼
∼
∠CM A and ∠F GE = ∠BM A. But ∠CM A and ∠BM A are supplementary
angles. By proposition 2.3.6 i, ∠BM A is congruent to the supplementary angle
of ∠HGE, and since ∠F GE ∼
= ∠BM A. Thus by uniqueness of C.4, the supplementary angle of ∠HGE is ∠F GE. Hence H ∗ G ∗ F .
Since ∠EHF ∼
= ∠ACB, then by SAS, 4ABC ∼
= 4EHF . And by definition of
congruent triangles, ∠BAC ∼
= ∠F EH. Proposition 5.3.13 (SSS) Given 4ABC and 4DEF . If AB ∼
= DE, BC ∼
=
EF , and AC ∼
= DF , then 4ABC ∼
= 4DEF .
Proof. Given 4ABC and 4DEF , AB ∼
= DE, BC ∼
= EF , and AC ∼
= DF .
←→
By corollary of SAS, on the opposite side of DE as F , there is a unique point,
say M such that 4ABC and 4DEM are congruent. Then BC ∼
= EM and
CA ∼
= DM , by CS.2 EF ∼
= EM and F D ∼
= DM .
←→
←→
Since F and M are in opposite sides of DE, F M intersects the line DE at a
point say P . Then either P = D, P = E, P ∗ D ∗ E, D ∗ E ∗ P , or D ∗ P ∗ E.
Case 1: If P = D, see the following figure.
Then by proposition 2.3.2 ∠EF M ∼
= ∠EM F . That is ∠DF E ∼
= ∠DM E. But
∠DM E ∼
∠ACB.
Then
by
CS.4,
we get ∠ACB ∼
=
= ∠DF E. Hence, by SAS
4ABC ∼
= 4DEF .
Case 2: If P = E, the proof is similar to case 1.
5.3. SPHERICAL GEOMETRY; AXIOMATIC APPROACH
127
Case 3. If P ∗ D ∗ E, see the following figure.
Then by proposition 2.3.2, ∠P F E ∼
= ∠P M E, and ∠P F D ∼
= ∠P M D. Thus by
∼
angle subtraction ∠DF E = ∠DM E. But ∠DM E ∼
= ∠ACB. Then by C.5, we
get ∠ACB ∼
= ∠DF E. By SAS, 4ABC ∼
= 4DEF .
Case 4: If D ∗ E ∗ P , the proof is similar to case 3.
Case 5: If D ∗ P ∗ E then F is in the interior of ∠DM E, and M in the interior
of ∠DF E.
By proposition 2.3.2, in the triangle 4F EM , we get ∠EF M ∼
= ∠EM F , and
128CHAPTER 5. AXIOM SYSTEM OF EUCLIDEAN 3-SPACE AND SPHERICAL GEOMETRY
in the triangle 4F DM , we get ∠DF M ∼
= ∠DM F . Then by angle addition,
we get that ∠DF E ∼
= ∠DM E. But ∠DM E ∼
= ∠ACB. The by CS.4, we get
∠ACB ∼
= ∠DF E. Hence, by SAS, 4ABC ∼
= 4DEF . Proposition 5.3.14 If l and m are two distinct lines, that are perpendicular
to a line n, then l and m intersect in a point, say P . And all segments from P
to the line n are perpendicular to n, and all such segments are congruent.
Proof: Given two distinct lines l and m that are perpendicular to a line n
at the points C and A respectively. By proposition 5.3.7 l and m intersect in
exactly two points,let P be one of them. Let P 0 be the opposite point of P , P 0
is the other common point of l and m by IS.2.
Let D be any point on n, we need to show that P D is perpendicular to n, and
PD ∼
= P A.
PA ∼
= P C, since in the triangle 4P AC, ∠P CA ∼
= ∠P AC, by proposition 5.3.9.
Also, by ASA 4P AC ∼
= 4P 0 AC. Hence, P A ∼
= P 0 A. Then 4P AD ∼
= 4P 0 AD,
by SAS.
Thus, ∠P DA ∼
= ∠P 0 DA, and hence, ∠P DA is a right angle.
Now, by proposition 5.3.7 4P DA is an isosceles triangle, that is P D ∼
= P A. Definition 5.3.9 The two opposite points, say P and P 0 at which two perpendiculars to a given line n intersect are called the poles of n, and n is called the
polar of P , and P 0 .
A segment from a pole to its polar is called a quadrant.
Proposition 5.3.15 All quadrant are congruent.
Proof. Let AB and EG be two quadrants, where A is the pole of the polar l,
and B at l. And E is the pole of the polar m, and G on m. Choose C on l and
F on m such that CB ∼
= F G. Since, ∠B, ∠C, ∠G, and ∠F are right angles
and hence are congruent, then by ASA 4ABC ∼
= 4EGF . Thus, AB ∼
= EG. Proposition 5.3.16 Let P be a point not on a line m, and let A and C be two
points on m, such that P A ∼
= P C,and are congruent to a quadrant, then P is
the pole of m.
Proof. Given a line m and a point P not on m, such that P A ∼
= P C, and are
congruent to a quadrant. Then the two perpendicular to m at A and at C meet
at a point say, P 0 , and are quadrants, since all quadrant are congruent, then
PA ∼
= P 0 A. By SSS, 4P AC ∼
= 4P 0 AC. Then ∠P AC ∼
= ∠P 0 AC. Thus, P is a
pole of m. 5.3. SPHERICAL GEOMETRY; AXIOMATIC APPROACH
129
Proposition 5.3.17 If three segments from a point to a line are congruent,
then they are quadrant.
Proof. Given 3 segments from a point P to a line m, that are congruent. Then
they are sides of two adjacent isosceles triangles, and hence are perpendicular.
See the figure. Proposition 5.3.18 Given any line l and any point P not on l. Then there
exists a perpendicular from P to l. If P is not a pole of l, then the perpendicular
from P to l is unique.
Proof. Given any line l and any point P not on l. Then by proposition 5.3.11,
there exists a perpendicular from P to l. To show uniqueness if P is not a pole.
Let AB be any segment on l. Let Q be a point on the other side of l from
P , such that ∠P AB ∼
= ∠QAB. Choose Q such that P A ∼
= QA. Then P Q
intersects l at a point, say S. Then by SAS, 4P AB ∼
= 4QAB. Then P Q is
perpendicular to l at S.
If there is another perpendicular from P to l, then by the previous definition,
P is a pole of l, which is a contradiction. −−→
Definition 5.3.10 A point B on a given ray P P 0 such that BP ∼
= BP 0 is called
the bisection-point of the ray.
A point P between A and C such that P A ∼
= P C, is called midpoint of the
segment AC.
−−→
−−→
−→
For any angle ∠BAC, if a ray AD is between the ray AB and the ray AC, such
−
−
→
that ∠CAD ∼
= ∠BAD, then AD is called a bisector of the angle angleBAC.
Lemma 5.3.1 Given any two opposite points P and P 0 , and r any ray emanating from P . Then there exists a point B ∈ r that is a bisection-point of
r.
Proof. Given any two opposite points P and P 0 , and r any ray emanating
from P . At any two points on r other than P there exists two perpendiculars
−−→
to r. These two perpendiculars meet at a point say Q. Let P C be another ray
−−→
emanating from O, and not part of the line through r. At any two points on P C
−−→
other than P there exists two perpendiculars to P C. These two perpendiculars
meet at a point say R. See the following figure.
130CHAPTER 5. AXIOM SYSTEM OF EUCLIDEAN 3-SPACE AND SPHERICAL GEOMETRY
←→
−→
The line RQ meets r at a point say, B. And meets AC at a point, say D.
←→
Since Q is a pole of the line containing r, and R is a pole of the line P C, then
the angle at B and the angle at D are right, and hence are congruent. Then by
ASA, 4P BD ∼
= 4P 0 BD. Thus, P B ∼
= P 0 B. Proposition 5.3.19 If P and P 0 are two opposites and A, C any two points
distinct from P and P 0 , and not both on the same line with P and P 0 . Then
∠AP C ∼
= ∠AP 0 C.
Proof. Let P and P 0 be two opposites and A, C any two points distinct from
−→
P and P 0 , and not both on the same line with P and P 0 . Bisect the ray P A at
−−→
the point B, and the ray P C at the point D. Then by SSS 4P BD ∼
= 4P 0 BD,
0
hence ∠AP C ∼
∠AP
C.
=
Proposition 5.3.20 In an isosceles triangle, the bisector of the angle between
two congruent sides of the triangle bisects the third side at right angle.
Proof. Given the isosceles triangle 4ABC, in which AC ∼
= AB. Let M be the
midpoint of BC. Then by SSS, 4ADB ∼
= 4ADC. Then, ∠BAD ∼
= ∠CAD,
and the two supplementary angles ∠ADC and ∠ADB are congruent and hence,
−−→
are right angles. Then AD is a bisector of the angle ∠BAC, and it bisects BC
at a right angle. Proposition 5.3.21 Given any two lines l and m with A and B as common
points, if n is a line through the poles of l and m, then n is the polar of A and
B.
Proof. Given any two lines l and m with A and B as common points, let n be
a line through the poles of l and m, which are P and Q respectively. Enough
to show that n is the polar of A.
Since A on l, and P is a pole of l, then P A is a quadrant. And, since A on
m and Q is a pole of m, then QA is also a quadrant. Thus, P A and QA are
←→
quadrant, and n = P Q, then by previous proposition, n is a polar of A. Corollary 5.3.2 The line through the poles of two lines is perpendicular to
both.
5.3. SPHERICAL GEOMETRY; AXIOMATIC APPROACH
131
Definition 5.3.11 Given two triangles 4ABC, and 4A0 B 0 C 0 , such that A0 is
a pole BC on the same side of BC as A, B 0 is a pole AC on the same side of
AC as B, and C 0 is a pole BC on the same side of AB as C . Then 4A0 B 0 C 0
is called the polar triangle of 4ABC.
Proposition 5.3.22 If 4A0 B 0 C 0 is the polar triangle of 4ABC, then 4ABC
is the polar triangle of 4A0 B 0 C 0 .
Proof. Let 4A0 B 0 C 0 be a polar triangle of 4ABC. Since B 0 is a pole of AC,
then B 0 A is a quadrant, and since C 0 is a pole of AB, then C 0 A is a quadrant.
Thus, A is a pole of B 0 C 0 . In the same manner, we conclude that, B is a pole
of A0 C 0 , and C is a pole of A0 B 0 .
Moreover, since by hypothesis A0 and A are on same side BC, and A0 is the
pole of BC, then A0 A is less than a quadrant. Therefore, A0 and A are on the
same side of B 0 C 0 . In the same manner, we show that B 0 and B are on the
same side of A0 C 0 , and C 0 and C are on the same side of A0 B 0 . Hence, 4ABC
is the polar triangle of 4A0 B 0 C 0 . Definition 5.3.12 Two segments AB and CD are called supplemental,if they
are congruent to two segments made by a point on a ray, say r with its endpoints.
←−→
That is, if r is a part of the line P P 0 , emanating from P , and G ∈ r, such that
AB ∼
= P G and CD ∼
= GP 0
Proposition 5.3.23 The supplements of congruent segments are congruent.
∼ CD, EF be the supplement of AB, and HJ be the supProof. Let AB =
−−→
−−→
plement CD. Then there exists a ray P P 0 , and a point G ∈ P P 0 , such that
−
−
→
−−→
AB ∼
= P G and EF ∼
= GP 0 . Also there exists a QQ0 , and a point on K ∈ QQ0 ,
such that QK ∼
= CD, and KQ0 ∼
= HJ. Then GP ∼
= QK. Enough to show
0
0
that P G ∼
= Q K. If not that is P 0 G > Q0 K or P 0 G < Q0 K, we should get a
contradiction. Consider the case P 0 G > Q0 K. Then there exists I such that
−−→
P 0 ∗ I ∗ G and GI ∼
= KQ0 . Let M and N be the bisection points of P P 0 , and
−−→0
QQ respectively.
If G = M then we are done.
If P ∗ G ∗ M , then since quadrants are congruent, GM ∼
= KN , then M I ∼
= Q0 N ,
which is a contradiction.
If P ∗ M ∗ G, then M G ∼
= KN , and since P 0 M ∼
= Q0 N , then M I ∼
= Q0 N , which
is a contradiction. Lemma 5.3.2 In a pair of polar triangles, any angle of either triangles intercepts, on the side of the other triangle which lies opposite it, a segment which
is supplement of that side.
Proof. Let 4ABC and 4A0 B 0 C 0 be two polar triangles. Let D and E be
−−→
−→
respectively the points where rays AB and AC meet B 0 C 0 . Since B 0 is the pole
of AC, then B 0 E is a quadrant, and since C 0 is a pole of AB, then C 0 D is a
quadrant. Since C 0 D and B 0 E are two quadrants, and B 0 E +C 0 D = B 0 C 0 +ED.
Thus ED is the supplement of B 0 C 0 . 132CHAPTER 5. AXIOM SYSTEM OF EUCLIDEAN 3-SPACE AND SPHERICAL GEOMETRY
Proposition 5.3.24 AAA For any two triangles, that have three angles of
respectively congruent to three angles of the other are congruent triangles.
Proof. Let 4ABC and 4DEF be two triangles in which ∠A ∼
= ∠D, ∠B ∼
= ∠E,
0 0 0
∼
and ∠C = ∠F . Let 4A B C be the polar triangle of 4ABC, and 4D0 E 0 F 0
be the polar triangle of 4DEF . See the figure.
Claim: If 4ABC and 4DEF are equiangular, then their polar triangles 4A0 B 0 C 0
and 4D0 E 0 F 0 are equilateral.
By SAS, congruent angles at the poles of lines, intercepts congruent segments
on those lines, since all quadrants are congruent. By the previous lemma, these
congruent segments are the supplements of corresponding sides of the polar triangles. Since the supplements of congruent segments are congruent, then these
polar triangles have three sides respectively congruent.
Since the polar triangles 4A0 B 0 C 0 and 4D0 E 0 F 0 are equilateral then by definition of congruent triangles, are equiangular. Thus, since the claim is true, their
polar triangles, which are 4ABC and 4DEF , are equilateral. Therefore are
congruent by SSS. •H
HH
HH
•
•
Appendix: Some results in Euclidean Geometry
Theorem 5.3.1 In Euclidean geometry, if the two triangles 4ABC and 4A0 B 0 C 0
are similar then the corresponding sides are proportional.
Proof. Without loss of generality, we suppose that A0 B 0 < AB. Let D ∈ AB
−→
such that AD ∼
= A0 B 0 and E ∈ AC such that AE ∼
= A0 C 0 . Since ∠B ∼
= ∠D, and
then the line through D and E is parallel to side BC and E is between A and
AD ∼ AE
C. Now the corollary above implies that
. And since AD ∼
=
= A0 B 0
AB
AC
5.3. SPHERICAL GEOMETRY; AXIOMATIC APPROACH
133
−−→
A0 B 0 ∼ A0 C 0
. Now consider F ∈ BC such that
and AE ∼
= A0 C 0 we have
=
AB
AC
BF ∼
= B 0 A0 . Similarly, one obtains that
= B 0 C 0 and D0 ∈ AB such that BD0 ∼
0
0
0
0
0
BF
BD
BC
AB
and hence
. =
=
BA
BA
BC
BC
Proposition 5.3.25 In Euclidean geometry, the shortest segment from a point
P to a line l is the perpendicular from P to l.
Proof. Given any line l and any point P not on l. Let P S be the shortest
segment from P to l. Let P Q be the perpendicular from P to l at the point Q.
Consider the triangle 4P QS. Then angle ∠P QS is right and hence the other
two angles are acute angles. Then the greater angle is opposite to greater side.
Then P S > P Q which is a contradiction since P S > P Q unless S = Q. Proposition 5.3.26 In Euclidean geometry, let γ be a circle with radius r and
centre O. Then a line t is tangent to γ at the point P if and only if OP ⊥ t at
P.
Proof. Given a circle γ with centre O and radius r, and given a line t.
=⇒ Let the line t be tangent to γ at the point P . If S is any other point on
t. Then S lies outside γ, that is OS > r = OP . That is, OP is the shortest
distance from O to the line t. And hence, OP ⊥ t.
⇐= Let OP ⊥ t at the point P . If t is not tangent to γ. Then t intersects
γ at other points say, R. Then 4P RO is a triangle with OP ∼
= OR, Then
∠OP R ∼
= ∠ORP . Hence this triangle has two right angles which is a contradiction. Proposition 5.3.27 In Euclidean geometry, if AB is a chord of a circle γ, and
C is the intersection of the tangent lines to γ at A and B. Then 4ABC is an
isosceles.
Proof. Exercise.
Proposition 5.3.28 Given a circle γ with centre O. If A, B and C are three
1
distinct points on γ, then (∠BAC)◦ = (∠BOC)◦ .
2
Proof. First, we consider the case when one side of the angle ∠BAC is a diameter, say AC. See the figure.
134CHAPTER 5. AXIOM SYSTEM OF EUCLIDEAN 3-SPACE AND SPHERICAL GEOMETRY
In the triangle 4OBA is an isosceles triangle, then ∠ABO ∼
= ∠OAB. By exterior angle theorem, (∠BAD)◦ = (∠ABO)◦ + (∠BAO)◦ = 2(∠BAO)◦ . Thus,
1
(∠BAC)◦ = (∠BOC)◦ .
2
Now, we suppose that neither of its sides is a diameter and consider a point
D ∈ γ such that AD is a diameter. Then, the first part of this proof implies
that
(∠BOD)◦ = 2(∠BAD)◦ (∠DOC)◦ = 2(∠DAC)◦ .
−−→
−→
−−→
−−→
−→
−−→
Now, either AD is between AC and AB, or AD is not between AC and AB.
−−→
−→
−−→
Case 1: Let AD is not between AC and AB. Without loss of generality, assume
−→
−−→
−−→
that AC is between AB and AD. See the figure.
1
Then, (∠BAD)◦ = (∠BOD)◦
2
1
◦
(∠CAD) = (∠COD)◦
2
1
1
(∠BAC)◦ = (∠BAD)◦ − (∠CAD)◦ = [(∠BOD)◦ − (∠COD)◦ ] = (∠BOC)◦
2
2
−−→
−→
−−→
Case 2: Assume that AD is between AC and AB. Then (∠BAC)◦ =
1
(∠BAD)◦ +(∠DAC)◦ . Therefore, (∠BAC)◦ = [(∠BOD)◦ +(∠DOC)◦ ]. Then
2
(∠COB)◦ = 2(∠BAC)◦ . Proposition 5.3.29 Given a circle γ with centre O. If A and B are two dis−−→
tinct points on γ. Let t be the tangent to γ at B. Let P ∈ t such that OP lies
−→
−−→
1
between the rays OA and OB. Then (∠P BA)◦ = (∠AOB)◦
2
Proof. Given a circle γ with centre O. If A and B are two distinct points on γ.
−−→
−→
Let t be the tangent to γ at B. Let P ∈ t such that OP lies between the rays OA
−−→
and OB. Since BO ⊥ P B, we have (∠P BA)◦ = 90◦ − (∠ABO)◦ . By previous
1
proposition, (∠ABO)◦ = (∠AOD)◦ . Since (∠AOD)◦ = 180◦ − (∠AOB)◦ ,
2
1
substituting above, we get that (∠P BA)◦ = (∠AOB)◦ . 2
Proposition 5.3.30 Given a circle γ with centre O. Let t be the tangent to γ
at A. Let BC be a chord in γ, and let P ∈ t such that C ∗ B ∗ P . Then 4P CA
is similar to 4P AB.
Proof. Given a circle γ with centre O. Let t be the tangent to γ at A. Let BC
be a chord in γ, and let P ∈ t such that C ∗ B ∗ P . See the figure.
5.3. SPHERICAL GEOMETRY; AXIOMATIC APPROACH
135
The two triangle have a common angle, hence, it is enough to find another
pair of congruent angles. We show that ∠ACP ∼
= ∠P AB. By previous propo1
◦
◦
sition, (∠P AB) = (∠AOB) , and by the other proposition (∠ACB)◦ =
2
1
(∠AOB)◦ . Thus, we get the result. 2
Proposition 5.3.31 Any three distinct non-collinear points determine a unique
circle.
Proof. Let A, B and C be three distinct non-collinear points. First, in the
triangle 4ABC we prove that the perpendicular bisectors of the sides of 4ABC
all meet at a single point.
Let l and m be the perpendicular bisector of segments AB and BC respectively.
We claim that l and m meet at some point O. In fact, suppose that l k m. Since
AB ⊥ l, then AB ⊥ m. Since BC ⊥ m, then alternate interior angle theorem
←→ ←→
implies that AB k BC, which is a contradiction, since they both meet at B.
Now, we claim that OA ∼
= OB ∼
= OC. Assume that l meet AB at L, and m
meet BC at M . Note that by SAS, 4AOL ∼
= 4BOL, and hence OA ∼
= OB.
Also, 4BOM ∼
= 4COM , by SAS, which implies that OB ∼
= OC. See the
figure.
←→
Let N be the midpoint of AC. To complete the proof we need to show that ON
←→
is perpendicular bisector of AC, that is, ON ⊥ OC. Since 4AON ∼
= 4CON
by SSS, we obtain
(∠ON C)◦ = (∠ON A)◦ = 90◦ .
Second, the circle with centre O and radius OA pass through the three points
A, B, and C.
Uniqueness: To show that there is a unique circle passing through A, B and
C. Assume there is another circle, say γ 0 with centre O0 and radius r0 , passing
through A, B and C. Then
r0 = O0 A = O0 B = O0 C
136CHAPTER 5. AXIOM SYSTEM OF EUCLIDEAN 3-SPACE AND SPHERICAL GEOMETRY
Since O0 A = O0 B, then O0 lies on the perpendicular bisector of AB.
Since O0 C = O0 B, then O0 lies on the perpendicular bisector of CB.
Since O0 A = O0 C, then O0 lies on the perpendicular bisector of AC. Hence the
three perpendicular bisectors are concurrent at O0 . Since this point is unique,
then O0 = O. Otherwise, the three perpendicular bisectors are equal, and hence
the lines containing the sides of the triangle are parallel, since they have a
common perpendicular, which is a contradiction. Proposition 5.3.32 Given a circle γ and a point P outside γ, then there is
two tangent lines to γ through P .
Proof. Given a circle γ with centre O and a point P outside γ. We want to
←→
find two points Q and R on γ such that the line P Q is tangent to γ, and the
←→
line P R is tangent to γ. Let M be the midpoint of the segment P O. Consider
the circle δ with centre M and radius M P . Then P lies outside the circle γ,
and the point O lies inside the circle γ. Then by circular continuity principle,
γ and delta intersects at two points, say Q and R. See the figure.
Now, ∠P QO and ∠P RO are right angles by previous proposition. Then both
lines are tangent to the circle γ. Let γ and δ be any two circles. Then only one of the followings occur:
i. γ ∩ δ = ∅,
ii. the two circles tangent to each other,
iii. the two circles intersect at two points,
iv. or γ = δ
5.3. SPHERICAL GEOMETRY; AXIOMATIC APPROACH
137