Euclid's Elements, from Hilbert's Axioms

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Euclid’s Elements, from Hilbert’s Axioms

THESIS

Presented in Partial Fulfillment of the Requirements for the Degree Master of Science in

the Graduate School of The Ohio State University

Peter James Ward

Graduate Program in Mathematics

The Ohio State University

2012

Master's Examination Committee:

Professor Rodica Costin, Advisor

Professor Ronald Solomon

Peter James Ward

2012

Abstract

This project is an exposition of Book I of Euclid’s Elements consistent with modern mathematical rigor. This text is designed to serve as a first introduction to geometry, building from Hilbert’s axioms the tools necessary for a thorough investigation of planar geometry.

Vita

May 2006 ...... Christian Brothers Academy

2010...... B.A. Mathematics, Rutgers University

2010 to present ...... Graduate Teaching Associate, Department

of Mathematics, The Ohio State University

Fields of Study

Major Field: Mathematics

iii

Table of Contents

Abstract ...... ii

Vita ...... iii

Table of Contents ...... iv

List of Figures ...... vi

0. Introduction ...... 1

I. Axioms of Incidence ...... 10

II. Axioms of Order ...... 13

III. Axioms of Congruence of Segments ...... 34

IV. Axioms of Congruence of Angles ...... 41

V. Triangles ...... 58

VI. Axiom of Circles ...... 80

VII. Axiom of Parallels ...... 84

VIII. Archimedes Axiom and Area ...... 94

IX. Axiom of Completeness ...... 116

References ...... 117

Appendix A: List of Axioms ...... 119

Appendix B: A Note to the Instructor ...... 121

List of Figures

FIGURE 1: FINITE GEOMETRIES ...... 12

FIGURE 2: O IS BETWEEN A AND B ...... 13

FIGURE 3: O IS BETWEEN ...... 21

FIGURE 4: LINE AND SEGMENT ...... 14

FIGURE 5: DIAGRAMS WHICH SATISFY AXIOM II.4 ...... 14

FIGURE 6: DIAGRAMS WHICH DON’T SATISFY AXIOM II.4 ...... 15

FIGURE 7: THEOREM 2, CASE 1 ...... 18

FIGURE 8: THEOREM 2, CASE 2 ...... 19

FIGURE 9: THEOREM 2, CASE 1 ...... 20

FIGURE 10: THEOREM 2, CASE 2 ...... 20

FIGURE 11: ON THE LEFT, A AND B ARE ON THE SAME SIDE OF A; ON THE RIGHT, A AND B ARE ON DIFFERENT SIDES OF A ...... 21

FIGURE 12: ON THE LEFT, A AND B ARE ON THE SAME SIDE OF O; ON THE RIGHT, A AND B ARE ON DIFFERENT SIDES OF O ..... 21

FIGURE 13: RAY ...... 21

FIGURE 14: INTERIOR AND EXTERIOR OF AN ANGLE ...... 22

FIGURE 15: THEOREM 3 ...... 25

FIGURE 16: COROLLARY TO THEOREM 3 ...... 25

FIGURE 17: THEOREM 4 ...... 26

FIGURE 18: THEOREM 5 ...... 27

FIGURE 19: THREE BROKEN LINES ...... 28

FIGURE 20: THREE POLYGONS. THE TWO POLYGONS ON THE RIGHT ARE SIMPLE POLYGONS...... 29

FIGURE 21: INTERIOR AND EXTERIOR OF A POLYGON ...... 29

FIGURE 22: CROSSBAR THEOREM ...... 30 vi

FIGURE 23: THEOREM 7 ...... 31

FIGURE 24: EXERCISE 5 ...... 33

FIGURE 25: AB ≡ CD ≡ EF, AND GH ≡ JK...... 35

FIGURE 26: AXIOM III.3 ...... 35

FIGURE 27: AB + CD ≡ AE ...... 36

FIGURE 28: AB – CD ≡ AE ...... 37

FIGURE 29: AB < CD < EF < GH < JK < LM ...... 38

FIGURE 30: COROLLARY TO THEOREM 8 ...... 40

FIGURE 31: ∡ABC ≡ ∡DEF AND ∡GHJ ≡ ∡JHK ≡ ∡KHL ...... 42

FIGURE 32: CONGRUENT TRIANGLES ...... 43

FIGURE 33: THEOREM 9 ...... 44

FIGURE 34: SUPPLEMENTARY ANGLES ...... 45

FIGURE 35: THEOREM 10 ...... 46

FIGURE 36: VERTICAL ANGLES ...... 47

FIGURE 37: COROLLARY TO THEOREM 10 ...... 47

FIGURE 38: THEOREM 11 ...... 49

FIGURE 39: ∡ABC + ∡DEF ≡ ∡ABG ...... 50

FIGURE 40: ∡ABG - ∡DEF ≡ ∡ABC ...... 51

FIGURE 41: ∡ABC < ∡DEF ...... 52

FIGURE 42: THEOREM 12 ...... 53

FIGURE 43: RIGHT ANGLES ...... 54

FIGURE 44: THEOREM 13 ...... 54

FIGURE 45: THEOREM 14 ...... 56

FIGURE 46: THE LEFT ANGLE IS ACUTE, WHILE THE RIGHT ANGLE IS OBTUSE ...... 56

FIGURE 47: ABC IS OBTUSE, DEF IS A RIGHT TRIANGLE, AND GHI IS ACUTE...... 59 vii

FIGURE 48: ABC IS EQUILATERAL, DEF IS AN ISOSCELES TRIANGLE, AND GHI IS SCALENE...... 59

FIGURE 49: THEOREM 17 ...... 61

FIGURE 50: THEOREM 18 ...... 63

FIGURE 51: THEOREM 18, CASE 3 ...... 64

FIGURE 52: COROLLARY TO THEOREM 18 ...... 65

FIGURE 53: THEOREM 19 ...... 65

FIGURE 54: THEOREM 20 ...... 66

FIGURE 55: THEOREM 22 ...... 68

FIGURE 56: EXTERIOR ANGLE ∡CBD ...... 70

FIGURE 57: THEOREM 23 ...... 70

FIGURE 58: THEOREM 24 ...... 72

FIGURE 59: THEOREM 25 ...... 73

FIGURE 60: THEOREM 26 ...... 74

FIGURE 61: THEOREM 27 ...... 75

FIGURE 62: G CANNOT BE EXTERIOR TO ∡DEF ...... 76

FIGURE 63: D CANNOT BE EXTERIOR TO ∡EFG ...... 76

FIGURE 64: THEOREM 27 ...... 77

FIGURE 65: THEOREM 29 ...... 78

FIGURE 66: THEOREM 30 ...... 81

FIGURE 67: THEOREM 31 ...... 83

FIGURE 68: THEOREM 32 ...... 85

FIGURE 69: COROLLARY TO THEOREM 32 ...... 86

FIGURE 70: THEOREM 33 ...... 87

FIGURE 71: THEOREM 34 ...... 87

FIGURE 72: AXIOM VII ...... 88

viii

FIGURE 73: COROLLARY TO AXIOM VII ...... 89

FIGURE 74: THEOREM 35 ...... 90

FIGURE 75: THEOREM 36 ...... 91

FIGURE 76: THEOREM 37 ...... 92

FIGURE 77: THEOREM 39, WITH ORDER AEDF ...... 96

FIGURE 78: THEOREM 39, WITH ORDER ADEF ...... 96

FIGURE 79: THEOREM 40 ...... 97

FIGURE 80: THEOREM 41 ...... 98

FIGURE 81: THEOREM 42 ...... 99

FIGURE 82: THEOREM 43 ...... 100

FIGURE 83: THEOREM 44 ...... 101

FIGURE 84: THEOREM 45 ...... 102

FIGURE 85: THEOREM 46 ...... 103

FIGURE 86: THEOREM 48 ...... 105

FIGURE 87: THEOREM 49 ...... 106

FIGURE 88: THEOREM 50 ...... 108

FIGURE 89: THEOREM 51 ...... 110

FIGURE 90: THEOREM 52 ...... 112

FIGURE 91: THEOREM 53 ...... 114

0. Introduction

There are two major goals of this book: to present and justify many geometric facts and theorems1 from a small set of axioms, and to develop the skill of logical reasoning. We develop these skills as we deduce those geometrical facts; by stating facts that we intuitively think should be true, and working out proofs for them in a logical and natural way.

Our inspiration is Euclid’s Elements,2 a work by the Greek Euclid of Alexandria, written around the year 300BC. Euclid began by assuming the following five postulates, or axioms:

1. A line can be drawn connecting any two points.

2. A line can be extended as far as desired.

3. A circle can be drawn with any center and any radius.

4. All right angles are equal.

5. If a line crosses two other lines and makes the internal angles on the same side

sum to less than two right angles, the two lines will meet on that side.

1 In mathematics, a theorem is a statement that can be proven true. 2 Joyce 1

From these five axioms, Euclid deduced an amazing number of interesting and useful geometric facts. The Elements would serve as the beginning (and in some sense, the culmination also) of the study of geometry until the 19th century. The significant advances in mathematics in the intervening 2200 years led to the discovery of geometries with properties different than that which Euclid deduced. While these new geometries rejected one (or more) of Euclid’s five axioms, the development of these geometries caused mathematicians to analyze more closely the implicit assumptions Euclid used in addition to his axioms. In 1899, David Hilbert presented a new axiomatization of

Euclidean geometry in the Foundations of Geometry;3 in order to fill the gaps in Euclid’s proofs, Hilbert required 20 axioms in all!

This text seeks to recreate the work of Euclid’s Elements, but beginning from

Hilbert’s axioms. We will concern ourselves only with the geometry of a plane; i.e. a flat surface with only two directions. Thus, we require only the 17 axioms of Hilbert’s planar geometry. We will introduce these axioms in groups, and then derive the theorems that follow.

The first four sections of this text will lay the groundwork of our geometry. We will derive results that allow us to add, subtract, and order segments and angles. These sections will mostly follow the outlines suggested by Hilbert and expanded on in the text

Geometry: Euclid and Beyond by Robin Hartshorne. Sections five, seven, and eight will then proceed in much the same manner as Euclid’s Elements; in many cases, the proofs

3 Hilbert pgs. 2-15 2 will be essentially Euclid’s, updated and made rigorous. Section six introduces circles, and completes the proofs of Euclid we skipped in section five.

This text contains a large number of proofs, which should serve as models on how to completely and rigorously prove a statement. In general, theorems can be presented as if-then statements: If P, then Q, where P is our hypothesis and Q is our conclusion. When theorems are not presented in this manner, you should translate the statement of the theorem into such a statement. Then, in every proof, we should begin by stating our hypothesis P, and stating the conclusion Q that we wish to work towards. Though we will often be repeating or paraphrasing the statement of the theorem exactly, it is still important to include as part of the proof. For one reason, we wish the proofs to stand alone; for another, it will serve as reminder of what exactly we know and what exactly we are trying to show.

Our primary method of proof is a proof by deduction. We begin by stating the hypothesis P of the problem, and proceed step-by-step, justifying each step by a stated hypothesis, an axiom, or a previously proved statement, until we arrive at our conclusion

Q. This is the same method we use to develop the entire text. We will state our axioms, which are principles we will assume to always be true. They may appear to be intuitively clear or obvious, but we must state them first. We will then use those axioms to define various geometric objects and to prove theorems about those objects. In this way, our axioms are just our hypotheses, and our theorems are our conclusions.

For an example of a proof by deduction, suppose that we take our set of axioms to be the rules of baseball, and the properties of addition. Now suppose that in certain baseball game, the home team is losing by one run in the bottom of the ninth inning, John is at bat, and there is one runner on base. We will prove the following theorem: If John hits a home run, the home team will win. So assume that John hits a home run (P). Since

John is batting in the bottom of the ninth inning, both he and the base runner play for the home team (according to the rules of baseball). Then John and the base runner each score one run (according to the rules of baseball). Since the home team had one run less than the visiting team, and they scored two more runs, the home team has one run more than the visiting team (according to the properties of addition). Then since the home team wins the game if it has scored more runs than the visiting team at any point in the bottom of the ninth inning (according to the rules of baseball), we conclude that the home team will win (Q). But that is exactly what we wanted to show, so the theorem is true.

There are additional methods of proof which we will occasionally use. One method of proof is the proof by contradiction, or reductio ad absurdum. We begin by taking our hypothesis P, and the opposite of the conclusion Q, and then use this assumption, along with the hypothesis, our axioms, and any theorems we have already proven, to show that this creates a contradiction. Whether it contradicts the hypothesis, or an axiom, or a theorem, or the conclusion that we assumed does not matter; in any case, we can conclude that the conclusion must follow. That is, we show that P being true and

Q being false cannot possibly happen; hence, if P is true, Q must also be true.

For example, suppose that we take our set of axioms to be the properties of arithmetic, and we will prove the following theorem: There is no largest natural number.

Note that we do not have any hypothesis P; our theorem is of the form “Q must be true”.

In order to derive a contradiction, suppose there is a largest natural number (our assumption), and call this number n. Then n + 1 is a natural number, and n + 1 > n (by properties of arithmetic). So n + 1 is a larger natural number than n, contradicting our assumption that n was the largest. Thus, our assumption must be false, and there is no largest natural number. But that is exactly what we wanted to show, so the theorem is true.

Another method of proof is the proof by induction. If we have a statement that depends on a number n, we can prove that it is true for every natural number (counting number) n by using induction. We make a statement S(n) in terms of a natural number n.

We show that the statement is true for some base case, usually the number 1. Then we make our induction assumption that S(n-1) is true, and use this to show that S(n) is also true. Once we have done this, we can conclude it is true for every number n greater than or equal to our base case.

For example, suppose that we take our set of axioms to be the properties of algebra, and we will prove the following theorem: The sum of the first n natural numbers is n(n+1)/2. Let S(n) be the statement “the sum of the first n natural numbers is n(n+1)/2; i.e. 1+2+3+…+n = n(n+1)/2.” For the base case, when n = 1, the statement becomes 1 =

1(1+1)/2, which is true. Now assume that S(n-1) is true. Then the sum of the first n

5 positive integers is 1+2+3+…+n-1+n = the sum of the first n-1 positive integers + n, so by the inductive assumption, = (n-1)((n-1)+1)/2 + n = (n-1)(n)/2 + 2n/2 = (n2-n+2n)/2 =

(n2+n)/2 = n(n+1)/2. Thus S(n) is true. By induction, S(n) is true for every natural number n.

One last method of proof worth mentioning here is proof by contraposition. As above, we are trying to show that if P is true, then Q is true. This is logically the same as showing that if Q is false, then P is false. This reordering of the theorem, if not Q then not

P, is called the contrapositive of the theorem. If we prove the contrapositive of a theorem, since that is logically the same as the original theorem, we have also proved the theorem.

For example, suppose that we take our set of axioms to be the properties of algebra, and we will prove the following theorem: If (x+1)2 is odd, then x is even. The contrapositive of the theorem is the statement: if x is not even, then (x+1)2 is not odd; that is, if x is odd, then (x+1)2 is even. So suppose that x is odd. Then x = 2k - 1 for some integer k (the definition of odd). Then x+1 = 2k, so (x+1)2 = (2k)2 = 4k2 = 2(2k2) (by properties of algebra). As 2k2 is an integer, (x+1)2 = 2(2k2) is even (the definition of even). We have shown the contrapositive is true, and thus the original theorem is also true.

Note that the contrapositive is not the same as the converse, or the reverse of the theorem. If a theorem is “if P, then Q”, the contrapositive is “if not Q, then not P”, while the converse is “if Q, then P”. The theorem and its contrapositive are logically equivalent;

6 proving one proves the other. The theorem and its converse are independent; both may be true, both may be false, or one may be true while the other is false.

For example, in the real numbers consider the theorem “if x > 1, then x2 > 1.”

Then the theorem is obviously true, as a basic algebraic fact. The converse “if x2 > 1, then x > 1” is false, which we can see by the counterexample x = -2. Then x2 = 4 > 1, but x = -

2 < 1.

Incidentally, the preceding example is the last instance of algebra in this book. In fact, it will be the last time we use numbers in this book for anything other than as labels.

Our study and development of geometry will be completely geometric, and will not rely on any prior mathematical knowledge.

The methods of proof discussed above are not restricted to the study of geometry.

They are valid and useful in all of mathematics, and more generally in any system which follows the laws of classical logic. In almost all instances, we accept that the real world obeys the laws of classical logic, so our proofs should be valid in any context. With these methods of proof in mind, we can begin our study of geometry.

We would like to begin by defining all the objects and terms that we will encounter. In any complete system, it turns out this is impossible, because we must define each term using other terms, and ultimately we will finish with circular definitions, or we will never finish. To avoid this, we begin with undefined objects: the point and the line.

Points will be denoted by capital letters A, B, C, … and lines will be denoted by bold lowercase letters a, b, c, …, or by two points on the line ��, ��, ��, …We think of a

7 point being a “shape” with no “dimension;” or as similar to how Euclid defined a point, a

“figure” with no “width, length, or height.” Similarly, we think of a line being a “shape” with only one “dimension;” or a collection of all the points that “lie straight with one another”. Obviously, the terms in quotations have not been defined yet, and when we ultimately do, they will include the use of the terms point and line in their definitions.

However, we can think of points and lines to be the objects we normally think of when we use the words point and line, and no harm will be done.

Similarly, we have four undefined relations: a point may be between two others, a line may contain a point, two segments may be congruent, and two angles may be congruent. We will be able to define the terms segment and angle in terms of points, lines, betweenness, and containment, after we have given the axioms which characterize those undefined terms. It is worth noting that we use the same word congruent to express two different relations, the congruence of segments and the congruence of angles, because they express similar ideas. These four relations are all undefined, and undefinable, because we do not have any other terms with which to define them. We will, however, define all the properties of these relations satisfy. Again, no harm will come from thinking of these relations as follow:

A point A is between two other points B and C if all three are on the same line, and B and C are on “opposite sides” of A.

A line a contains a point A if A “lies on” a, or if A is “a part of” a.

Two segments are congruent if they have the same “length.”

Two angles are congruent if they have the same “angle measure.”

We next will assume the truth of 17 axioms, taken in groups. These axioms may seem obviously true in the real world, but in fact they are just defining the properties of the four undefined relations we listed above. From these 17 axioms, we will develop the geometry of the plane. We use these axioms to prove various theorems which constitute

Euclidean geometry; the geometry of the plane which we are most familiar with. Many of the early theorems will also seem obvious, but in order to be complete we will prove all of these theorems. Proving these early theorems is important, both because we will use these theorems in later proofs, and also because it will offer opportunities to practice using our methods of proof.

I. Axioms of Incidence

The first axioms we present are the Axioms of Connection, or of Incidence. These axioms give meaning to the relation of containment. We may think of containment to mean a line contains a point if the point is a part of the line, and similarly a point contains a line if the line passes through the point, but the formal definition of containment is any relation that satisfies the following three axioms.

Axiom I.1. Two points A and B determine a unique line a. We write �� = a or �� = a.

We will often use other terminology to refer to the same relation of containment; for instance, A lies upon a, A is a point of a, a contains A, a goes through A and through B, a joins A and B, etc.

Axiom I.2. Any two points of a line determine that line; that is, if �� = a and ��= a, where B ≠ C, then also �� = a.

Axiom I.3. Every line contains at least two points, and there are at least three points not all on the same line.

From these three axioms, we can state and prove our first theorem.

Theorem 1- Any two lines have at most one point in common.

Proof- Suppose the two lines a and b, meet at two distinct points A and B, and we show a = b. As two points determine a line, and A and B lie on a, �� = a (I.1). But similarly A and B lie on b, so �� = b (I.1). Thus a = �� = b, and a and b are the same line. □4

Exercise 1- A geometry is a system of points and lines. Throughout the rest of the text we will be defining and studying one possible geometry, Euclidean geometry; however, there are other possible geometries that we could define, some of which have practical applications. There are many possibly geometries that satisfy the axioms I.1, I.2, and I.3.

We know much too little to say exactly what our geometry looks like. Our geometry could even be a finite geometry; a geometry with only a finite number of points and lines.

Examine each finite geometry (collection of points and lines) below, and verify that they satisfy all three axioms of connection, or state which axioms they fail to satisfy.

4 The symbol □ is used to denote the end of a proof. 11

Figure 1: Finite Geometries

II. Axioms of Order

The next axioms we present are the Axioms of Order. These axioms give meaning to the relation of between. We may think of between to mean a point O is between two other distinct (different) points A and B if all three are on the same line, and the line passes first through A, then through O, and finally through B. Again, the formal definition of between is any relation that satisfies the following four axioms.

Figure 2: O is between A and B

F i Axiom II.1. If A, B, O are three distinct points of a line, then O is between A and B if g u and only if O is also between B and A. Note that we require all three points to be distinct; r e thus a point A cannot be between A and any other point B.

Axiom II.2. If A and C are two distinct points of a line, then there is at least one point B O lying between A and C and at least one point D with C between A and D. i s

b Axiom II.3. Given three distinct points of a line, one and only one lies between the other e t two. w e

e n 13

With this concept of betweenness we define new objects. A segment AB of the line �� = a is the collection of all points C on the line a with C between A and B, along with the points A and B. Note that AB = BA (II.1). The endpoints of the segment AB are the points A and B.

Figure 4: Line and Segment

Axiom II.4. Let A, B, C be three points not lying on the same line, and let a be a line that does not pass through any of the points A, B, C. Then if the line a passes through a point of the segment AB, it also passes through a point of the segment BC or through a point of the segment AC, but not both.

Figure 5: Diagrams which satisfy Axiom II.4 Each of the above diagrams satisfy II.4. In the first, a passes through a point of

AB and through a point of BC. In the second, a passes through a point of AB and through a point of AC. In the third, a does not pass through any points of any of the segments AB,

AC, or BC.

Figure 6: Diagrams which don’t satisfy Axiom II.4 Neither of these two diagrams satisfy II.4. In the first, a passes through two points of the segment AB. In the second, a passes through a point of AB but does not pass through any points of the segments AC or BC.

The previous axiom, II.4, is known as Pasch’s Axiom.5 It is equivalent to saying our geometry is two dimensional. Everything is occurring in one plane; there are no other planes or dimensions. Thus we can represent our geometry by diagrams on a sheet of paper without adding any ambiguity and without losing any information (so long as we imagine our sheet of paper to be infinitely long and infinitely wide).

Theorem 26 (Plane Separation Theorem)- Every line a divides the points of the plane not on the line a into two sides having the following properties: If A is on one side and B is on the other side, then the segment AB contains a point of the line a. If A and B are on the same side, then the segment AB contains no points of the line a.

5 Moritz Pasch developed the first modern axiomatic system of geometry in 1882. 6 This theorem is often taken in place of Axiom IV.4; the two can be shown equivalent, assuming the other 16 axioms. 15

The proof of this theorem we give is due to Hartshorne.7 The proof is rather long and detailed, and relies heavily on the use of an equivalence relation. Recall that a relation ~ is an equivalence relation if it satisfies the following conditions: It is reflexive; that is, for every point A, A ~ A. It is symmetric; that is, for every points A and B, if A ~

B, then also B ~ A. It is transitive; that is, for every points A, B, and C, if A ~ B and B ~

C, then also A ~ C. An equivalence relation partitions the set into disjoint subsets, called equivalence classes.

Equivalence relations are extremely important in all areas of mathematics. If a set of elements has an equivalence relation, then every element of the set belongs to one and only one equivalence class, and every element in an equivalence class is equivalent to every other element of that class. Here we will define a relation ~ on all the points not lying on a line a, and we will show the relation is an equivalence relation which splits all the points not on the line a into two equivalence classes, called the sides. Then every point not on the line a will belong to one, and only one, of the two sides, and every point on one side of the line will be equivalent to every other point on the same side of the line a.

Idea of the Proof of Theorem 2- Given a line a, we will define a relation ~ on the points not on a, with A ~ B if A = A or if the segment AB does not contain a point of a. It will be easy to show that ~ is reflexive and symmetric. To show ~ is transitive, we will split into two cases: A ~ B and B ~ C with A, B, and C not all in the same line (Case

7 This proof is from Hartshorne 16

1), and A ~ B and B ~ C with A, B, and C all in the same line (Case 2). In the first case it will be easy to show A ~ C. In the second case, we will construct a new point E that is not on any of the lines ��, ��, ��. Then we will apply the first case to each of the three sets of three points A, B, E; A, C, E; and B, C, E to show A ~ C. We will have shown that

~ is an equivalence relation. We will then show that if A ≁ C and B ≁ C, then A ~ C, again in two cases, in a similar manner to the first step. We will then have shown there are only two equivalence classes, which we will call the two sides of the line a.

Proof of Theorem 2- Let a be a line. We define a relation ~ on the set of points that are not on the line a by A ~ B if A = A or if the segment AB does not contain a point of a. As A = A, by definition A ~ A, or ~ is reflexive. As AB = BA, if A ~ B, AB contains a point of a, so BA contains a point of a, and B ~ A, or ~ is symmetric. To show that ~ is transitive, suppose A, B, and C are points not on the line a, and suppose that A ~

B and B ~ C.

Case 1- If A, B, and C are not all in the same line, we have that a does not pass through A, B, or C, so if a passes through AC, then a passes through a point of AB or a point of BC (II.4). As A ~ B and B ~ C, a does not pass through a point of AB or through a point of BC (definition of ~). Thus a cannot pass through any point of AC, so A ~ C

(definition of ~).

Figure 7: Theorem 2, Case 1 Case 2- If A, B, and C are all in the same line b, we will find another point D that will allow us to use our proof of Case 1. Since A, B, and C are not on the line a, a and l are distinct lines. Then a and b meet in at most one point (Theorem 1). But every line contains at least two points, so there is a point D on the line a that is not on the line b

(I.3).

As A is not on the line a, A and D are two distinct points, so they determine a line

�� (I.1). Then there is a point E on the line �� with A between E and D (II.2). Note that

E must be a different point the D. Then if E were on the line a, then both E and D would be points on the lines a and AD, but that can’t happen because two lines can only meet in one point, so E is not on the line a (Theorem 1). As A is between E and D, D is not between A and E; in other words, D is not on the segment AE (II.3). As D is the only point on the line �� that is also on the line a, and as AE is a segment of the line ��, the segment AE does not pass through any points of a. Thus, E ~ A (definition of ~).

Figure 8: Theorem 2, Case 2

If E were on the line b, then as A and E are both on b, we would have ��= b

(I.1). But then as D is on ��, D would be on b, contradicting our choice of D. Thus we know that E is not on the line b. Now we have three points A, B, E that are not all on the same line, with E ~ A and A ~ B, so E ~ B (Case 1). We also have three points B, C, E that are not all on the same line, with E ~ B and B ~ C, so E ~ C (Case 1). Finally, we have the three points A, C, E that are not all on the same line, with A ~ E and E ~ C, so A

~ C (Case 1). We conclude that ~ is transitive.

We have shown that ~ is an equivalence relation. Now we show that there are two, and only two, equivalence classes (sides of the line), from which it follows that two points A, B are on opposite sides of the line a if the segment AB passes through a point of the line a. There is a point not on the line a, so there is at least one equivalence class

(I.3). Let A be a point not on the line a, and let D be a point on the line a. Then there is a point C with D between A and C (II.2). Then A and C are not equivalent, so there are at least two classes. Now we must show that there are at most two classes, which we prove by showing if A ≁ C and B ≁ C, then A ~ B. We do this again in two cases.

Case 1- If A, B, C are not all in a line. As A ≁ C, AC passes through a point of a

(definition of ~). As B ≁ C, BC passes through a point of a (definition of ~). Then AB cannot pass through a point of a (II.4). So A ~ B (definition of ~).

Figure 9: Theorem 2, Case 1 Case 2- If A, B, C are all in the same line b. By the exact same construction as in

Case 2 above, we can find a point E with E ~ A. If C ~ E, as E ~ A, we would have C ~ A

(~ is transitive), which would contradict the assumption C ≁ A. Thus, C ≁ E. Then we have three points B, C, E not all in a line, with B ≁ C and C ≁ E. Then B ~ E (Case 1).

Now we have B ~ E and E ~ A, so B ~ A (~ is transitive).

Figure 10: Theorem 2, Case 2

We conclude that there are only two equivalence classes. Thus, every line divides the plane into two sides. □

If there are two points A, B and a line a which does not pass through A or B, we say that A and B are on the same side of a if A ~ B, and A and B are on different sides of a if A ≁ B.

Figure 11: On the left, A and B are on the same side of a; on the right, A and B are on different sides of a

If A, B, O are points on a line a, and b is any other line passing through O, we say that A and B are on the same side of O if, compared to the line b, A ~ B, and A and B are on different sides of O if, relative to the line b, A ≁ B.

Figure 12: On the left, A and B are on the same side of O; on the right, A and B are on different sides of O

The ray �� is the portion of the line �� which is on the same side of O as B, along with the point O. The point O is the origin of the ray ��. The portion of the line

�� on the opposite side of O as B, along with O, is the ray opposite ��.

Figure 13: Ray

If A, O, B are three points not lying on the same line a, the angle ∡AOB is the union of the two rays �� and ��. Where there are only two other points A and B under consideration, we will refer to the angle ∡AOB as ∡O. The point O is the vertex of the angle. The interior of the angle ∡AOB is the collection of points C where C is on the same side of the line �� as B and on the same side of the line �� as A. The exterior of the angle ∡AOB is the remainder of the points C where C is not on the angle ∡AOB and

C is not in the interior of the angle.

Figure 14: Interior and exterior of an angle

Exercise 2- Let A, B, C, D, and O be five points. a. Prove that if O is between A and B, then O and B are on the same side of A, and O

and A are on the same side of B. Prove the converse also: if O and B are on the same

side of A, and O and A are on the same side of B, then O is between A and B. b. Prove that if O is between A and B, then the ray �� together with the ray �� is the

line �� ( = �� = ��). Prove the converse also: if the ray �� together with the ray

�� is the line ��, then O is between A and B. c. Prove that if O is between A and B, then O is the only point in common to the two

rays �� and ��. Show that the converse may fail: find an example of three points A,

B, and O where O is the only point in common to the rays �� and ��, but the rays

together do not form the line ��. d. Prove that if O is between A and B, then �� = �� If �� = ��, what can we conclude

about the three points? e. Prove that the points in common to both rays �� and �� are the points of the

segment AB. f. If A is between B and O, and C is between D and O, prove that ∡AOC = ∡AOD =

∡BOC = ∡BOD

Theorem 3- Let A, B, C, D be four points.

1) If B is between A and C, and C is between B and D, then B is also between A and D, and C is also between A and D.

2) If B is between A and D, and C is between B and D, then B is also between A and C, and C is also between A and D.

Proof- Let A, B, C, D be four points.

1) Let B be between A and C, and let C be between B and D, and we will show that B is between A and D, and C is between A and D. As B is between A and C, all three points lie on the same line �� (II.1). As C is between B and D, D also lies on the same line �� (II.1). Thus, the four points all lie on the same line (I.1).

As B is between A and C, A and B are on the same side of C (Exercise 2a). As C is between B and D, B and D are on opposite sides of C (definition of side). Then A and

D are on opposite sides of C (Theorem 2), so C is between A and D (definition of side).

Similarly, as C is between B and D, C and D are on the same side of B (Exercise

2a). As B is between A and C, A and C are on opposite sides of B (definition of side).

Then A and D are on opposite sides of B (Theorem 2), so B is between A and D

(definition of side).

2) Let B be between A and D, and let C be between B and D, and we will show that B is between A and C, and C is between A and D. As in 1), we can see that all four points lie on the same line.

As C is between B and D, C and D are on the same side of B (Exercise 2a). As B is between A and D, A and D are on opposite sides of B (definition of side). Then A and

C are on opposite sides of B (Theorem 2), so B is between A and C (definition of side).

Now as B is between A and C, A and B are on the same side of C (Exercise 2a).

As C is between B and D, B and D are on opposite sides of C (definition of side). Then A and D are on opposite sides of C (Theorem 2). □

Note that the previous theorem (Theorem 3) can be restated in the following way-

Any four points of a line can always be so labeled A, B, C, D so that B is between A and

C and also between A and D, and that C is between A and D and also between B and D.

The only other labeling possessing this property is the reverse labeling D, C, B, A.

Figure 15: Theorem 3 This theorem is known as Pasch’s Theorem. Hilbert originally included this theorem as an axiom, but in 1902 E.H. Moore and R.L. Moore showed that it is dependent; i.e. that it can be proved from the other axioms.8

Corollary9- Any finite number of points on a line can always be labeled in a sequence

A1, A2, A3, …, An, so that A2 is between A1 and A3, A4, …, An; A3 between A1, A2 and

A4, A5,…, An; etc. The only other labeling possessing this property is the reverse order

An,…, A3, A2, A1.

Figure 16: Corollary to Theorem 3

Idea of the Proof- Use induction on the statement of the Corollary, and Theorem

4, to obtain the result. We omit the full proof.

Theorem 4- Between any two distinct points of a line, there are an unlimited number of points.

8 Moore 9 In mathematics, a corollary is a theorem which is an immediate consequence of another theorem. 25

Figure 17: Theorem 4 Remark- This theorem may seem more intuitive than the preceding theorems 2 and 3, and we may wish to include it earlier. The idea of the proof is simple: starting with the two endpoints of a segment, say A0 and B, we know that there is at least one point between them, say A1 (II.2). Then looking at the points A1 and B, we know there is a point between them, say A2 (II.2). We then would repeat this process to find as many points as we want. However, without Theorem 3 (Pasch’s Theorem), we know that A2 is between A1 and B, but we don’t know that A2 is between A0 and B, as we wish to claim.

And we can’t prove Theorem 3 without Theorem 2 (Plane Separation Theorem). This highlights the need to be careful in our proofs not to use any unstated assumptions. This is particularly important to remember when using diagrams; we include diagrams with our proofs to aid our intuition, but they do not constitute part of the proof. We must justify each step through the use of axioms and theorems, not just by appealing to a diagram.

Proof of Theorem 4- Suppose A0 and B are points of a line, and we will use induction to show there are an unlimited number of points between them. Our induction hypothesis is: there are at least n points between A0 and B. Our base case: between A0 and B, there is at least one point between them, say A1 (II.2). Now assume we have already found points A1, A2, …, An-1, with An-1 between A0 and B (Induction hypothesis).

Then there is at least one point between An-1 and B, say An (II.2). As An-1 is between A0 26 and B, and An is between An-1 and B, An is between A0 and B (Theorem 3). Thus by induction, for any natural number n, we can find points A1, A2,…, An all between A0 and

B. □

Theorem 5- Through any point, there are an unlimited number of lines.

Idea of the Proof of Theorem 5- Let A be a point. We will find a line not containing A, find an unlimited number of points on that line, draw new lines between A and each of those points, and show that each line is distinct.

Figure 18: Theorem 5

Proof- Let A be a point. Since there are at least three points not all in a line, there are points B and C not both in the same line as A (I.3). Then the points B and C determine a line (I.1). Then for any number n, there are n points D, E, …, Y between B and C (Theorem 4). Now each of these points, together with A, determines a line (I.1).

We have found n lines ��, ��, …, �� that pass through A, but we still need to show that each line is distinct.

So suppose two lines are the same; by relabeling, say �� = ��. Then D is a point on the line ��, so �� = �� = ��, and the point A is on the line �� (I.2). On the other hand, D and E are two points on the line ��, so �� = �� (I.2). So A is on the line ��, contradicting the assumption that A, B, and C are not all on the same line. Thus each �� is a distinct line passing through A, and we have an unlimited number of lines passing through A. □

We have now shown that every line contains an unlimited number of points, and ever point lies on an unlimited number of lines. Thus, our geometry cannot be finite. In particular, none of the finite geometries from the exercise at the end of Section I satisfy the axioms of this section. We will no longer draw a geometry as a collection of points and lines; there are too many to draw. When we draw diagrams, we will only include the points and lines that are being highlighted in the theorem or proof.

If D, E, F, …, X, Y are distinct points (except possibly D = Y), then the collection of segments DE, EF, FG, …, XY is called the broken line DEF…XY.

Figure 19: Three broken lines

If a broken line begins and ends at the same point, i.e. if D = Y, then the broken line is called a polygon DE…X. The points D, E, …, X are the vertices of the polygon, and the segments DE, EF, …, XD are the sides of the polygon. If no two sides of the polygon have a point in common, other than two consecutive sides meeting at an endpoint, then DE…X is called a simple polygon.

Figure 20: Three polygons. The two polygons on the right are simple polygons.

A simple polygon with 3 sides is a triangle. A simple polygon with 4 sides is a quadrilateral. A simple polygon with 5 sides is a pentagon. A simple polygon with 6 sides is a hexagon. A simple polygon with n sides is an n-gon.

Given any simple polygon, the following theorem tells us that we can define an interior region and exterior region to the polygon. We omit the proof, but note that it could be proved in a similar manner to Theorem 2.

Figure 21: Interior and exterior of a polygon

Theorem 6- Every simple polygon divides the points of the plane on a side of the polygon into two regions, an interior and an exterior, having the following properties: If

A is a point of the interior region and B is a point of the exterior region, then any broken line joining A and B pass through at least one point on a side of the polygon. If A1, A2 are two points of the interior and B1, B2 two points of the exterior, then there are always broken lines joining A1 with A2 and B1 with B2 without passing through any points on the sides of the polygon. There are lines which lie entirely outside of the given polygon, but there are none which lie entirely within it.

Theorem 710 (Crossbar Theorem) - Let A, B, C be three points not lying on the same line and let D be a point in the interior of angle ∡BAC. Then the ray �� passes through the segment BC.

Figure 22: Crossbar Theorem

Note that Theorem 7 is very similar to Pasch’s Axiom (II.4). Theorem 7 says “if a line passes through some point in the interior of a triangle, and through one vertex of the triangle, it must pass through the side not containing that vertex”. Using our newly

10 The theorem and its proof are due to Hartshorne 30 defined terminology, Pasch’s Axioms says “if a line passes through one side of a triangle, and does not pass through any vertices, it must pass through a second side”. Combining these two statements, we have the following:

Corollary- If a line passes through any point in the interior of a triangle, then it passes through two sides of the triangle, or through one side and one vertex.

Idea of the Proof of Theorem 7- We will construct a triangle EBC with A on the segment EC. Then the line �� meets the segment CE at the point A, so �� must meet the segment BE or the segment BC (II.4). We then show that �� does not meet the segment

BE using the Plane Separation Theorem (Theorem 2), so �� meets BC. Finally, we will show that the ray originating at A and pointing away from D does not meet BC, again using the Plane Separation Theorem (Theorem 2). We conclude that the ray �� meets

BC, as desired.

Figure 23: Theorem 7

Proof of Theorem 7- Let A, B, C be three points not lying on the same line and D be a point in the interior of angle ∡BAC. Let E be a point on the line �� with A between

E and C (II.2). Then the line �� meets the segment CE at the point A. We will show that

�� does not meet the segment BE, so �� must meet the segment BC (II.4).

First we show that BE does not meet the ray ��. As the segment BE meets the line �� at the point E, it cannot meet the line at any other points (Theorem 1). Thus, every point of the segment BE besides B is on the same side of the line �� (definition of side). But C and E are on the opposite sides of ��, as A is between C and E (definition of side). As D is in the interior of angle BAC, every point of the ray �� besides A is on the same side of �� as C (definition of side). Thus, the segment BE does not meet the ray

�� (Theorem 2).

Now we show that BE does not meet the ray opposite ��. Let F be a point on that ray. Then as �� meets �� at A, F is on the opposite side of �� as D (definition of side).

As the segment BE meets the line �� at the point E, it cannot meet the line at any other points (Theorem 1). Thus, every point of the segment BE besides E is on the same side of the line �� as B (definition of side). As B and D are on the same side of ��, every point of the segment BE beside E is on opposite side from F of the line �� (Theorem 2), so F is not on the segment BE.

Thus we have shown that the segment BE does not meet the either side of the line

��, so BE does not meet ��. So the segment BC meets the line �� at some point G

(II.4). As G is between B and C, B and G are on the same side of �� (definition of side).

As D is in the interior of the angle ∡BAC, B and D are on the same side of �� (definition

32 of interior). Thus, D and G are on the same side of �� (Theorem 2). That is, G is on the ray �� (definition of ray). We conclude that G is a point on the ray �� and the segment

BC. □

Exercise 3- Find an example of a polygon that does not divide the points of the plane into interior and exterior regions.

Exercise 4- Find an example of a simple polygon and points A and B in the interior region such that the segment AB has a point in common with the polygon. Find a broken line joining A and B that has no points in common with the polygon. Do the same with points C, D in the exterior region.

Exercise 5- How many different polygons can you form by connecting the following points? Draw all the simple polygons you can form. Identify the interiors and exteriors.

Figure 24: Exercise 5

III. Axioms of Congruence of Segments

The next three axioms we present give meaning to the relation congruence of segments. We may think of two segments as being congruent if they have the same length.

Axiom III.1.11 If A, B are distinct points on a line a, and if C is a point on a line c, then on each side of C on the line c there is one and only one point D such that the segment

AB is congruent to the segment CD. We indicate this relation by writing AB ≡ CD. Every segment is congruent to itself; that is, we always have AB ≡ AB.

Axiom III.2. If a segment AB is congruent to the segment CD and also to the segment

EF, then the segment CD is congruent to the segment EF; that is, if AB ≡ CD and AB ≡

EF, then CD ≡ EF.

Remark- From these axioms we can conclude that congruence is an equivalence relation on the collection of segments. It is reflexive (III.1.). We show it is symmetric and transitive. Suppose AB ≡ CD. We also have AB ≡ AB (III.1.). Then CD ≡ AB (III.2.), so

11 This axiom is Proposition 2 of Book I in the Elements. We will reference Euclid’s propositions as follows: Euclid I.2 34

≡ is symmetric. Now suppose AB ≡ CD and CD ≡ EF. As ≡ is symmetric, CD ≡ AB.

Thus AB ≡ EF (III.2.), so ≡ is transitive.

In diagrams, we may depict that segments are congruent by marking dashes into the segments: all segments with one dash are congruent to one another, similarly, all segments with two dashes are congruent to one another, and so on as needed.

Figure 25: AB ≡ CD ≡ EF, and GH ≡ JK.

Axiom III.3.12 Let AB and BC be two segments of a line a which have no points in common other than the point B, and let DE and EF be two segments of a line d which have no points in common other than the point E. Then if AB ≡ DE and BC ≡ EF, we have AC ≡ DF.

Figure 26: Axiom III.3

12 Euclid I.3 35

We would like to be able to add and subtract segments together. Axiom III.3 provides a way to combine segments that are on the same line and share only one endpoint. Using segments arranged as in III.3, we have AB + BC = AC. However, we will often want to combine segments which are not on the same line, or which do not share an endpoint, or which overlap in more than just an endpoint. In order to be completely rigorous with our definitions, we will define addition and subtraction on equivalence classes of congruent segments, rather than on the segments themselves.

Let AB and CD be two segments. Then by III.3, there is a point E on the line ��, with B between A and E, and with BE congruent to CD. The sum of the class of the segment AB and the class of the segment CD is the class of the segment AE. We often omit references to equivalence classes, and say the sum of AB and CD is AE. We write

AB + CD ≡ AE.

Figure 27: AB + CD ≡ AE Let AB and CD be two segments, and suppose there is a point E between A and B with EB congruent to CD. The difference of the class of the segment AB and the class of the segment CD is the class of the segment AE. We again often omit references to equivalence classes, and say the difference of AB and CD is AE. We write AB - CD ≡

AE.

Figure 28: AB – CD ≡ AE Thus addition and subtraction of segments is formally defined on equivalence classes of segments; in practice, however, we will “add” and “subtract” segments.

Exercise 6- It is important to check that these definitions for addition and subtraction are well-defined. Axiom III.3 says that addition is well defined. Prove that subtraction is well-defined; that is, if AB ≡ CD and EF ≡ GH, and there is a point J between A and B with BJ congruent to EF, then AB - EF ≡ CD – GH.

Exercise 7- Verify that the usual rules for addition and subtraction are valid. That is, show the following are true for any segments AB, CD, EF: a. AB + CD ≡ CD + AB b. (AB + CD) + EF ≡ AB + (CD + EF) c. AB – (CD + EF) ≡ (AB – CD) – EF if the subtractions are defined

We would also like to be able to compare different segments, and say which of two segments is greater than or less than the other. Again, to be completely rigorous, we will define the relations less than and greater than on equivalence classes of segments, but as with addition and subtraction, we will often omit the mention of the equivalence classes.

We say that AB is less than CD if there is a point E between C and D with AB congruent to CE. That is: AB < CD if there is a point E between C and D with AB ≡ ED. We say also that CD is greater than AB if the same relation holds, and write CD > AB.

Figure 29: AB < CD < EF < GH < JK < LM

We can restate our definition of subtraction of segments in the following way. If

AB and CD are segments with AB < CD, then subtraction is defined and CD - AB ≡ CE, where E is the point between C and D with ED ≡ AB. If AB ≥ CD, then subtraction is not defined (for if it were, we would have “negative” or “zero” segments).

Exercise 8- Verify that the usual rules for < are valid. That is, show the following are true for any segments AB, CD, EF: a. AB < CD, AB > CD, or AB ≡ CD b. If AB < CD and CD < EF, then AB < EF c. AB < AB + CD d. AB > AB – CD, if the subtraction is defined e. If AB < CD, then EF + AB < EF + CD f. If AB < CD, then EF – AB > EF – CD, if the subtractions are defined g. If AB < CD, then AB – EF < CD – EF, if the subtractions are defined

If two pairs of points A, B and C, D are such that AB ≡ CD, we say that the pair

A, B is congruent to the pair C, D. Similarly, two sequences of points A, B, C, D, …, K,

L on the line a and U, V, W, X, …, Y, Z on the line z are congruent if every corresponding pair of segments are congruent, i.e. if AB ≡ UV, AC ≡ UW, …, AL ≡ UZ;

BC ≡ VW, BD ≡ VX, …, BL ≡ VZ; CD ≡ WX, …, CL ≡ VZ; etc.

Theorem 8- Let A, B, C be three points on the same line a, and let D, E, F be three points on the same line b, with AB ≡ DE, BC ≡ EF, and AC ≡ DF. Then if B is between A and

C, E is between D and F.

Proof- A, B, C be distinct points of a line a and let D, E, F be distinct points of a line b with AB ≡ DE, AC ≡ DF, and BC ≡ EF. Suppose B is between A and C, and we show that E is between D and F. In order to derive a contradiction, suppose not. Then either D is between E and F, or F is between D and E (II.3).

If D is between E and F, DF < EF (definition of <). But AC ≡ DF and BC ≡ EF, so BC > AC. On the other hand, B is between A and C, BC < AC (definition of <). Thus we have a contradiction. If F is between D and E, we can derive a contradiction in a similar manner, by relabeling D, F and A, C.

Corollary- Let A, B, C, E, …, K, L and U, V, W, X, …, Y, Z be two congruent sequences. Suppose that B is between A and C, E, …, K, L; C is between A, B and E, …,

K, L; etc. Then the corresponding points in the second sequence are arranged in a similar manner: V is between U and W, X, …, Y, Z; W is between U, V, and X, …, Y, Z; etc.

Figure 30: Corollary to Theorem 8

Proof- By repeated applications of Theorem 8, we can arrive at the conclusion. □

IV. Axioms of Congruence of Angles

The next three axioms we present give meaning to the relation congruence of angles. Recall the definition of angle (pg. 10): If A, O, B are three points not lying on the same line a, the angle ∡AOB is the union of the two rays �� and ��. We may think of two angles as being congruent if the angle measures between the two rays of each angle are equal.

Axiom IV.1. Let ∡AOB be an angle and let DE be a line. Then on a given side of the line

DE there is one and only one ray �� such that the angle AOB is congruent to the angle

∡CDE and all interior points of ∡CDE lie on the given side of the line ��. We use the same notation for congruence of angles as for congruence of segments: ∡AOB ≡ ∡CDE.

Every angle is congruent to itself: ∡AOB ≡ ∡AOB.

Axiom IV.2. If the angle ∡A is congruent to the angle ∡B and to the angle ∡C, then the angle ∡B is congruent to the angle ∡C; that is, if ∡A ≡ ∡B and ∡A ≡ ∡C, then ∡B ≡ ∡C.

Remark- From these axioms we can conclude that congruence is an equivalence relation on the collection of angles. It is reflexive (IV.1). We show it is symmetric and transitive. Suppose ∡A ≡ ∡B. We also have ∡A ≡ ∡A (IV.1). Then ∡B ≡ ∡A (III.2), so ≡ 41 is symmetric. Now suppose ∡A ≡ ∡B and ∡B ≡ ∡C. As ≡ is symmetric, ∡B ≡ ∡A. Thus

∡A ≡ ∡C (IV.2), so ≡ is transitive.

In diagrams, we may depict that angles are congruent by marking arcs into the angles: all angles with one arc are congruent to one another, all angles with two arcs are congruent to one another, and so on as needed.

Figure 31: ∡ABC ≡ ∡DEF and ∡GHJ ≡ ∡JHK ≡ ∡KHL

Axiom IV.3. If, for the two triangles ABC and DEF, the congruencies AB ≡ DE, AC ≡

DF, ∡A ≡ ∡D, hold, then the congruencies ∡B ≡ ∡E and ∡C ≡ ∡F also hold.

Exercise 9- If OA, BC are segments and ∡DEF is an angle, prove that on a given side of

OA there is one and only one point G with ∡AOG ≡ ∡DEF and AG ≡ BC.

As we did with segments, we would like to define addition, subtraction, and less than. However, we do not yet have the ability to ensure that these relations would be well-defined. In order to do so, we must develop a few more definitions and theorems.

First we extend the notions of congruence of segments and congruence of angles to a new notion of congruence of triangles.

If ABC and DEF are triangles, with corresponding sides congruent and corresponding angles congruent, the triangles are said to be congruent. That is; ABC ≡

DEF if AB ≡ DE, AC ≡ DF, BC ≡ EF, ∡A ≡ ∡D, ∡B ≡ ∡E, and ∡C ≡ ∡F. Note that for instance ∡A is another name for ∡BAC, since in a triangle there are only two points different from A to define the rays that compose the angle.

Figure 32: Congruent Triangles

Theorem 913 (SAS)- If, for the two triangles ABC and DEF, the congruencies AB ≡ DE,

AC ≡ DF, ∡A ≡ ∡D, hold, then the two triangles are congruent.

Proof14- Suppose ABC and DEF are two triangles with AB ≡ DE, AC ≡ DF, and

∡A ≡ ∡D, and we show that ABC and DEF are congruent. We need to show BC ≡ EF,

∡B ≡ ∡E, and ∡C ≡ ∡F. But ∡B ≡ ∡E, and ∡C ≡ ∡F follows directly from IV.3, so we

13 Euclid I.4 14 The proof is due to Hilbert 43 only need to show BC ≡ EF. We know that there is a point G on the line ��, on the same side of E as F, with BC ≡ EG (III.1).

Figure 33: Theorem 9

Then the triangles ABC and DEG have AB ≡ DE, ∡B ≡ ∡E, and BC ≡ EG, so we can conclude ∡A ≡ ∡EDG (IV.3). But we assumed ∡A ≡ ∡EDF, so ∡EDG ≡ ∡EDF

(IV.2). Then the rays �� and �� must be the same ray (IV.1). Then as �� meets �� in at most one point (Theorem 1), and �� meets �� at both points F and G, F and G must be the same point. Thus, EG = EF, and we can conclude from BC ≡ EG that BC ≡ EF. □

The previous theorem is the first theorem that allows us to determine if two triangles are congruent. We will later return to this topic and find other ways to determine if two triangles are congruent. We will name the theorems based on the conditions necessary on adjacent segments or angles. For instance, the theorem we just proved says that if two triangles have congruent angles between pairs of congruent segments, the triangles are congruent. We will call this SAS, or side-angle-side, because it requires a pair of sides to be congruent, then a pair of adjacent angles to be congruent, then the pair of sides adjacent on the opposite ends of the angles to be congruent. We will similarly 44 have a theorem for AAS, or angle-angle-side; a theorem for ASA, or angle-side-angle; and a theorem for SSS, or side-side-side. We will see, however, that ASS, or angle-side- side, fails to prove the congruence of two triangles.

If A, O, B are three points of a line with O between A and B, and C is some point not on that line, then the angles ∡AOC and ∡BOC are supplementary angles.

Figure 34: Supplementary angles

Theorem 1015- If two angles are congruent, their supplementary angles are also congruent.

Idea of the Proof16- From two congruent angles ∡ABC and ∡DEF, we will construct two congruent triangles. We will also construct triangles on the supplementary angles ∡GBC and ∡HEF. We will then use Theorem 9 three times, on the triangles ABC and DEF, on the triangles GAC and HDF, and on the triangles BGC and EHF.

15 Euclid I.13 16 The proof is again due to Hilbert 45

Figure 35: Theorem 10

Proof- Let ∡ABC ≡ ∡DEF be congruent angles, with AB ≡ DE and BC ≡ EF

(III.1). Then the triangles ABC and DEF are congruent (SAS Theorem 9); in particular,

AC ≡ DF and ∡BAC ≡ ∡EDF.

Now there are points G, H with B between A and G, E between D and H, and BG

≡ EH (III.1). As AB ≡ DE, BG ≡ EH, B is between A and G, and E is between D and H, we have AG ≡ DH (III.3). Moreover, as B and G are on the same ray from A, ∡BAC =

∡GAC, and as E and H are on the same ray from D, ∡EDF = ∡HDF (definition of angle).

Then ∡BAC ≡ ∡EDF is the same as ∡GAC ≡ ∡HDF. Thus we have AG ≡ DH, ∡GAC ≡

∡HDF, and AC ≡ DF, so the triangles GAC and HDF are congruent (SAS Theorem 9); in particular, CG ≡ FH and ∡AGC ≡ ∡DHF.

We have BC ≡ EF and CG ≡ FH. As A and B are on the same ray from G, ∡AGC

= ∡BGC, and as D and E are on the same ray from H, ∡DHF = ∡EHF (definition of angle). Then ∡AGC ≡ ∡DHF is the same as ∡BGC ≡ ∡EHF. Then the triangles BGC and

EHF are congruent (SAS Theorem 9); in particular, ∡GBC ≡ ∡HEF. □

If A, O, B are three points of a line, and C, O, D are three points of another line, then ∡AOC and ∡BOD are vertical angles. Similarly, ∡AOD and ∡BOC are also vertical angles.

Figure 36: Vertical angles

Corollary17-Vertical angles are congruent.

Figure 37: Corollary to Theorem 10

Proof- Let A, O, B be three points of a line, and let C, O, D be three points of another line, and we show that ∡AOC ≡ ∡BOD. Note ∡AOC and ∡COB are supplementary angles (definition). Similarly, ∡BOD and ∡COB are supplementary angles (definition). Thus ∡AOC and ∡BOD are supplementary angles of the same angle

∡COB. Clearly ∡COB ≡ ∡COB (IV.1), so ∡AOC ≡ ∡BOD (Theorem 10). □

17 Euclid I.15 47

We return to our attempt to define relations addition, subtraction, and less than on angles. Axiom III.3 essentially says that the sum of two pairs of congruent segments is congruent. We do not have a similar axiom for the addition of angles, but we can now prove an analogous theorem.

Theorem 11- Let ∡ABC be an angle, let D be a point in the interior of the angle, and let

∡ABD ≡ ∡EFG and ∡DBC ≡ ∡GFH be congruent angles, with E and H on opposite sides of the line ��. Then ∡ABC ≡ ∡EFH, and G is in the interior of the angle ∡EFH.

Proof18- Let ∡ABC be an angle, let D be a point in the interior of the angle, and let ∡ABD ≡ ∡EFG and ∡DBC ≡ ∡GFH be congruent angles. Then the ray BD meets the segment AC at a point X (Crossbar Theorem 7); using Exercise 2f if necessary, relabel this point X as D, and we can relabel new points E, G, and H so that the rays ��, ��, and

�� are unchanged (and hence the angles between them also are unchanged), with BA ≡

FE, BD ≡ FG, and BC ≡ FH (III.1).

As ∡DBC ≡ ∡GFH, BD ≡ FG, and BC ≡ FH, the triangles DBC and GFH are congruent (SAS Theorem 9); in particular, DC ≡ GH, ∡CDB ≡ ∡HGF, and ∡DCB ≡

∡GHF. Similarly as ∡ABD ≡ ∡EFG, BA ≡ FE, and BD ≡ FG, the triangles ABD and

EFG are congruent (SAS Theorem 9); in particular, AD ≡ EG and ∡ADB ≡ ∡EGF.

18 The proof is due to Hartshorne 48

Figure 38: Theorem 11

Now let J be a point on the line �� with G between E and J (III.1). As G is between E and J, and F is not on the same line, ∡EGF and ∡JGF are supplementary angles (definition). Similarly, as D is between A and C, and B is not on the same line,

∡ADB and ∡CDB are supplementary angles (definition). But ∡ABD ≡ ∡EFG, so ∡CDB

≡ ∡JGF (Theorem 10). We now have ∡CDB ≡ ∡HGF and ∡CDB ≡ ∡JGF, so ∡HGF ≡

∡JGF (IV.2). As both angles share the same vertex G, and are on the same side of the line

��, they are the same angle (IV.1). Thus, G, H, and J are on the same line ��. Now as E,

G, and H are all on the same line, and E and H are on opposite sides of the line ��, G is between E and H (definition of side).

We have shown AD ≡ EG, DC ≡ GH, D is between A and C, and G is between E and H. Then AC ≡ EH (III.3). We have also shown ∡DCB ≡ ∡GHF, ∡DCB = ∡ACB, and

∡GHF = ∡EHF, so ∡ACB ≡ ∡EHF. We began with the assumption BC ≡ FH, so we conclude ∡ABC ≡ ∡EFH (IV.3).

It remains to show that ∡EFH is actually an angle, and that G is in the interior of the angle. But E, G, and H are all on the line ��, and ∡EFG is an angle, so F is not on the

49 line ��. Thus, E, F, and H are not all on the same line (I.1), and ∡EFH is an angle

(definition). As G is between E and H, G and H are on the same side of E, and G and E are on the same side of H (Exercise 2a), so G is in the interior of ∡EFH (definition). □

As we did with addition and subtraction of segments, to be completely rigorous, we will define addition and subtraction of equivalence classes of angles. In practice, however, we will “add” and “subtract” angles.

Let ∡ABC and ∡DEF be angles. Then we can choose a point G on the opposite side of the line �� from A with ∡CBG ≡ ∡DEF. If ∡ABG is an angle with C in the interior, the sum of the class of the angle ∡ABC and the class of the angle ∡DEF is the class of the angle ∡ABG. We will often omit referencing equivalence classes and say the sum of the angle ∡ABC and the angle ∡DEF is the angle ∡ABG. We write ∡ABC +

∡DEF ≡ ∡ABG.

Figure 39: ∡ABC + ∡DEF ≡ ∡ABG Let ∡ABG and ∡DEF be two angles, and suppose there is a point C in the interior of the angle ∡ABG with ∡CBG congruent to ∡DEF. The difference of the class of the angle ∡ABG and the class of the angle ∡DEF is the class of the angle ∡ABC. We will

50 often omit referencing equivalence classes and say the difference of the angle ∡ABG and the angle ∡DEF is the angle ∡ABC. We write ∡ABG - ∡DEF ≡ ∡ABC.

Figure 40: ∡ABG - ∡DEF ≡ ∡ABC Exercise 10- It is important to check that these definitions for addition and subtraction are well-defined. Theorem 11 says that addition is well defined. Prove that subtraction is well-defined; that is, if ∡A ≡ ∡B, and ∡C ≡ ∡D, then ∡A - ∡C ≡ ∡B - ∡D.

Exercise 11- Verify that the usual rules for addition and subtraction are valid. That is, show the following are true for any angles ∡A, ∡B, ∡C: a. ∡A + ∡B ≡ ∡B + ∡A if the addition is defined b. (∡A + ∡B) + ∡C ≡ ∡A + (∡B + ∡C) if the additions are defined c. ∡A – (∡B + ∡C) ≡ (∡A – ∡B) – ∡C if the addition and subtractions are defined

We also define a relation less than on equivalence classes of angles. As with addition and subtraction, we will often omit the mention of the equivalence classes. We say that ∡ABC is less than ∡DEF if there is a point G in the interior of ∡DEF with ∡GEF

≡ ∡ABC, and write ∡ABC < ∡DEF. We say also that DEF is greater than ∡ABC if the same relation holds, and write ∡DEF > ∡ABC. 51

Figure 41: ∡ABC < ∡DEF

We can restate our definition of subtraction of angles as: If ∡DEF and ∡ABC are segments with ∡ABC < ∡DEF, then subtraction is defined and ∡DEF - ∡ABC = ∡DEG, where G is a point in the interior of ∡DEF with ∡GEF ≡ ∡ABC. If ∡ABC ≥ ∡DEF, then subtraction is not defined (for if it were, we would have “negative” or “zero” angles).

Exercise 12- Verify that the usual rules for ∡ are valid. That is, show the following are true for any angles ∡A, ∡B, ∡C: a. ∡A < ∡B, ∡A > >B, or ∡A ≡ ∡B b. If ∡A < ∡B and ∡B < ∡C, then ∡A < ∡C c. ∡A < ∡A + ∡B, if the addition is defined d. ∡A > ∡A – ∡B, if the subtraction is defined e. If ∡A < ∡B, then ∡C + ∡A < ∡C + ∡B, if the additions are defined f. If ∡A < ∡B, then ∡C – ∡A > ∡C – ∡B, if the subtractions are defined g. If ∡A < ∡B, then ∡A – ∡C < ∡B – ∡C, if the subtractions are defined

Theorem 1219 (ASA)- If one side and the two adjacent angles of a triangle are congruent to the respective side and angles of a second triangle, the two triangles are congruent.

Proof- Suppose that ABC and DEF are two triangles with ∡A ≡ ∡D, AB ≡ DE, and ∡B ≡ ∡E, and we show that the triangles are congruent. We will do this by showing that BC ≡ EF, and then applying Theorem 9 (SAS) to arrive at our conclusion. So in order to derive a contradiction, suppose BC and EF are not congruent. Then either BC <

EF or BC > EF (Exercise 8a) ; by relabeling, we may assume BC < EF. Then there is a point G on the line �� between E and F with BC ≡ EG. Note that ∡DEF = ∡DEG, as ��

= �� (Exercise 2f).

Figure 42: Theorem 12

In the triangles ABC and DEG, we have AB ≡ DE, ∡ABC ≡ ∡DEG, and BC ≡ EG, so

∡BAC ≡ ∡EDG (IV.3). But ∡BAC ≡ ∡EDF, so ∡EDG ≡ ∡EDF (IV.2). Thus as G and F are on the same side of the line ��, the angles are the same angle, and hence the line �� and the line �� are the same (Exercise 2f). Then both the points F and G are on the line

�� = �� and the line ��, so F and G are the same point (Theorem 1). But this

19 Euclid I.26 53 contradicts the assumption that G is between E and F (II.1). So BC ≡ EF, and the two triangles are congruent (SAS Theorem 9). □

An angle which is congruent to its supplementary angle is a right angle. In diagrams, this may be depicted by drawing a small square in the angle.

Figure 43: Right angles

Theorem 1320- If O is a point on the line ��, the rays �� and �� lie on opposite sides of the line ��, and ∡AOB and ∡AOC are supplementary angles, then B, O, and C lie on a line.

Figure 44: Theorem 13

Proof- Let �� be a line with �� and �� on opposite sides of the line ��, and suppose that ∡AOB and ∡AOC are supplementary angles, and we show that B, O, and C

20 Euclid I.14 54 line on a line. We know there is a point D on the line �� with O between B and D (II.2), so D and B are on opposite sides of �� (definition of side), and hence D and C are on the same side of �� (Theorem 2). Then as B, O, and D are on a line, ∡AOB and ∡AOD are supplementary angles (definition). As ∡AOB and ∡AOC are also supplementary angles,

∡AOC ≡ ∡AOD (Theorem 10). But ∡AOC and ∡AOD both lie on the same side of ��, so ∡AOC = ∡AOD (IV.1). Thus, �� = �� and C is on the line ��. □

Theorem 14- All right angles are congruent to one another.

21 Proof - Let ∡ABC and ∡DEF be right angles, and we show ∡ABC ≡ ∡DEF. Let

∡ABC and ∡DEF have supplementary angles ∡CBG and ∡FEH, respectively. Then

∡ABC ≡ ∡CBG and ∡DEF ≡ ∡FEH (definition of supplementary angle). In order to derive a contradiction, suppose ∡ABC is not congruent to ∡DEF. Then either ∡ABC <

∡DEF or ∡ABC > ∡DEF (Exercise 12a), so by relabeling, we may assume ∡ABC >

∡DEF. Thus there is a point J in the interior of the angle ∡ABC with ∡ABJ ≡ ∡DEF

(definition of <). Now as A, B, and G are points of a line and J is a point not on the line,

∡ABJ and ∡GBJ are supplementary angles (definition). But ∡DEF and ∡FEH are also supplementary angles, and as ∡ABJ ≡ ∡DEF, ∡GBJ ≡ ∡ FEH (Theorem 10). But we assumed ∡DEF ≡ ∡FEH, so ∡GBJ ≡ ∡DEF (IV.2). But ∡ABJ ≡ ∡DEF also, so ∡ABJ ≡

∡GBJ (IV.2).

21 The proof is due to Hilbert 55

Figure 45: Theorem 14

As ∡ABC ≡ ∡CBG, and J is a point in the interior of ∡ABC, there is a point K in the interior of ∡CBG with ∡GBK ≡ ∡ABJ and ∡JBC ≡ ∡CBK. Then as ∡ABJ ≡ ∡GBJ and ∡GBK ≡ ∡ABJ, ∡GBK ≡ ∡ GBJ (IV.2). But K and J are on the same side of the line

��, so ∡GBK = ∡GBJ and the ray �� is the same ray ��. But K and J were chosen on opposite sides of the ray ��, which is impossible. Thus, ∡ABC ≡ ∡DEF. □

An angle that is less than a right angle is an acute angle. An angle that is greater than a right angle is an obtuse angle.

Figure 46: The left angle is acute, while the right angle is obtuse

Exercise 13- If ∡AOB and ∡BOC are supplementary angles, prove that either ∡AOB and

∡AOC are both right angles, or one is an acute angle and one is an obtuse angle.

Exercise 14- Prove that the sum of two acute angles is an angle, that is, if ∡ABC and

∡DEF are both acute angles, then there is a point G with ∡DEG ≡ ∡ABC + ∡DEF, with

F in the interior of the angle ∡DEG.

V. Triangles

In this section, we will not present any new axioms. Instead, we will continue to develop geometry from the axioms and theorems previously presented. In a certain sense, the theorems we have so far can be viewed as tools to assist in developing Euclidean geometry. In fact, Theorems 1 through 8 and Theorem 11 were all implicit assumptions in Euclid’s Elements, which we have instead proved. Moreover, Theorem 14, which says all right angles are congruent, was one of Euclid’s 5 axioms. Thus, we are not much further now than where Euclid was when he began his work. However, we have been completely rigorous, and we will be able to proceed much more easily now. Most importantly, we will now proceed to much more “interesting” geometry.

Much of this section will concern properties of triangles. We can classify triangles by the types of angles that compose the triangle, or by the segments that compose the triangle.

A triangle with an obtuse angle is an obtuse triangle. A triangle with a right angle is a right triangle. We will see later that a triangle can have at most one right or obtuse angle, so a triangle cannot be both obtuse and right. A triangle with every angle acute is an acute triangle. A triangle with all angles congruent is an equiangular triangle.

Figure 47: ABC is obtuse, DEF is a right triangle, and GHI is acute.

A triangle in which every side is congruent is an equilateral triangle. A triangle in which two sides are congruent, and the third side is not congruent to the other two, is an isosceles triangle. If no side of a triangle is congruent to another side, the triangle is a scalene triangle.

Figure 48: ABC is equilateral, DEF is an isosceles triangle, and GHI is scalene.

Theorem 1522- The angles opposite congruent sides in an isosceles triangle are congruent.

22 Euclid I.5 59

Proof23- Let ABC be an isosceles triangle, with AB ≡ AC, and we show ∡B ≡

∡C. Consider the triangle as two triangles ABC and ACB. As ∡A = ∡A, ∡A ≡ ∡A

(IV.1). By assumption, AB ≡ AC and thus also AC ≡ AB, so the two triangles are congruent (SAS Theorem 9); in particular, ∡B ≡ ∡C. □

Corollary- An equilateral triangle is equiangular.

Proof- Let ABC be an equilateral triangle, i.e. AB ≡ BC ≡ AC, and we show ∡A

≡ ∡B ≡ ∡C. As AB ≡ AC, ∡B ≡ ∡C (Theorem 15), and as AB ≡ BC, ∡A ≡ ∡C (Theorem

15). Now ∡B ≡ ∡C and ∡A ≡ ∡C, so ∡B ≡ ∡A (IV.2). Thus, ∡A ≡ ∡B ≡ ∡C. □

Theorem 1624- If a triangle has two angles congruent it is an isosceles triangle.

Proof- Let ABC be a triangle with ∡B ≡ ∡C, and we show AB ≡ AC. Consider the triangle as two triangles ABC and ACB. As BC = BC, BC ≡ BC (III.1). By assumption, ∡B ≡ ∡C and thus also ∡C ≡ ∡B, so the two triangles are congruent

(Theorem 12 ASA); in particular, AB ≡ AC. □

23 This proof is much simpler than that given by Euclid, and is attributed to Pappus (circa 320) 24 Euclid I.6 60

Note that Theorem 16 is the converse of Theorem 15. Putting these together, we know that a triangle has two congruent sides if and only if it has two congruent angles.

We also have the converse to the corollary to Theorem 16.

Corollary- An equiangular triangle is equilateral.

Proof- Exercise 15. □

Theorem 1725- Every segment is the base of an isosceles triangle.

Figure 49: Theorem 17

Proof- Let AB be a segment and we show there is an isosceles triangle with base

AB. Let C be a point not on the line �� (I.3). If ∡ABC ≡ ∡BAC, ABC is isosceles and we are done, so suppose not. Then ∡ABC < ∡BAC or ∡ABC > ∡BAC (Exercise 12a); by relabeling if necessary, we may assume ∡ABC < ∡BAC. Then there is a point D in the interior of ∡BAC with ∡BAD ≡ ∡ABC (definition of <). The ray �� passes through the vertex A and the point D in the interior of the angle ∡BAC, so �� meets the segment BC

25 We follow the lead of Hartshorne, who uses this theorem in place of Euclid I.1, which asserts the existence of equilateral triangles (Theorem 30). 61 at a point E (Crossbar Theorem 7). As ∡BAD = ∡BAE and ∡ABC = ∡ABE (Exercise

2f), ∡BAE ≡ ∡ABC. Thus, AE ≡ BC (Theorem 15), and BAE is an isosceles triangle

(definition). □

Theorem 1826 (SSS)- If two triangles have all three sides congruent to the corresponding sides, the two triangles are congruent.

Proof- Let ABC and DEF be two triangles with AB ≡ DE, AC ≡ DF, and BC ≡

EF, and we show that the triangles are congruent. Note there is a point G on the opposite side of the line �� with ∡BAC ≡ ∡GDF and AB ≡ DG (Exercise 9). From those two congruences, and as AC ≡ DF, the two triangles ABC and DGF are congruent (Theorem

9 SAS); in particular BC ≡ GF. Since BC ≡ EF, GF ≡ EF (IV.2). Note also AB ≡ DG and

AB ≡ DE, so DG ≡ DE (III.2).

Now there are three cases to consider: D is on the line �� (Case 1), F is on the line �� (Case 2), or neither D nor F is on the line �� (Case 3).

26 Euclid I.8 62

Figure 50: Theorem 18

Case 1- As D is on the line ��, F is not on the same line (or else D, E and F are all on a line, contradicting that DEF is a triangle), so EFG is a triangle. Then since GF ≡

EF, EFG is an isosceles triangle (definition), and ∡GEF ≡ ∡EGF (Theorem 15). But the angle ∡FEG is ∡FED and the angle ∡EGF is ∡DGF (Exercise 2f), so ∡DEG ≡ ∡FED.

Case 2- As F is on the line ��, D is not on the same line (or else D, E and F are all on a line, contradicting that DEF is a triangle), so EDG is a triangle. Then since DG ≡

DE, EDG is an isosceles triangle (definition), and ∡DEG ≡ ∡DGE (Theorem 15). But the angle ∡DEG is ∡DEG and the angle ∡DGE is ∡DEF (Exercise 2f), so ∡DEG ≡ ∡FED.

Case 3- Neither D nor F is on the line ��, so EDG and EFG are triangles. Then since GF ≡ EF, EFG is an isosceles triangle (definition), and ∡FEG ≡ ∡FGE (Theorem

16). Similarly, since DG ≡ DE, EDG is an isosceles triangle (definition), and ∡DEG ≡

∡DGE (Theorem 15).

If D and F are on opposite sides of the line ��, by adding congruent angles,

∡DGF ≡ ∡DGE + ∡EGF ≡ ∡DEG + ∡GEF ≡ ∡DEF (addition of angles). If D and F are

63 on the same side of the line ��, then either ∡DGE ≡ ∡FGE, ∡DGE > ∡FGE, or ∡DGE <

∡FGE (Exercise 8a). But we cannot have ∡DGE ≡ ∡FGE, because then D = F (IV.1), contradicting that DF is a segment. If ∡DGE > ∡FGE, by subtracting congruent angles,

∡FGD ≡ ∡FGE - ∡DGE ≡ ∡FEG - ∡DEG ≡ ∡FED; if ∡DGE < ∡FGE we get the same congruence by subtracting in the opposite order.

Figure 51: Theorem 18, Case 3

In all three cases, we have ∡DEG ≡ ∡FED, DE ≡ DG, and EF ≡ GF. Thus, ∡GDF

≡ ∡EDF (IV.3). We began by assuming ∡BAC ≡ ∡GDF, so ∡BAC ≡ ∡EDF (IV.2). Now

AB ≡ DE, AC ≡ DF, and ∡BAC ≡ ∡EDF, so we conclude ∡ABC ≡ ∡DEF and ∡ACB ≡

∡DFE (IV.3), and thus the triangles are congruent. □

Corollary27- If AB, CD, and EF are segments, there is at most one point G on a given side of the line �� with AG ≡ CD and BC ≡ EF.

27 Euclid I.7 64

Figure 52: Corollary to Theorem 18

Proof- Let AB, CD, and EF be segments, and suppose there are two points G and

H on the same side of the line �� satisfying AG ≡ AH ≡ CD and BG ≡ BH ≡ EF. Then the triangles ABG and ABH are congruent (Theorem 18 SSS); in particular, ∡BAG ≡

∡BAH. Thus, the rays �� and �� are the same ray (IV.1). As AG ≡ AH, we conclude G

= H (III.1). □

Theorem 1928- Every angle is the sum of two congruent angles.

Figure 53: Theorem 19

Proof29 - Let ∡AOB be an angle, and we show there is a point D in the interior of

∡AOB with ∡AOD ≡ ∡BOD. We know there is a point C on the ray �� with OA ≡ OC

28 Euclid I.9 65

(III.1). Then there is an isosceles triangle ACD with AD ≡ CD, with base AC, and with O and D on opposite sides of �� (Theorem 17). As OD = OD, OD ≡ OD (III.1). From these three congruences and the fact that ∡BOD = ∡COD, we conclude ∡AOD ≡ ∡BOD (SAS

Theorem 9). By construction, ∡AOD + ∡BOD = ∡AOB. □

Theorem 2030- Every segment is the sum of two congruent segments.

Figure 54: Theorem 20

Proof- Let AB be a segment, and we show there is a point M between A and B with AM ≡ BM. We know there is an isosceles triangle ABC with base AB and AC ≡ BC

(Theorem 17). Then there is a point D with ∡ACD ≡ ∡BCD and ∡ACD + ∡BCD =

∡ACB (Theorem 19). As �� passes through the vertex C and the point D in the interior of the angle ∡ACB, �� must meet the segment AB in a point M (Crossbar Theorem 7).

As CM = CM, CM ≡ CM (III.1), and hence AM ≡ BM (SAS Theorem 9). By construction, AM + BM = AB. □

29 Beginning with this proof, the remaining proofs follow closely to Euclid’s original proofs 30 Euclid I.10 66

For an angle ∡AOB, the ray �� which gives ∡AOD ≡ ∡BOD is the angle bisector. We write ∡AOD ≡ ½×∡AOB or ∡AOB ≡ 2×∡AOD. For a segment AB, the point M between A and B which gives AM ≡ BM is the midpoint. We write AM ≡ ½ AB or AB ≡ 2AM.

A line at right angles to a line or segment is a perpendicular to the line or segment. If a perpendicular passes through the midpoint of a segment, it is the perpendicular bisector.

Exercise 16- Show that the line �� constructed in the proof of Theorem 20 is the perpendicular bisector of AB.

Theorem 2131- There is a perpendicular through any point on a line.

Proof- Let M be a point on the line m, and we find a perpendicular to m passing through M. Let A be any other point on the line m (I.3). Then there is a point B on the opposite side from M as A, with AM ≡ BM (III.1). We now proceed exactly as in

Theorem 20.

We know there is an isosceles triangle ABC with base AB and AC ≡ BC

(Theorem 17). Then there is a point D with ∡ACD ≡ ∡BCD and ∡ACD + ∡BCD =

∡ACB (Theorem 19). As �� passes through the vertex C and the point D in the interior

31 Euclid I.11 67 of the angle ∡ACB, �� must meet the segment AB in a point E (Crossbar Theorem 7).

As CE = CE, CE ≡ CE (III.1), and hence AE ≡ BE (SAS Theorem 9).

From the same congruence of triangles, we also conclude ∡AEC ≡ ∡BEC. But A,

B, and E are on a line m, so ∡AEC and ∡BEC are supplementary angles (definition), and hence right angles (definition). Thus, �� is a perpendicular to m (definition). Since M and E are both on the line m, and both are between A and B, with AM ≡ BM and AE ≡

BE, we conclude E = M (III.1). Thus, �� = �� is the perpendicular to m passing through the point M. □

Theorem 2232- There is a perpendicular to a line through any point not on the line.

Figure 55: Theorem 22

Proof- Let m be a line, and let A be a point not on the line m, and we find a perpendicular to m passing through A. Pick any point O on the line m (I.3). Then there are points B and C on the line m on either side of m (II.1, II.2 and definition of side), and

32 Euclid I.12 68

∡AOB and ∡AOC are supplementary angles (definition of supplementary angle). If

∡AOB ≡ ∡AOC, then both angles are right angles (definition of right angle), and the line

�� is perpendicular to m (definition of perpendicular) and passes through A, as desired.

So suppose ∡AOB and ∡AOC are not congruent. Then one of ∡AOB and ∡AOC is acute

(Exercise 13), so by relabeling, we may assume that ∡AOB is acute.

Then there is a point D on the opposite side of the line m from A, with ∡AOB ≡

∡DOB, and AO ≡ DO (Exercise 9). As ∡AOD is an angle (Exercise 14) and AO ≡ DO,

AOD is an isosceles triangle (definition), so ∡OAD ≡ ∡ODA (Theorem 15). As A and D are on opposite sides of m, the segment AD meets m at a point E (definition of side). As

E and B are both in the interior of the angle ∡AOD on the line m, �� = �� (Exercise 2a and 2d); in particular, ∡AOB = ∡AOE and ∡DOB = ∡DOE, so ∡AOE ≡ ∡DOE. Then as

OE = OE, OE ≡ OE (III.1), and as AO ≡ DO, the triangles AOE and DOE are congruent

(Theorem 10); in particular, ∡OEA ≡ ∡OED. But A, E, and D are in a line, so ∡OEA and

∡OED are supplementary angles (definition of supplementary angle). Thus, ∡OEA and

∡OED are right angles (definition), and the line ��is perpendicular to m and passes through A. □

For a triangle ABC, if we extend the ray �� to a point D, the angle ∡CBD is an exterior angle to the triangle.

Figure 56: Exterior angle ∡CBD

Theorem 2333- Any exterior angle of a triangle is greater than each of the opposite interior angles.

Proof (Euclid)- Let ABC be a triangle with exterior angle ∡CBD, and we show that ∡CBD > ∡CAB and ∡CBD > ∡ACB. Let E be the midpoint of the segment BC

(Theorem 20), so CE ≡ BE. Then there is a point F with E between A and F and AE ≡ EF

(III.1).

Figure 57: Theorem 23

Note ∡AEC and ∡FEB are vertical angles, and thus congruent (Corollary to

Theorem 10). Then ∡ACE ≡ ∡FBE (IV.3). F is in the interior of the angle ∡CBD

(Exercise 17 below), so as ∡CBD = ∡DBE (Exercise 2f), F is in the interior of the angle

∡DBE. Thus, ∡DBE > ∡FBE (definition of >). But ∡ACE = ∡ACB (Exercise 2f), and

33 Euclid I.16 70

∡ACE ≡ ∡FBE, so ∡FBE ≡ ∡ACB. We conclude ∡DBE > ∡ACB, i.e. ∡CBD > ∡ACB, as desired. To show ∡CBD > ∡CAB, we repeat the same process, but start with E as the midpoint of the segment AB. We omit the details.□

Exercise 17- Use Theorem 2 to justify that F is in the interior of the angle ∡CBD in the proof above.

Corollary34- Any angle in a triangle is less than the supplementary angle of any other angle in the triangle.

Proof- Let ABC be a triangle, and we show that ∡ABC is less than the supplementary angle of ∡ACB (and thus by the symmetry of the statement, that ∡ABC is less than the supplementary angle of ∡CAB). Extend the ray �� to a point D (II.2). Then

∡CBD > ∡ACB (Theorem 23). But A, B, and D lie on a line with B between A and D, so

∡CBD and ∡ABC are supplementary angles (definition). Thus, ∡ABC > ∡ACB. □

Theorem 2435- In any triangle, if one side is greater than another, the angle opposite the greater side is greater than the angle opposite the lesser side.

34 Euclid I.17 35 Euclid I.18 71

Figure 58: Theorem 24

Proof- Let ABC be a triangle with AC > AB, and we show that ∡ABC > ∡ACB.

As AC > AB, there is a point D between A and C with AB ≡ AD (definition of >). Then

ABD is an isosceles triangle, so ∡ABD ≡ ∡ADB (Theorem 15). Note that ∡ADB is an exterior angle to the angle ∡BDC in the triangle BDC, so ∡ADB > ∡DCB (Theorem 23).

But D is between A and C, so ∡DCB = ∡ACB and ∡ABD = ∡ABC (Exercise 2f). Thus,

∡ABD > ∡ACB. □

Corollary36- In any triangle, if one angle is greater than another, the side opposite the greater angle is greater than the side opposite the lesser angle.

Proof- Let ABC be a triangle with ∡B > ∡C, and we show AC > AB. We know either AC < AB, AC ≡ AB, or AC > AB (Exercise 8a). If AC < AB, then ∡C > ∡B

(Theorem 24), contradiction our assumptions. Similarly, if AC ≡ AB, then ABC is an isosceles triangle and ∡B ≡ ∡C (Theorem 15), contradiction our assumption. Thus, AC >

AB. □

36 Euclid I.19 72

Theorem 2537- In any triangle, the sum of any two sides is greater than the third.

Proof- Let ABC be a triangle, and we show that AB + BC > AC (and thus by the symmetry of the statement, also that AB + AC > BC, and that AC + BC > AB). If either

AB or BC ≥ to AC alone, then clearly AB + BC > AC (Exercise 8b and 8c), so we may assume AC > AB and AC > BC (Exercise 8a). We know there is a point D, with B between A and D, such that BC ≡ BD (III.1).

Figure 59: Theorem 25

Then BCD is an isosceles triangle, so ∡BCD ≡ ∡BDC (Theorem 15). As B is between A and D, B is in the interior of the angle ∡ACD, and hence ∡ACD > ∡BCD

(definition of >). As ∡BCD ≡ ∡BDC, and as B is between A and D, ∡BDC = ∡ADC

(Exercies 2f). Thus, ∡ACD > ∡ADC, so AD > AC (Theorem 24). But AD ≡ AB + BC, so

AB + BC > AC. □

Theorem 2638- If D is an interior point of the triangle ABC, then AB + AC > DB + DC, and ∡BAC < ∡BDC.

37 Euclid I.20 73

Proof- Let ABC be a triangle, let D be a point in the interior of the triangle, and we show that AB + AC is greater than DB + DC, and ∡BAC is less than ∡BDC. As D is in the interior of the triangle ABC, the ray �� meets the segment AC in a point E

(Crossbar Theorem 7). Then B, A, E; and C, E, D, respectively are not all in a line, so

BAE and CED are triangles.

Figure 60: Theorem 26 Considering triangle BAE, we know AB + AE > BE (Theorem 25). Adding EC to both sides of the inequality, AB + AE + EC > BE + EC (Exercise 8e). Then as E is between A and C, AE + EC ≡ AC (definition of addition), we have AB + AC > BE + EC.

Similarly, considering triangle CED, we know ED +EC > DC (Theorem 25). Adding DB to both sides of the inequality, DB + ED + EC > DB + DC (Exercise 8e). Then as D is between E and B, DB + ED ≡ BE (definition of addition), we have BE + EC > DB + DC.

Combining the two inequalities AB + AC > BE + EC and BE + EC > DB + DC, we conclude AB + AC > DB + DC.

38 Euclid I.21 74

Now ∡BDC is an external angle to the triangle CED, and not adjacent to ∡CED, so ∡BDC > ∡CED (Theorem 23). Similarly, ∡CEB is an external angle to the triangle

BAE, and not adjacent to ∡BAE, so ∡CEB > ∡BAE (Theorem 23). But D is between B and E, so ∡CED = ∡CEB (Exercise 2f), and E is between A and C, so ∡BAE = ∡BAC

(Exercise 2f). Thus, ∡BDC > ∡CEB and ∡CEB > ∡BAC, and we conclude ∡BDC >

∡BAC (Exercise 12b). □

Theorem 2739- If two triangles ABC and DEF have AB ≡ DE, AC ≡ DF, and ∡BAC >

∡EDF, then BC > EF.

Figure 61: Theorem 27

Proof- Let ABC and DEF be triangles with AB ≡ DE, AC ≡ DF, and ∡BAC >

∡EDF, and we show BC > EF. As ∡BAC > ∡EDF, there is a point G on the same side of

�� as F with ∡BAC ≡ ∡EDG and AC ≡ DG (Exercise 9). Then ABC and DEG are congruent triangles (SAS Theorem 9).

Now D, F, and G are not all on a line, so DFG is a triangle. Moreover, AC ≡ DF, and AC ≡ DG, so DF ≡ DG (III.2), and DFG is an isosceles triangle (definition). Thus,

39 Euclid I.24 75

∡DFG ≡ ∡DGF (Theorem 15). We show that G is in the interior of ∡DEF, and that D is in the interior of ∡EFG.

Figure 62: G cannot be exterior to ∡DEF In order to derive a contradiction, first suppose G is not in the interior of ∡DEF.

By construction, F and G are on the same side of ��, so D and G must be on opposite sides of �� (definition of interior). Then DG meets �� at a point H (Theorem 2). As

∡EDH = ∡EDG, and ∡EDG > ∡EDF, we must have F between H and E (definition of >).

Note ∡DFH is an exterior angle to the triangle FGH, so ∡DFH > ∡FGH (Theorem 23).

But H is between D and G, so ∡FGH = ∡FGD (Exercise 2f) and ∡DFH > ∡FGD. Again as H is between D and G, H is in the interior of the angle ∡DFG, so ∡DFH < ∡DFG

(definition of <). But ∡DFG ≡ ∡DGF, so we have both ∡DFH < ∡DGF and ∡DFH >

∡DGF, contradicting Exercise 12a.

Figure 63: D cannot be exterior to ∡EFG Again, in order to derive a contradiction, suppose D is not in the interior of

∡EFG. We know D and G are on the same side of ��, so D and E must be on opposite

76 sides of �� (definition of interior). Then DE meets �� at a point H (Theorem 2). As G and F are on the same side of �� and G is in the interior of ∡DEF, G must be between F and H (definition of interior). Then G is in the interior of the triangle DEF; in particular,

∡EDG < ∡EDF, contradicting our choice of G.

Figure 64: Theorem 27

We now have ∡DFG ≡ ∡DGF, G is in the interior of ∡DEF, and D is in the interior of ∡EFG. Thus ∡DFG > ∡EGF and ∡EFG > ∡DFG (definition of >), and hence

∡EFG > ∡EGF (Exercise 12b). Then in the triangle EFG, ∡EFG is a greater angle than

∡EGF, so the opposite side EG is greater than EF (Theorem 27). But EG ≡ BC, and we conclude BC > EF. □

Theorem 2840- If two triangles ABC and DEF have AB ≡ DE, AC ≡ DF, and BC > EF, then ∡A > ∡D.

Proof- Let ABC and DEF be triangles with AB ≡ DE, AC ≡ DF, and BC > EF, and we show ∡A > ∡D. We know that one of the following must hold: ∡A < ∡D, ∡A ≡

∡D, or ∡A > ∡D (Exercise 12a).

40 Euclid I.25 77

But if ∡A ≡ ∡D, the two triangles are congruent (SSS Theorem 18); in particular,

BC ≡ EF, contradicting our assumptions. Similarly, if ∡A < ∡D, then BC < EF (Theorem

27), again contradicting our assumptions. Thus, ∡A > ∡D. □

Theorem 2941 (AAS)- If two angles and one side of a triangle are congruent to the respective angles and side of a second triangle, the two triangles are congruent.

Proof- Let ABC and DEF be triangles with two angles and one side congruent respectively, and we show that the two triangles are congruent. If the congruent side is between the congruent angles, ASA Theorem 12 applies, and the triangles are congruent, so we can assume that the congruent side is not between the congruent angles. In particular, let ∡BAC ≡ ∡EDF, ∡ABC ≡ ∡DEF, and AC ≡ DF. We will show AB ≡ DE, and hence the two triangles are congruent (SAS Theorem 9).

Figure 65: Theorem 29 In order to derive a contradiction, suppose AB is not congruent to DE. Then AB <

DE or AB > DE (Exercise 8a), by relabeling if necessary, we may assume AB < DE.

Then there is a point G between D and E with AB ≡ DG (definition of <). As G is

41 Euclid I.26 78 between D and E, ∡EDF = ∡GDF (Exercise 2f), and hence ∡BAC ≡ ∡GDF (IV.2). Now as AC ≡ DF, ∡BAC ≡ ∡GDF, and AB ≡ DG, we have ∡ABC ≡ ∡DGF (IV.3), and as

∡ABC ≡ ∡DEF, we have ∡DGF ≡ ∡DEF (IV.2).

Note that GEF is a triangle, and ∡DGF is exterior to the triangle and not adjacent to ∡GEF, so ∡DGF > ∡GEF (Theorem 23). But G is between D and E, so ∡GEF = ∡DEF

(Exercise 2f). Thus, ∡DGF > ∡DEF. But we already have ∡DGF ≡ ∡DEF, contradicting

Exercise 8a. Thus, AB ≡ DE. □

VI. Axiom of Circles

We present one axiom concerning circles. Given two distinct points O and A, the circle with center O and radius OA is the collection of all points B with OA ≡ OB. We denote the circle (O,OA).

Exercise 18- Prove if (O,OA) = (P,PA), then O = P. Show that (O,OA) = (O,OB) does not mean A = B.

A circle (O,OA) separates the points in the plane not lying on the circle into two sides. Since the center of a circle (O,OA) is well-defined, we can define the two sides as follows. If OD < OA (or O = D), then D is in the interior of the circle, and if OD > OA, then D is in the exterior of the circle.

Note that the circle does not contain the points interior to it. The circle (O,OA) together with the points interior to the circle is the disk with center O and radius OA.

Axiom VI- If (O,OA) and (B,BC) are two circles, and (O,OA) contains both a point interior to and exterior to (B,BC), then (O,OA) and (B,BC) meet at a point.

It is often an important problem in mathematics to show that a certain object exists. It is possible to show that a given object satisfies certain properties, without actually knowing that the object exists. Such is the case with equilateral triangles; we have already proven that equilateral triangles are equiangular (Corollary to Theorem 15), and that equiangular triangles are equilateral (Corollary to Theorem 16), and hence the two are the same. It is still possible that equilateral triangles do not exist, however.

Euclid’s first Proposition asserted the existence of equilateral triangles; we delayed the proof until this much later stage because we require the Circle Axiom to complete the proof. In fact, there are other geometries, which satisfy all the axioms we have previously stated besides the Circle axiom, in which no such triangle exists.

Theorem 3042- Every segment is the base of an equilateral triangle.

Figure 66: Theorem 30

Proof- Let AB be a segment, and we show there is an equilateral triangle with base AB. Consider the two circles (A,AB) and (B,AB). There is a point X on the ray opposite �� with AB ≡ BX (III.1), so X is on the circle (B,AB). Then AX = AB + BX

42 Euclid I.1 81

(definition of addition), so AX > AB (definition of >), and X is exterior to the circle

(A,AB). Clearly, A is on the circle (B,AB) and interior to the circle (A,AB) (definition of interior). Then (A,AB) and (B,AB) meet at a point C. As C is on the circle (A,AB), AB ≡

AC. As C is on the circle (B,AB), AB ≡ AC. Thus, ABC is an equilateral triangle. □

Similarly, we can show that triangles exist with sides congruent to any given segments, with one caveat. Recall that Theorem 25 says the sum of any two sides of a triangle is greater than the third side. Thus if we wish to show there is a triangle with sides congruent to given segments, we must have the sum of any two segments greater than the third. Theorem 31 below says that in such instances, there is such a triangle.

Theorem 3143- Given any three segments, with the sum of any two greater than the third, there is a triangle with sides congruent to the three given segments.

Proof- Let FG, HJ, and KL be three segments with the sum of any two greater than the third, and we show there is a triangle with sides congruent to the three segments.

By relabeling if necessary, we may assume FG ≥ HJ. Then on a line, there are points A,

B, C, and D with B between A and C, with C between B and D, and with AB ≡ FG, BC ≡

HJ, and CD ≡ KL (III.1).

43 Euclid I.22 82

Figure 67: Theorem 31 Consider the circles (B,AB) and (C,CD). Note as HJ + KL > FG, we have BC +

CD > AB. But AB > BC, so there is a point X between C and D with AB ≡ BX (III.1) and

X is on the circle (B,AB). But X is between C and D, so XD < CD (definition of <); that is, X is interior to the circle (C,CD) (definition of interior). Similarly, as FG + HJ > KL, we have AB + BC > CD; that is, A is exterior to the circle (C,CD). Thus as (B,AB) contains points both interior and exterior to the circle (C,CD), so (B,AB) and (C,CD) meet at a point, say, E (VI).

Then E is on the circle (B,AB); in particular BE ≡ AB (definition of circle). But

AB ≡ FG, so BE ≡ FG (III.2). Similarly, E is on the circle (C,CD); in particular CE ≡ CD

(definition of circle). But CD ≡ KL, so CE ≡ KL. And we began by assuming BC ≡ HJ.

Thus the triangle BCE is as required. □

VII. Axiom of Parallels

The next axiom we present is the Axiom of Parallels, which says that parallel lines exist. Two lines are parallel if they do not have any points in common. Two segments are parallel if they are segments of parallel lines. Before we present the axiom, however, we will deduce a number of theorems concerning the properties of parallel lines. We will prove these theorems without assuming the parallel axiom.

Theorem 3244- If two lines are cut by a third line, and the alternate interior angles on opposite sides are congruent, the two lines are parallel.

Proof- Let �� and �� be two distinct lines cut by a third line ��, with E between

A and B and F between C and D, and let ∡BEF ≡ ∡CFE, and we show that �� and �� are parallel.

44 Together with the Corollary, this is Euclid I.27 84

Figure 68: Theorem 32

In order to derive a contradiction, suppose �� and �� are not parallel. Then �� meets �� in a point G (definition of parallel). By relabeling if necessary, we may assume that G is on the same side of �� as B and D. Now G is on the distinct lines �� and ��, so EFG is a triangle. Then ∡CFE is exterior to EGF, and thus ∡CFE > ∡FEG (Theorem

23). But ∡CFE ≡ ∡FEG, contradicting Exercise 12a. Thus, no such G exists, and �� and

�� are parallel. □

Corollary- If two lines are cut by a third line, and an exterior angle is congruent to the alternate interior angle on the same side of the third line, the two lines are parallel.

Proof- Let �� and �� be two distinct lines cut by a third line ��, with A and C on the same side of ��, with E between A and B, and with F between C and D. Let G be a point of �� with E between F and G, let ∡AEG ≡ ∡CFE, and we show �� and �� are parallel.

Figure 69: Corollary to Theorem 32

Note ∡AEG and ∡BEF are vertical angles, so ∡AEG ≡ ∡BEF (Corollary to

Theorem 10). Then ∡BEF ≡ ∡CFE (IV.2), and ∡BEF and ∡CFE are alternate interior angles, so the two lines �� and �� are parallel (Theorem 32). □

Theorem 3345- If two lines are cut by a third line, and the alternate interior angles on the same side of the third line are supplementary, the two lines are parallel.

Proof- Let �� and �� be two distinct lines cut by a third line ��, with A and C on the same side of ��, with E between A and B, and with F between C and D. Let

∡AEF and ∡CFE be supplementary angles, and we show �� and �� are parallel.

45 Euclid I.28 86

Figure 70: Theorem 33

As A, E, and B are in a line, with E between A and B, ∡AEF and ∡BEF are also supplementary angles (definition of supplementary). Then ∡BEF ≡ ∡CFE (Theorem 10).

But ∡BEF and ∡CFE are alternate interior angles, so �� and �� are parallel lines

(Theorem 32). □

Theorem 3446- Through a given point not on a given line, there is a line parallel to the given line.

Figure 71: Theorem 34

Proof- Let b be a line, and let A be a point not on the line b, and we show there is a line parallel to b passing through A. Let B and C be distinct points on the line b (I.3).

46 Euclid I.31 87

Then there is a point D on the opposite side of �� from C, with ∡ABC ≡ ∡BAD (IV.1).

Then ∡ABC and ∡BAD are congruent alternate interior angles, so b and �� are parallel

(Theorem 32). □

Axiom VII. Through any point A not lying on a line a, there is at most one line parallel to a.

Figure 72: Axiom VII

Note that combining Theorem 34 and Axiom VII, we see that through any point A not lying on a line a, there is exactly one line parallel to a. It is worth noting that there are other geometries which satisfy all the axioms we have previously stated besides the

Parallel axiom. Of particular interest is hyperbolic geometry, a type of geometry developed in the 19th century which is part inspiration for the geometry of Einstein’s

General Theory of Relativity.47 In hyperbolic geometry, through any point A not on a line a, there are infinitely many lines parallel to a.

47 Greenberg pg. xxvi 88

Corollary48- If two lines are each parallel to a third line, then they are parallel to each other.

Figure 73: Corollary to Axiom VII

Proof- Suppose a and b are two lines, neither of which meets a third line c, and suppose that a and b meet in a point, say A, and we will derive a contradiction. Then a and b would both be lines passing through A that do meet c. This contradicts axiom V, so a cannot meet b. □

Exercise 19 - Suppose a and b are parallel, and let c be a line that meets a. Show that c meets b also.

Theorem 3549- If two parallel lines are cut by a third line, the alternate interior angles on opposite sides of the third line are congruent, an external angle is congruent to the alternate interior angle on the same side of the third line, and the alternate interior angles on the same side of the third line are supplementary.

48 Euclid I.30 49 Euclid I.29 89

Proof50- Let �� and �� be parallel lines cut by a third line ��, with A and C on the same side of ��, with E between A and B, and with F between C and D. Let G be a point of �� with E between F and G, let H be a point of �� with F between E and H, and we show that ∡AEF ≡ ∡EFD, ∡AEG ≡ ∡CFE, and that ∡AEF and ∡CFE are supplementary angles.

Figure 74: Theorem 35

In order to derive a contradiction, suppose that ∡AEF is not congruent to ∡EFD.

Then there is a point X on the same side of �� as A, with ∡XEF ≡ ∡EFD (IV.1), and X is not on the line ��. Thus �� is distinct from ��, and both lines are parallel to ��

(Theorem 32). But this contradicts VII, so ∡AEF must be congruent to ∡EFD.

Similarly, in order to derive a contradiction, suppose that ∡AEG is not congruent to ∡CFE. Then there is a point X on the same side of �� as A, with ∡XEG ≡ ∡CFE

(IV.1), and X is not on the line ��. Thus �� is distinct from ��, and both lines are parallel to �� (Corollary to Theorem 32). But this contradicts VII, so ∡AEG must be congruent to ∡EFD.

50 The proof is due to Hartshorne. 90

Once more, in order to derive a contradiction, suppose that ∡AEF and ∡CFE are not supplementary angles. Then there is a point X on the same side of �� as A, with

∡XEF and ∡CFE supplementary angles (IV.1), and X is not on the line ��. Thus �� is distinct from ��, and both lines are parallel to �� (Theorem 33). But this contradicts

VII, so ∡AEF and ∡CFE must be supplementary angles. □

Theorem 3651- In a triangle, the sum of two angles is supplementary to the third angle.

Figure 75: Theorem 36

Proof- Let ABC be a triangle, and we show that ∡BAC + ∡ABC is supplementary to ∡BCA. Let D be a point with C between B and D (II.2), and let E be a point on the same side of �� as A, with �� parallel to �� (Theorem 34). Then ∡BAC and ∡ACE are alternate interior angles on opposite sides of the line ��, so ∡BAC ≡

∡ACE (Theorem 35). Likewise, ∡ABC is an alternate interior angle on the same side of

�� as the exterior angle ∡ECD, so ∡ABC ≡ ∡ECD (Theorem 35). But ∡ACE + ∡ECD ≡

∡ACD, and hence ∡BAC + ∡ABC ≡ ∡ACD. As B,C, and D all lie on a line, ∡ACD and

51 Euclid I.32 91

∡BCA are supplementary angles (definition of supplementary). Thus, ∡BAC + ∡ABC is supplementary to ∡BCA. □

Theorem 3752- If parallel congruent segments are joined at the endpoints by segments which do not meet, the segments are also parallel and congruent.

Figure 76: Theorem 37

Proof- Let AB and CD be parallel segments with AB ≡ CD, such that AC and BD do not meet, and we show AC ≡ BD and the two are parallel. Then ∡ABC and ∡BCD are alternate interior angles on opposite sides of the line ��, so ∡ABC ≡ ∡BCD (Theorem

35). Now AB ≡ CD, and as BC = BC, BC ≡ BC (III.1), so the triangles ABC and BCD are congruent (SAS Theorem 9); in particular, AC ≡ BD and ∡ACB ≡ ∡CBD. But ∡ACB and ∡CBD are alternate interior angles on opposite sides of the line ��, so AC and BD are parallel (Theorem 32). Thus we have shown AC ≡ BD, and the two segments are parallel. □

52 Euclid I.33 92

The objects constructed in Theorem 37 have special names. A quadrilateral in which the opposite sides are parallel is a parallelogram. The segment BC constructed in the proof of Theorem 37 is a diagonal. More generally, a diagonal of a polygon is a segment joining any two nonadjacent vertices.

Theorem 3853- In a parallelogram, the opposite sides and angles are congruent, and a diagonal forms two congruent triangles.

Proof- Let AB and CD be parallel segments, and let AC and BD be parallel segments, and we show AB ≡ CD, AC ≡ BD, and the triangles ABC and DCB are congruent. As AB and CD are parallel, with ∡ABC and ∡DCB alternate interior angles on opposite sides of the line ��, ∡ABC ≡ ∡DCB (Theorem 35). Likewise, as AC and BD are parallel with ∡ACB and ∡DBC alternate interior angles on opposite sides of the line

��, ∡ACB ≡ ∡DBC (Theorem 35). As BC = BC, BC ≡ BC (III.1), and hence the triangles ABC and DCB are congruent (ASA Theorem 12); in particular, AB ≡ CD and

AC ≡ BD. □

53 Euclid I.34 93

VIII. Archimedes Axiom and Area

The next axiom we present, attributed to Archimedes, essentially says that given any two segments, if we add enough copies of the first segment to itself, we will eventually have a segment that is greater than the second segment.

Axiom VIII. Let A1 be any point upon a line between the arbitrarily chosen points A and

B. Take the points A2, A3, A4, …, so that A1 lies between A and A2, A2 between A1 and

A3, A3 between A2 and A4 etc. Moreover, let the segments AA1, A1A2, A2A3, A3A4, … be congruent to one another. Then, among this series of points, there always exists a certain point An such that B lies between A and An.

This axiom allows us to define a concept of area.54 In order to do so, however, would require a much longer and more advanced treatment than this text allows. Instead, we will treat area as an undefined relation, in addition to the undefined relations of incidence, betweenness, congruence of segments, and congruence of angles.

Recall that a simple polygon is a collection of distinct points A, B, C, …, X and segments AB, BC, CD, …, XA, beginning and ending at the same point, where no two segments have a point in common other than two consecutive sides meeting at an

54 Greenberg pg. 142 94 endpoint. Then area is an ordered equivalence relation on all simple polygons satisfying the following conditions: 55

Area Property 1. Congruent triangles have equal area.

Area Property 2. Sums of polygons with equal area have equal area.

Area Property 3. Differences of polygons with equal area have equal area, so long as we are subtracting the polygon with lesser area from the polygon with greater area.

Area Property 4. Halves of polygons with equal area have equal area.

Area Property 5. The area of the whole is greater than area of a part.

Area Property 6. Squares with equal area have equal sides.

We denote the area of a polygon ABC…X by area(ABC…X).

Theorem 3956- Parallelograms on the same base between the same parallels have equal areas.

Proof- Let a and b be parallel lines, let ABCD and EBCF be parallelograms, let

A, D, E, and F be on the line a, and let B and C be on the line b, and we show area(ABCD) = area(EBCF). Clearly, if A = E and D = F are the same points, ABCD and

EBCF are the same parallelogram, and thus have the same area. If the parallelograms are distinct, by relabeling if necessary, we may assume the four points on a are in the order

55 The conditions necessary to treat area as a primitive notion are due to Hartshorne 56 Euclid I.35 95

A, E, D, F (or with E = D) or the order A, D, E, F (Theorem 3). Both cases proceed in the same manner initially.

Since ABCD and EBCF are parallelograms, AD ≡ BC and EF ≡ BC (Theorem

38), and thus AD ≡ EF (III.2). Then adding DE to both sides of the congruence, AD + DE

≡ EF + DE. But AD + DE ≡ AE, and EF + DE ≡ DF, so AE ≡ DF (III.3). As ABCD is a parallelogram, we also have AB ≡ DC (Theorem 38). Moreover, AB and DC are parallel with ∡EAB is the alternate interior angle to the exterior angle ∡FDC on the same side of

��, so ∡EAB ≡ ∡FDC (Theorem 35). Thus, the triangle EAB and FDC are congruent

(SAS Theorem 9). Then the area(EAB) = area(FDC) (Area property 1).

Figure 77: Theorem 39, with order AEDF Suppose A, E, D, F are in such an order (or with E = D between A and F). Clearly area(ABCF) = area(ABCF) (Area property 1). Subtracting the congruent FDC and EAB from ABCF, we conclude area(ABCD) = area(EBCF) (Area property 3), as desired.

Figure 78: Theorem 39, with order ADEF

Otherwise, we have A, D, E, F in such an order. Then we can show BE and DC meet at a point G (Crossbar Theorem 7). Subtracting DEG from both EAB and FDC, we have area(ABGD) = area(EGCF) (Area property 3). Then adding BCG to both ABGD and EGCF, we conclude area(ABCD) = area(EBCF) (Area property 2). □

Theorem 4057- Parallelograms with congruent bases between the same parallels have equal areas.

Proof- Let a and b be parallel lines, let ABCD and EFGH be parallelograms, let

A, D, E, and H be on the line a, let B, C, F, and G be on the line b, and let BC ≡ FG, and we show area(ABCD) = area(EFGH).

Figure 79: Theorem 40 As FG and EH are opposite sides of a parallelogram, FG ≡ EH (Theorem 38). But

BC ≡ FG, so BC ≡ EH (III.2). As BC and EH are congruent parallels, BCEH is a parallelogram (Theorem 37). But BCEH has the same base BC as ABCD, so area(ABCD) = area(BCEH) (Theorem 39). Similarly, as BCEH has the same base EH as

EFGH, area(BCEH) = area(EFGH). (Theorem 39). Thus, area(ABCD) = area(EFGH). □

57 Euclid I.36 97

Theorem 4158- Triangles on the same base between the same parallels have equal areas.

Proof- Let a and b be parallel lines, let ABC and DBC be triangles, let A and D be on the line a, let B and C be on the line b, and we show area(ABC) = area(DBC).

Clearly, if A = D are the same points, then the ABC and DBC are the same triangle, and thus have the same area. Otherwise, there is a line parallel to AC passing through B

(Theorem 34); since it is not the line b, it is not parallel to a (VII), and thus meets a at a point E. Similarly, there is a line parallel to BD passing through C (Theorem 34); since it is not the line b, it is not parallel to a (VII), and thus meets a at a point F.

Figure 80: Theorem 41 Now EBCA and DBCF are parallelograms on the same base BC between the same parallels a and b, and thus area(EBCA) = area(DBCF) (Theorem 39). But AB is a diagonal of the parallelogram EBCA separating ABC from a congruent triangle, so ABC is half of EBCA (Theorem 38). Similarly, CD is a diagonal of the parallelogram DBCF separating DBC from a congruent triangle, so DBC is half of DBCF (Theorem 38). We conclude area(ABC) = area(DBC) (Area property 4). □

58 Euclid I.37 98

Theorem 4259- Triangles on congruent bases between the same parallels have equal areas.

Proof- Let a and b be parallel lines, let ABC and DEF be triangles, let A and D be on the line a, let B, C, E, and F be on the line b, and let BC ≡ EF, and we show area(ABC) = area(DEF). Now there is a line parallel to AC passing through B (Theorem

34); since it is not the line b, it is not parallel to a (VII), and thus meets a at a point G.

Similarly, there is a line parallel to DE passing through F (Theorem 34); since it is not the line b, it is not parallel to a (VII), and thus meets a at a point H.

Figure 81: Theorem 42 Now ACBG and DEFH are parallelograms on the congruent bases BC ≡ EF between the same parallels a and b, so area(ACBG) = area(DEFH) (Theorem 40). But

AB is a diagonal of the parallelogram ACBG separating ABC from a congruent triangle, so ABC is half of ACBG (Theorem 38). Similarly, DF is a diagonal of the parallelogram

DEFH separating DEF from a congruent triangle, so DEF is half of DEFH (Theorem 38).

We conclude area(ABC) = area(DEF) (Area property 4). □

59 Euclid I.38 99

Theorem 4360- Triangles with equal areas, on the same base, on the same side, are between the same parallels.

Proof- Let ABC and DBC be triangles, with area(ABC) = area(DBC), and with A and D on the same side of BC, and we show �� is parallel to ��.

Figure 82: Theorem 43 In order to derive a contradiction, suppose not. Then there is a line a parallel to

�� passing through A (Theorem 34). If a was also parallel to ��, then �� and �� would be parallel (Corollary to VII), contradicting the fact they meet at B, so a is not parallel to

�� and thus meets �� at a point E. Then ABC and EBC are triangles on the same base

BC between the same parallels �� and a, so area(ABC) = area(EBC) (Theorem 41). As E is on the line ��, one of DBC or EBC is part of the other, so area(EBC) ≠ area(DBC)

(Area property 5). Thus, area(DBC) ≠ area(ABC), contradicting our assumption. Thus, no other line besides �� passing through A can be parallel to ��. Therefore �� is parallel to �� (Theorem 34). □

60 Euclid I.39 100

Theorem 4461- Triangles with equal areas, on congruent bases, on the same side, are between the same parallels.

Proof- Let ABC and DEF be triangles, with area(ABC) = area(DEF), with B, C,

E, and F on a line b, with A and D on the same side of the line b, and let BC ≡ EF, and we show �� is parallel to b.

Figure 83: Theorem 44 In order to derive a contradiction, suppose not. Then there is a line a parallel to

�� passing through A (Theorem 34). If a was also parallel to ��, then �� and b would be parallel (Corollary to VII), contradicting the fact they meet at E, so a is not parallel to

�� and thus meets �� at a point G. Then ABC and GEF are triangles on the congruent bases BC ≡ EF between the same parallels b and a, so area(ABC) = area (GEF) (Theorem

41). As G is on the line ��, one of DEF or GEF is part of the other, so area(GEF) ≠ area(DEF) (Area property 5). Thus area(ABC) ≠ area(DEF), contradicting our assumption. Thus, no other line besides �� passing through A can be parallel to b.

Therefore �� is parallel to b (Theorem 34). □

61 Euclid I.40 101

Theorem 4562- A parallelogram on the same base between the same parallels as a triangle has double the area of the triangle.

Proof- Let a and b be parallel lines, let ABCD be a parallelogram, let BCE be a triangle, let A, D, and E be on the line a, and let B and C be on the line b, and we show area(ABCD) = 2×area(BCE).

Figure 84: Theorem 45 Then ABC and EBC are triangles with the same base BC between the same parallels a and b, so area(ABC) = area(EBC) (Theorem 41). But AC is a diagonal of the parallelogram ABCD separating ABC from a congruent triangle, so ABC is half of

ABCD (Theorem 38). Thus, area(ABCD) = 2×area(ABC), and therefore area(ABCD) =

2×area (BCE). □

Theorem 4663- Given a triangle and an angle, there is a parallelogram with area equal to that of the triangle and with one angle congruent to the given angle.

62 Euclid I.41 63 Euclid I.42 102

Proof- Let ABC be a triangle, let ∡D be an angle, and we show there is a parallelogram with area equal to that of the triangle ABC and with one angle congruent to

∡D. We know there is a line a passing through A and parallel to �� (Theorem 34). We also know there is a point E on �� with BE ≡ EC (Theorem 20). Let �� be the ray originating at E, on the same side of �� as A, making ∡FEC ≡ ∡D (IV.1). Then as �� passes through E and is not the line ��, �� meets a at a point (VII); by relabeling, we may assume F is on the line a. Now there is a line c passing through C and parallel to ��.

Again as c passes through C and is not the line ��, c meets a at a point G (VII). Then

ECGF is a parallelogram.

Figure 85: Theorem 46 Note ABE and AEC are triangles on congruent bases BE ≡ EC between the same parallels a and ��, so area(ABE) = area(AEC) (Theorem 42). Thus, area(ABC) = 2×area

(AEC) (Area property 4). But ECGF is a parallelogram on the same base EC and between the same parallels a and �� as the triangle AEC, so area(ECGF) = 2×area(AEC)

(Theorem 45). Thus, area(ECGF) = area(ABC). □

Theorem 47- Any two perpendicular segments between two parallel lines are congruent.

103

Proof- Suppose that two parallel lines a and b are joined by two perpendicular segments AD and BC, with both A and B on the line a, and we show AD ≡ BC. Since

AD and BC are perpendicular to a, ∡BAD and ∡ABC are right angles (definition of perpendicular), and thus supplementary angles (definition of right angle). But ∡BAD and

∡ABC are alternate interior angles on the same side of the line a, and hence AD and BD are parallel (Theorem 33). Then ABCD is a parallelogram and AD ≡ BC (Theorem 38). □

Note that using Theorem 47, we can define the distance between two parallel lines a and b as the (congruence class of a) segment AD perpendicular to a (and hence also perpendicular to b) between a and b. Since Exercise 9 and SAS Theorem 9 allow us to construct congruent triangles on any line at any point we wish, and since congruent triangles have equal areas, we can reinterpret Theorems 39 through Theorems 47 and replace the phrase “between the same parallels” by “between parallels the same distance apart.”

Theorem 4864- Given a parallelogram and an interior point on a diagonal, if the point is connected to the sides by lines parallel to the sides of the parallelogram, the two parallelograms formed on either side of the diagonal have equal areas.

Proof- Let ABCD be a parallelogram and let K be an interior point on the diagonal AC. Let E be between A and B, let F be between C and D, let G be between B

64 Euclid I.43 104 and C, and let H be between A and D, such that �� passes through K and is parallel to

�� (and hence also to ��) and �� passes through K and is parallel to �� (and hence also to ��), and we show area(EGBK) = area(HKFD).

Figure 86: Theorem 48 As AC is a diagonal of the parallelogram ABCD, the triangles ABC and CDA are congruent (Theorem 38); in particular, area(ABC) = area(CDA). Similarly as AK is a diagonal of the parallelogram AEKH, the triangles AEK and KHA are congruent

(Theorem 38); in particular, area(AEK) = area(KHA). Again, as KC is a diagonal of the parallelogram KGCF, the triangles KGC and CFG are congruent (Theorem 38); in particular, area(KGC) = area(CFG).

Then subtracting AEK and KGC from ABC and subtracting KHA and CFG from

CDA gives area(EGBK) = area(HKFD) (Area property 3). □

Theorem 4965- Given a segment, an angle, and a triangle, there is a parallelogram with the given side, with one angle congruent to the given angle, and with area equal to that of the given triangle.

65 Euclid I.44 105

Proof- Let AB be a segment, let ∡C be an angle, and let DEF be a triangle, and we show there is a parallelogram with side AB, with one angle congruent to ∡C, and with area equal to that of DEF.

We know there is a point G with B between A and G such BG ≡ DE (III.1). Let

EX be a segment perpendicular between DE and the line parallel to DE passing through F

(Theorem 21). Let BY be such that BY ≡ EX, with BY perpendicular to AB (Theorem

21), and let y be the line parallel to AB passing through Y (Theorem 34). Now there is a ray �� originating at B, such that ∡GBH ≡ ∡C (IV.1); since �� is not parallel to AB,

BH is not parallel to y (Exercise 19), so let H be on the line y. Then the triangles GBH and DEF are triangles on the same base between parallels the same distance apart, and thus area(GBH) = area(DEF) (Theorem 42).

Figure 87: Theorem 49

Let J be a point of y, on the same side of �� as G, with HJ ≡ BG (III.1). Then JH and BG are congruent parallel segments, so JHBG is a parallelogram (Theorem 37). But 106

JHBG is a parallelogram on the same base GB and between the same parallels y and �� as the triangle GBH, so area(JHBG) = 2×area(GBH) (Theorem 45).

Let K be the midpoint of HB and let L be the midpoint of JG (Theorem 20); that is, HK ≡ KB and JL ≡ LG. Then HKLJ and LKBG are parallelograms (Theorem 37) with congruent bases HK ≡ KB, and hence area(HKLJ) = area(LKBG). But HKLJ and LKBG are halves of JHBG; thus, area(JHBG) = 2×area(LKBG). Thus we have area(LKBG) = area(DEF).

Now there is a line a parallel to �� passing through A (Theorem 31). As �� is parallel to �� and passes through A, a is not parallel to �� (VII), and thus meets �� at a point M. As �� is parallel to a and passes through M, the line �� is not parallel to ��

(VII), and thus meets �� at a point N. Again there is a line n parallel to �� passing through N (Theorem 31). As �� is a line parallel to both �� and a passing through N, n is not parallel to �� or a (VII), and thus meets �� at a point O and a at a point P.

Note MLNP is a parallelogram with diagonal MN, with LKBG and BAPO internal parallelograms opposite the point B. Then area(LKBG) = area(BAPO) (Theorem

48), and thus as area(LKBG) = area(DEF), we conclude area(BAPO) = area(DEF).

Now ∡GBK ≡ ∡C. Moreover, �� and n are parallel, and ∡BON is the alternate interior angle to the exterior angle ∡GBK on the same side of ��, so ∡GBK ≡ ∡BON

(Theorem 35). But �� and a are parallel, and ∡APO is the alternate interior angle to the exterior angle ∡BON on the same side of n, so ∡BON ≡ ∡APO (Theorem 35). Thus, ∡E

≡ ∡APO (IV.2), and the parallelogram BAPO satisfies the desired properties. □

107

Theorem 5066- Given a quadrilateral and an angle, there is a parallelogram with area equal to that of the quadrilateral and with one angle congruent to the given angle.

Proof- Let ABCD be a quadrilateral, let ∡E be an angle, and we show there is a parallelogram with area equal to that of ABCD and with one angle congruent to ∡E. Note

DB is a diagonal of the quadrilateral ABCD and cuts ABCD into two triangles ABD and

CBD. Let FGHJ be a parallelogram with area(FGHJ) = area(ABD) and with ∡FGH ≡ ∡E

(Theorem 46), and let JHKL be a parallelogram constructed on the segment JH, on the opposite side of �� as FGHJ, with ∡JHK ≡ ∡E, and with area(JHKL) = area(CBD)

(Theorem 49).

Figure 88: Theorem 50

Now as ∡FGH ≡ ∡E and ∡JHK ≡ ∡E, we have ∡FGH ≡ ∡JHK (IV.2). As JL and

HK are parallel and ∡LJH and ∡JHK are alternate interior angles on the same side of the line ��, ∡LJH and ∡JHK are supplementary angles (Theorem 35). Then ∡LJH and

∡FGH are also supplementary angles (Theorem 10). But ∡FGH and ∡FJH are opposite

66 Euclid I.45 108 angles in the parallelogram FGHJ, so ∡FGH ≡ ∡FJH (Theorem 38). Thus, ∡LJH and

∡FJH are supplementary angles (Theorem 10), and F, J, and L all lie on a line (definition of supplementary angles). An identical argument shows that G, H, and K all lie on a line.

We have FL and GK are parallel segments, with J between F and L, and with H between G and K. As FGHJ is a parallelogram, FJ ≡ GH (Theorem 38), and similarly as

JHKL is a parallelogram, JL ≡ HK (Theorem 38). Thus, FL ≡ GK (III.3), and as the segments are parallel, FGKL is a parallelogram (Theorem 37). Now as ABCD is the sum of ABD and CBD, and FGKL is the sum of FGHJ and JHKL, with area(ABD) = area(FGHJ) and area(CBD) = area(JHKL), we conclude area(ABCD) = area(FGKL).

Moreover, ∡FGK = ∡FGH (Exercise 2f), and ∡FGH ≡ ∡E, and thus ∡FGK ≡ ∡E. □

Corollary- Given any polygon and an angle, there is a parallelogram with area equal to that of the polygon and with one angle congruent to the given angle.

Proof- Let A1A2…An be a polygon, let ∡B be an angle, and we show there is a parallelogram with area equal to that of A1A2…An and with one angle congruent to ∡B, by induction on n = the number of sides of the polygon. If n = 3 or n = 4, we have

Theorem 46 and Theorem 49, respectively. Suppose the Corollary is true for and polygon with n-1 sides, and we show it is true for any polygon with n sides.

We omit the details, but give an outline of the proof. Note A3A1 is a diagonal of the polygon, separating A1A2…An into a triangle A1A2A3 and a polygon A1A3A4…An with n-1 sides. By the induction hypothesis, there is a parallelogram CDEF with area

109 equal to that of A1A3A4…An and with one angle congruent to ∡B. Using a similar method of construction as in Theorem 50, we add a new parallelogram EFGH with area equal to A1A2A3 to CDEF and get a parallelogram CDGH that satisfies the desired properties. □

Recall that a parallelogram has adjacent angles supplementary and opposite angles congruent. In the special case where two adjacent angles of a parallelogram are congruent, we can conclude that all four angles are congruent, and since they are congruent to their supplementary angles, they are right angles. A parallelogram in which all four angles congruent is a rectangle. A rectangle in which all four sides are also congruent is a square.

Theorem 5167- Every segment is the base of a square.

Figure 89: Theorem 51

67 Euclid I.46 110

Proof- Let AB be a segment, and we show there is a square with side AB. Let b be a line perpendicular to �� passing through B (Theorem 21). Then there is a point C on b with BC ≡ AB (III.1). Now let c be a line parallel to �� passing through C (Theorem

34). Then there is a point D on c with CD ≡ AB (III.1). As AB and CD are parallel congruent segments, ABCD is a parallelogram (Theorem 37). But ∡ABC is a right angle.

As adjacent angles of a parallelogram are supplementary (Theorem 33), ∡BAD and

∡BCD are both supplementary angles to the right angle ∡ABC, and hence right angles

(definition). As opposite angles of a parallelogram are congruent (Theorem 38), ∡CDA ≡

∡ABC and ∡CDA is also a right angle. Thus, ABCD is a square. □

In a right triangle, the side opposite the right angle is the hypotenuse of the triangle, and the sides adjacent the right angle are the legs of the triangle.

Theorem 5268- In a right triangle, the sum of the areas of squares on the legs is equal to the area of a square on the hypotenuse.

Proof- Let ABC be a triangle with ∡BAC a right angle. Then let BDEC be a square with D and E on the opposite side of �� as A, let AFGB be a square with F and G on the opposite side of �� as C, and let AHJC be a square with H and J on the opposite side of �� as B (Theorem 51), and we show area(AFGB) + area(AHJC) = area(BDEC).

68 Euclid I.47 111

Figure 90: Theorem 52

We know there is a line a parallel to �� (and hence also parallel to ��) passing through A (Theorem 34). As �� and �� are distinct lines passing through D, and �� is parallel to a, �� is not parallel to a (VII), so �� meets a at a point K. In order to derive a contradiction, suppose K is not between D and E. Then either D is between E and K, or E is between D and K; both cases follow the same argument, so suppose D is between E and K. Then A and K are on opposite sides of the line ��, so the segment AK meets ��

(Theorem 2). But as A and K are points of the line a, this contradicts a is parallel to ��.

Thus, the segment DE meets a at a point K. Now there is a point L between A and K with

KL ≡ DB (III.1); then DB and LK are congruent parallel segments, so BDKL is a parallelogram and L is between B and C (Theorem 37).

As ∡CBD and ∡ABG are right angles, ∡CBD ≡ ∡ABG (Theorem 14). Adding

∡ABC to both sides of the congruence, ∡CBD + ∡ABC ≡ ∡ABD, and ∡ABG + ∡ABC ≡ 112

∡GBC (definition of addition), so ∡ABD ≡ ∡GBC. But AFGB and BDEC are squares, so

AB ≡ GB and BD ≡ BC (definition of square). Thus, the triangles ABD and GBC are congruent (SAS Theorem 9), and area(ABD) = area(GBC) (Area property 1).

Note ∡BAC and ∡BAF are right angles, with F and C on opposite sides of the line ��, and thus ∡BAC and ∡BAF are supplementary angles, and F, A, and C are all points of the same line (definition of supplementary angles). Then AFGB is a parallelogram on the same base GB and between the same parallels �� and �� as the triangle GBC, so area(AFGB) = 2×area(GBC) (Theorem 45). But as area(ABD) = area(GBC), we conclude area(DBKL) = area(AFGB). Similarly, BDKL is a parallelogram on the same base BD and between the same parallels �� and �� as the triangle ABD, so area(DBKL) = 2×area(ABD) (Theorem 45).

In the same manner, we can conclude that area(LKEC) = area(AHJC). Then as

BDEC is the sum of BDKL and LKEC, area(BDEC) = area(AFGB) + area(AHJC). □

Exercise 20- Complete the proof of Theorem 52 by showing area(LKEC) = area(AHJC), by starting with the construction from the end of the second paragraph, and repeating the remaining argument.

Exercise 21- Let ABCD and EFGH be squares. a. If AB ≡ EF, show that area(ABCD) = area(EFGH). b. If area(ABCD) = area(EFGH), show AB ≡ EF.

113

Theorem 5369- In a triangle, if the sum of the areas of squares on two sides is equal to the area of a square on the third side, the triangle is a right triangle.

Proof- Let ABC be a triangle, let ADEB, CFGB, and AHJC squares constructed on the given segments (Theorem 51), let area(ADEB) + area(AHJC) = area(CFGB), and we show ∡BAC is a right angle.

Figure 91: Theorem 53

Let K be a point on the opposite side of �� as B, with AK ≡ AB, such that ∡CAK is a right angle (Exercise 9). Let KLMA be a square. As AK ≡ AB, area(KLMA) = area(ADEB) (Exercise 21a). Adding the area of square AHJC to both sides, area(KLMA)

+ area(AHJC) = area(ADEB) + area(AHJC) (Area property 2). But ∡CAK is a right

69 Euclid I.48 114 angle, so area(KLMA) + area(AHJC) = area(CKNO), where CKNO is a square constructed on CK (Theorem 52). And we assumed area(ADEB) + area(AHJC) = area(CFGB), so area(CKNO) = area(CFGB). As both polygons are squares, CK ≡ CB

(Exercise 21b). Note AC = AC, and hence AC ≡ AC (III.1). Thus, the triangles ABC and

AKC are congruent (Theorem 18); in particular, ∡CAB ≡ ∡CAK. As ∡CAK is a right angle, ∡CAB is also a right angle. □

115

IX. Axiom of Completeness

The final axiom we present is an Axioms of Completeness. The axiom is not a geometric statement, per se, asserting only that it is impossible to extend our geometry.

The consequence of this axiom lies beyond the scope of this text. We make no use of it, and present it only for completeness.

Axiom IX. To a system of points and straight lines, it is impossible to add other elements in such a manner that the system shall form a new geometry obeying all the axioms previously stated.70

70 This axiom was added later by Hilbert, to allow for a development of the real numbers within the geometry. 116

References

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Euclid. The Elements. Trans. D.E. Joyce. 1996. 26 November 2012 .

Eves, Howard. A Survey of Geometry. Boston: Allyn and Bacon, Inc., 1963.

Faber, Richard. Foundations of Euclidean and Non-Euclidean Geometry. New York: Marcel Dekker, Inc., 1983.

Fishback, W.T. Projective and Euclidean Geometry. New York: John Wiley & Sons, Inc., 1969.

Greenberg, Marvin Jay. Euclidean and Non-Euclidean Geometries: Development and History. New York: W. H. Freeman and Company, 2008.

Hartshorne, Robin. Euclid and Beyond. New York: Springer-Verlag New York, Inc., 2000.

Hilbert, David. The Foundations of Geometry. Trans. E. J. Townsend. La Salle, Il: The Open Court Publishing Company, 1950. Digitized by Joshua Hutchinson, Roger Frank, and David Starner. 26 November 2012. .

Moore, E. H. “On the projective axioms of geometry.” Transactions of the American Mathematical Society, vol. 3, 1902, 142-158.

Pasch, Moritz. Vorlesungen uber neuere Geometrie. Berlin: Springer, 1882. Digitized by Google. 26 November 2012.

Tarski, Alfred and Jan Tarski. Introduction to Logic and to the Methodology of Deductive Sciences (4 ed.). Oxford: Oxford University Press, 1994. 117

Tarski, Alfred and Steven Givant. “Tarski’s System of Geometry.” The Bulletin of Symbolic Logic, vol. 5, 1999, 175-214.

Venema, Gerard A. Foundations of Geometry. Upper Saddle River, NJ: Pearson Prentice Hall, 2005.

118

Appendix A: List of Axioms

Axiom I.1. Two points A and B determine a line a. We write �� = a or �� = a. We will often use other terminology to refer to the same relation of containment; for instance, A lies upon a, A is a point of a, a contains A, a goes through A and through B, a joins A and B, etc.

Axiom I.2. Any two points of a line determine that line; that is, if �� = a and ��= a, where B ≠ C, then also �� = a.

Axiom I.3. Every line contains at least two points, and there are at least three points not all on the same line.

Axiom II.1. If A, B, O are three distinct points of a line, then O is between A and B if and only if O is also between B and A. Note that we require all three points to be distinct; thus a point A cannot be between A and any other point B.

Axiom II.2. If A and C are two distinct points of a line, then there is at least one point B lying between A and C and at least one point D with C between A and D.

Axiom II.3. Given three distinct points of a line, one and only one lies between the other two.

Axiom III.1. If A, B are distinct points on a line a, and if C is a point on a line c, then on each side of C on the line c there is one and only one point D such that the segment AB is congruent to the segment CD. We indicate this relation by writing AB ≡ CD. Every segment is congruent to itself; that is, we always have AB ≡ AB.

Axiom III.2. If a segment AB is congruent to the segment CD and also to the segment EF, then the segment CD is congruent to the segment EF; that is, if AB ≡ CD and AB ≡ EF, then CD ≡ EF.

119

Axiom III.3. Let AB and BC be two segments of a line a which have no points in common other than the point B, and let DE and EF be two segments of a line d which have no points in common other than the point E. Then if AB ≡ DE and BC ≡ EF, we have AC ≡ DF.

Axiom IV.1. Let ∡AOB be an angle and let DE be a line. Then on a given side of the line DE there is one and only one ray �� such that the angle AOB is congruent to the angle ∡CDE and all interior points of ∡CDE lie on the given side of the line ��.

Axiom IV.2. If the angle ∡A is congruent to the angle ∡B and to the angle ∡C, then the angle ∡B is congruent to the angle ∡C; that is, if ∡A ≡ ∡B and ∡A ≡ ∡C, then ∡B ≡ ∡C.

Axiom IV.3. If, for the two triangles ABC and DEF, the congruencies AB ≡ DE, AC ≡ DF, ∡A ≡ ∡D, hold, then the congruencies ∡B ≡ ∡E and ∡C ≡ ∡F also hold.

Axiom VI- If (O,OA) and (B,BC) are two circles, and (O,OA) contains both a point interior to and exterior to (B,BC), then (O,OA) and (B,BC) meet at a point.

Axiom VII. Through any point A not lying on a line a, there is at most one line parallel to a.

Axiom VIII. Let A1 be any point upon a line between the arbitrarily chosen points A and B. Take the points A2, A3, A4, …, so that A1 lies between A and A2, A2 between A1 and A3, A3 between A2 and A4 etc. Moreover, let the segments AA1, A1A2, A2A3, A3A4, … be congruent to one another. Then, among this series of points, there always exists a certain point An such that B lies between A and An.

Axiom IX. To a system of points and straight lines, it is impossible to add other elements in such a manner that the system shall form a new geometry obeying all the axioms previously stated.

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Appendix B: A Note to the Instructor

The goal of this text is to present Euclidean Geometry, specifically Book I of Euclid’s Elements, in a manner consistent with modern mathematical rigor. To do so, we will base our study in the axioms David Hilbert gave in his 1899 Foundations of Geometry. When possible, we have kept as close as possible to the Elements. But before embarking on this study of geometry, three questions arise naturally: What is geometry? Why study geometry? And why study Euclid’s geometry, with Hilbert’s axioms? The first two we can answer quite briefly; the third requires some consideration.

The word geometry originates from the Greek words for earth and measurement, as geometry initially developed with the intent specifically to measure land, and generally to measure any objects. There is evidence that as early as the 3rd millennia BC, ancient cultures in places widespread such as Egypt, Babylon, India, and China were aware of and recorded various geometric facts.71 These ancient records generally refer to the measurements of land to levy the proper tax, and to the measurements of volumes to facilitate in commerce. In many regards, the geometry known was quite sophisticated and significant; for instance, by the year 2000 BC, the Babylonians knew how to compute the areas of a rectangle, a right triangle, an isosceles triangle, the trapezoid with one side perpendicular to the parallels, and the circle, using the value of 3 for �. They knew the volumes of a rectangular parallelepiped, a right prism with special trapezoidal base, and a right circular cylinder. They knew that corresponding sides of similar right triangles were proportional, that the altitude through the vertex of an isosceles triangle bisects the base, that an angle inscribed in a semi-circle is right, and the Pythagorean Theorem.72 Slightly more recent artifacts reveal Babylonian tables of Pythagorean triples, as well as a method for determining new triples.73 Book I of the Elements (and the present text) culminates in a proof of the Pythagorean Theorem and its converse; thus, the facts contained in Euclid’s Elements significantly predate the Elements itself. These early studies of geometry are focused on observation and experimentation, however. The concept of “proof” as we understand it today developed much later. For example, Middle Kingdom Egyptian (2000-1800 BC) texts give proofs by checking that an answer satisfies the properties.74 “Pre-Hellenistic mathematics was little more than a

71 Coolidge pgs. 5-23 72 Eves pg. 4 73 Faber pgs. 12-14 74 Faber pgs. 31-32 121 practically workable empiricism—a collection of rule-of-thumb procedures that gave results of sufficient acceptability for the simple needs of those early civilizations.”75 In this regard, the study of geometry was merely the study of a collection of facts and theorems. This is a practically useful approach, but it is a limited one. Unlike the ancient civilizations that began the science, we concern ourselves with more than just observations about land and objects. Instead, we consider idealized objects; namely points without any dimension, lines without any width, and planes without any volume. The study of geometry then is the study of how these idealized objects relate to one another. In so doing, we seek to learn facts which are true in every instance which can be modeled by geometry, rather than to observe facts that are true in a particular instance or situation. While there are a number of interesting and useful geometric facts we will encounter, in a certain sense, learning geometric facts is not the primary goal in studying geometry. Particularly now when most facts can be looked up on the internet, the importance of memorizing or cataloging geometric theorems is less important. Rather, geometry is a subject which allows us to study the nature of deduction and proof. We concern ourselves foremost with the notion of a logical, sound argument. In our study of geometry, we begin with a small set of basic principles and deduce a large and varied set of theorems. We do so by justifying every step of the argument with one of our basic principles. This method is employed in every branch of mathematics and science, but it is particularly evident in geometry.

This approach to geometry, characterized by logically deduced proofs of general theorems about abstract objects, was first employed by the ancient Greeks. Geometry transformed from a practice of observation to the science of developing and deducing theorems about abstract, idealized objects.76 The Greek interpretation of geometry did not depend upon real world objects, though these abstract geometrical entities were approximations or representations of real world objects. Thus for instance a line is truly an infinite length of points having no breadth, even if we have never encountered an object with infinite length or an object with no breadth. Using these abstract entities, the Greeks strove to prove statements without appealing to empirical results or observations, instead deducing statements from the characteristic properties of these abstract objects. As Plato wrote in the Republic, “the knowledge at which geometry aims is knowledge of the eternal…geometry will draw the soul toward truth, and create the spirit of philosophy.”77 The characteristic properties of these abstract objects are the postulates, or axioms, which we agree to take as true. The first attempt to study geometry by first presenting a set of axioms, and then deducing a chain of propositions which follow logically, was by the Pythagorean Hippocrates of Chios. Leon, Theudius, and others soon

75 Eves pg. 6 76 Greenberg pg. 3 77 Hartshorne pg. 8 122 followed,78 but the Greek process of developing and improving upon this axiomatic and logical derivation of geometry culminated in Euclid’s Elements around 300 BC. The Elements starts from a base of definitions, five postulates, and five common notions, and proceeds to justify essentially the totality of planar geometry known to the Greeks. Indeed, the Elements presents a substantial portion of the planar geometry which can be derived by purely geometric methods even today. Dating back to immediately after Euclid wrote them, the Elements has been considered the standard for the manner in which a logical argument and careful thought should progress. Indeed, the Elements serves as model for the presentation of nearly all modern mathematical and philosophical theories. Thus we know what geometry is, why we study it, and why we follow Euclid’s Elements, but we have yet to answer why we use Hilbert’s axioms. In answering that question, we first return to the Elements.

Particularly admired is Euclid’s ability to recognize a small set of propositions from which all others could be concluded.79 The small set of assumptions Euclid set out to prove all of Greek geometry from consists of five postulates and five common notions. The five common notions were meant to be self-evident truths common to all subjects: 1. Things which equal the same thing also equal one another. 2. If equals are added to equals, then the wholes are equal. 3. If equals are subtracted from equals, then the remainders are equal. 4. Things which coincide with one another equal one another. 5. The whole is greater than the part. The five postulates, or axioms, were to be facts deduced from real world observations that everyone could agree with: 1. To draw a straight line from any point to any point. 2. To produce a finite straight line continuously in a straight line. 3. To describe a circle with any center and radius. 4. That all right angles equal one another. 5. That, 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.80

Euclid’s first four postulates are readily apparent, the fifth less so. The earliest record we have of Greek mathematics, including Euclid’s Elements, is Proclus’s Eudemian Summary from the 5th century. By that early date, Proclus was already unsatisfied with Euclid’s choice of the fifth postulate. Proclus went so far as to describe the fifth postulate as “alien to the special character of postulates,” and believed that the fifth postulate should be a theorem instead.81 In his commentary on the Elements, Proclus

78 Eves pg. 10 79 Venema pg. 3 80 Joyce 81 Faber pg. 127 123 even included a proof of the fifth postulate from the first four, which would later be shown mistaken.82 In the 2300 years that have passed since Euclid, many other mistaken proofs of the fifth postulate have been proposed. After the fall of the Roman Empire, the Elements was lost to the European world, but faulty proofs of the fifth postulate were provided by the Arab mathematicians al-Gauhary, al-Haytham, Omar Khayyam, and Nasir ad-Din at- Tusi. After Euclid’s Elements was first translated into Latin by Gherardo of Cremona, and then printed in 1482,83 the Elements has been studied and translated more than any book besides the Bible,84 and has most likely inspired more mistaken proofs than any other book. Particularly famous faulty proofs of the fifth postulate were provided in the th th 17 and 18 century by Wallis, Saccheri, Lambert, and Legendre, who alone spent over 40 years working on proofs of the fifth postulate. Perhaps as many as 10,000 different proofs of the fifth postulate have been published. 85 In his dissertation 1763, Klugel first suggested that proof of the fifth postulate might be impossible.86 By 1824, Gauss privately accepted the possibility of non- Euclidean geometry, i.e. a geometry in which the fifth postulate failed, but did not publish any results.87 In 1829, Lobachevsky published the first article on a non-Euclidean geometry, hyperbolic geometry.88 In 1832, Bolyai published the independence of the fifth postulate, proving that any attempted proof of the fifth postulate from the other four is impossible.89 In 1868, Beltrami constructed a detailed model of hyperbolic geometry within Euclidean geometry, and showed that hyperbolic geometry is relatively consistent with Euclidean geometry. That is, if Euclidean geometry accepting the fifth postulate is valid, then so is non-Euclidean geometry rejecting the fifth postulate.90 These later mathematicians, like Euclid, viewed geometry as the study of the logical consequences of a set of assumptions, as opposed to the mathematical description of the real world.91 Taking this approach, and using the more advanced mathematics which had been developed in the intervening years, they were able to vindicate Euclid in his decision to make the fifth postulate a postulate, and not a theorem. This brief history the development of non-Euclidean geometry may seem out of place in a book about Euclidean geometry. However, the history of modern geometry is intricately tied to this history; for in the process of developing other consistent geometries, modern mathematicians began to reconsider Euclid’s axioms and proofs more rigorously. As Howard Eves wrote in A Survey of Geometry, “It is no exaggeration to say that the historical consideration of the independence of Euclid’s parallel postulate

82 Fishback pg. 29 83 Eves pg. 47 84 Eves pg. 19 85 Fishback pg. 29 86 Faber pg. 147 87 Faber pgs. 156-159 88 Faber pg. 163 89 Faber pg. 161 90 Greenberg pg. 292 91 Venema pg. 162 124 is responsible for initiating the entire study of properties of postulate sets and hence for shaping much of the modern axiomatic method.”92

And though the accomplishments of Euclid’s Elements are quite impressive, the work is in no way perfect. This was first noticed by Gauss while in the process of studying the possibility of a non-Euclidean geometry. In 1831, Gauss observed to Bolyai that Euclid failed to state explicitly all the assumptions necessary in the Elements, albeit in private correspondence and without mentioning which assumptions were lacking.93 Riemann identified the first such unstated assumption in his 1854 dissertation, noting that Euclid did not distinguished between the boundlessness and infinitude of straight lines.94 For instance, a “line” on a sphere is a great circle which has no end, but finite length. A second axiom to be found unstated in the Elements is a continuity axiom. In the proof of the very first proposition, Euclid constructs two circles which cross each other, and concludes that they much intersect, although this conclusion does not follow from any axiom. In 1871, Dedekind proposed an axiom which accounted for this; a modified version of the axiom became the basis for the “Dedekind cut” so pivotal in the development of the real number system.95 A third omission of the Elements is the failure to define any axioms for superposition, a modern term for the method Euclid uses in the proof of his fourth proposition.96 In his proof, Euclid interprets his fourth common notion “Things which coincide with one another equal one another” to mean that an object is unchanged under superposition. Using the modern language of transformations, this might be interpreted as saying all geometric properties are invariant under rigid motions of the Euclidean group on the plane; that is, reflections, rotations, translations, and glide reflections.97 This is, however, a property unique to Euclidean geometry. In fact, in his Erlanger Programme of 1872, Klein defined a geometry as a study of properties that remain unchanged under a given group of transformations;98 thus, this unstated axiom can in fact be seen as a defining property of Euclidean geometry. In 1882 Pasch noticed a further omission from the Elements; the lack of any properties of linear order. To rectify this, he introduced what has come to be known as Pasch’s Postulate, which essentially says that in a plane, if a line passes through one side of a triangle, it must pass through one other side or vertex. Using this axiom, Pasch formed the first complete axiomatization of Euclidean geometry.99 His axioms, unlike Euclid’s, also guaranteed that points and lines exist, that not all points are collinear, and

92 Eves pg. 398 93 Greenberg pg. 104 94 Eves pg. 374 95 Greenberg pg. 134 96 Fishback pg. 5 97 Venema pg. 285 98 Greenberg pgs. 122-123 99 Eves pg. 380 125 that every line has at least two points,100 which Euclid probably did not think necessary to justify. There is one final flaw to be mentioned about the Elements, though it is not a flaw or omission in the axioms themselves: the Elements begins with a list of definitions. Euclid recognized the impossibility of proving everything; certain postulates must be assumed true, and from those postulates other propositions can be proven. The same holds with defining objects and relations to use in our mathematics; if we don’t assign some undefined terms, or primitive notions, we will ultimately define everything in a circular manner, or we will never finish defining everything. Thus all modern axiom systems, dating back to Pasch’s, begin with a list of undefined terms, or primitive notions. These are usually terms, with which we are familiar, and of which we hold intuitive interpretations as to their meanings; strictly speaking, however, the axioms themselves define the properties of these undefined terms, and the theorems we deduce are independent of our preconceived interpretations101. In addition to his 14 Axioms, Pasch included the use of 4 primitive notions in his 1882 axiom set: point, line segment, planar section, and between.102

Concurrent with the axiomatization of geometry, the entire spectrum of mathematics underwent a similar axiomatization. In 1879 Frege introduced quantifiers into logic and developed the predicate calculus. He was followed in the 1880s by an Italian group of logicians led by Giuseppe Peano and Mario Pieri, who used a formal symbolic language to make mathematics entirely precise. The first major presentation using this formal language came from Peano in 1889, when he gave explicit axioms for the development of the natural numbers,103 and around the same time, gave a new treatment of Euclidean geometry. Peano combined Pasch’s axiomatization and the newly developed formal symbolism to develop a completely formal and constructed Euclidean geometry.104 In 1899, Pieri continued this tradition and developed a different approach to Euclidean geometry, based on a minimal set of only 2 primitive notions, but requiring 24 axioms.105 Each of these systems became in some ways simpler than the previous, relying on fewer undefined terms; in other ways, however, the axiomatic systems became more complicated, relying on more axioms. Hilbert continued in this new tradition of axiomatizing geometry, and in 1899 published his Foundations of Geometry, employing 17 axioms and 7 primitive terms to characterize planar geometry. Hilbert’s axioms, like those of Pasch, Peano, and Pieri, correct the errors listed above, yet more closely continues in the tradition of Euclid, taking the same approach to the subject of geometry.106 Hilbert went far beyond the previous authors, however, in a number of ways. Most significantly, the Foundations of

100 Greenberg pg. 103 101 Faber pg. 99 102 Pasch 103 Greenberg pg. 67 104 Eves pg. 382 105 Eves pgs. 383-384 106 Fishback pg. 6 126

Geometry was unique in demonstrating the independence of each axiom. That is, for each axiom, he exhibited explicitly geometries that satisfied all other axioms except the given one107. In this way he showed that every axiom was necessary in order to recover Euclidean Geometry. In no way was Hilbert’s the last axiomatic system of geometry to be developed. Veblen, in 1904, developed another system with only the 2 primitive notions point and order. In 1913, Robinson combined Veblen’s and Hilbert’s postulates in yet another axiomatic system.108 It should be noted that the axiomatic systems so far discussed can be considered revisions or extensions of Euclid’s axioms. They preserve the spirit of Euclid, while setting a basis for geometry consistent with the rigor of modern mathematics. They differ mainly in the choice and number of primitive notions, and hence also in what axioms must be assumed about each notion and in what order the theorems can be deduced. From among these axiomatic systems, we choose in this text to use that of Hilbert, for a number of reasons. One reason for this choice is tradition; the significance of the Foundations of Geometry was such that Hilbert’s system quickly became influential and widely used. A second reason is methodological; Hilbert’s axioms allow for a clearer, more intuitive development of the geometry. A final reason is that Hilbert’s axioms and work “most faithfully preserve the spirit of Euclid’s original work while building geometry on a completely rigorous foundation.”109 There are a number of other possible axiomatic systems which we must consider, however, and to do so we again consider the historical background.

In the late 16th century, Viete began the development of symbolic algebra. In the 17th century, Descartes and Fermat continued this development began the joining of algebra and geometry.110 It was Descartes who first identified points in the plane with coordinates, and founded what has come to be known as analytic geometry, as opposed to the axiomatic variety of synthetic geometry. From that time forward, numbers became another tool in the study of geometry, although for centuries, mathematicians favored synthetic arguments over analytic arguments. In the Foundations of Geometry, Hilbert showed that the real number plane was a model for his axiom system. He later added to the Foundations of Geometry a completeness axiom, which says the system of geometry cannot be expanded in any way. This axiom is completely unnecessary for the development of Euclidean geometry, but allows a way to assign a unique real number to represent angle measures and segment lengths.111 In fact, in 1960 Borsuk and Szmielew showed Hilbert’s axioms are categorical; that is, there is only one possible model of the axioms.112 Thus, as the real number plane is a model of his axioms, it is the only model of

107 Eves pgs. 384-385 108 Eves pgs. 387-388 109 Venema pg. 53 110 Greenberg pg. 35 111 Eves pg. 386 112 Greenberg pg. 135 127 his axioms, and therefore any theorem of geometry proven analytically would necessarily be true. In 1932, Birkhoff developed an entirely new approach to axiomatizing geometry, essentially working in the reverse direction from the historical, and also Hilbert’s, approach. Published in the Annals of Mathematics, Birkhoff began by assuming the existence and properties of the real numbers, and incorporates the real numbers into the axioms themselves.113 His axioms establish a one-to-one correspondence between segment lengths on a given line and the real numbers, and between angle measures from a given ray and the real numbers (mod 2�). In assuming the real numbers, Birkhoff was able to have a minimal set of axioms, containing only 4 axioms. In Birkhoff’s presentation, “the axiomatic, synthetic approach of Euclid and Hilbert is to some extent merged with the analytic approach of Descartes and Fermat.”114 Birkhoff preferred his development to Hilbert’s for educational purposes because he believed them to be simple, reasonable, and to makes proofs more intuitive.115 However, they require a full understanding and development of the real number system. While his 4 axioms appear to be a much smaller set from which to build geometry, the development of the real numbers requires at least 8 more axioms,116 which are not at all intuitive or easy. Indeed, the rigorous development of the real numbers traditionally depends upon Dedekind cuts, which were developed as generalizations of the Dedekind postulate developed for Euclidean geometry. Thus the smaller set of axioms is in fact an illusion, and requires more advanced mathematics than the geometry they are used to develop. When we compare this to Hilbert’s axioms, which “are like Euclid’s in that they are completely synthetic and do not rely on any previous knowledge about the real number system,” 117 Hilbert’s is particularly attractive. Following in Birkhoff’s lead, a number of other axiomatic systems were developed which incorporated properties of the real numbers into the axioms. In 1959, Sanders MacLane followed the example of Birkhoff but used directed angles to simplify the properties of angle addition.118 MacLane’s axioms are based on a metric which measures distances between points.119 In the 1960s, Edwin Moise and the School Mathematics Study Group (SMSG) similarly followed Birkhoff but introduced redundancy to increase clarity;120 that is, they included many unnecessary axioms. The University of Chicago School Mathematics Project UCSMP also introduced an axiom system similar to that of the SMSG, the main difference being the assumption of an axiom of reflection or transformation in place of the SAS axiom (Axiom IV.3 of this text).121 These two latest axiomatic systems include many more axioms, which are all

113 Venema pg. 400 114 Venema pgs. 54-55 115 Birkhoff pgs. 39-40 116 Tarski pgs. 201-208 117 Venema pg. 398 118 Venema pg. 401 119 Venema pg. 55 120 Venema pg. 402 121 Venema pg. 405 128 consequences of Birkhoff’s axioms, to allow the geometry to proceed to more interesting results more quickly.122 Thus in every sense except ease of development, these presentations are inferior to Birkhoff’s.

There is one additional axiomatic system which deserves mention. In 1927, the logician Tarski developed a system for Euclidean geometry based on one primitive object and two primitive relations,123 and can in fact be defined in terms of one object and one relation.124 Tarski’s system is unique in that it can be presented in first order logic;125 that is, the axioms do not require quantification over sets of objects. This has immense theoretical benefits: for instance, all of Tarski’s geometry can be programmed and generated by computer. Moreover, as he proved in 1930, his system admits quantifier elimination. Thus every statement in the language can be phrased without quantifying over any variables. As a consequence, his geometry is complete, decidable, and can be proven consistent by a constructive proof.126 That is, every statement that can be written in the language of first order logic can either be proven true, or its negation can be proven true.127 In addition, Tarski’s axioms system has the shortest presentation, in a technical logical sense. But while Tarski’s system has great benefits, there are also serious drawbacks to the system, particularly for the educational development of geometry. First, and perhaps most significantly, Tarski’s axioms cannot be used to develop the same geometry as Euclid’s. While Hilbert’s geometry is categorical and thus has only one model, Tarski’s geometry has many different models. Models of Tarski’s system are Cartesian 2-space over real closed fields, or Euclidean ordered fields in which every polynomial of odd degree with coefficients from the field has a root.128 The differences in each model are differences which cannot be characterized by first order statements, that is, statements which require quantifying over sets of sets of points, as opposed to quantifying over sets of points. Thus, strictly fewer geometric theorems can be derived from Tarski’s axioms. Second, and perhaps equally significantly, Tarski’s system requires an infinite number of axioms. Tarski’s system consists of 20 axioms and 1 infinite axiom schema, which gives an axiom for every formula of first order logic. This can be resolved by replacing the infinite axiom schema with one second order axiom,129 but then Tarski’s geometry loses its first order properties, completeness, and decidability. This modified axiom system has a unique model, the familiar geometry of Hilbert and Euclid.130 A third and final drawback to Tarski’s axioms is a result of their simplicity. His axioms are the simplest in that they require the fewest variables, fewest logical symbols,

122 Venema pg. 55 123 Tarski pg. 177 124 Tarski pg. 203 125 Tarski pg. 175 126 Tarski pg. 175 127 Tarski pg. 195 128 Tarski pg. 191 129 Tarski pgs. 188-189 130 Tarski pg. 191 129 and fewest quantifiers in their presentation. In fact, his axiom system with replaced axiom schema is the simplest axiom system possible to categorize Euclidean geometry, in this strict logical sense. To achieve this simplicity, his axioms sacrifice any appeal to intuition. Thus while his axiom system is preferable from a logicians standpoint, they are not recommended to develop the structure of logical and geometric argument from a pedagogic standpoint. In any case, Tarski’s presentation of geometry from axioms to theorems follows the same pattern as Hilbert’s work.131

We summarize the various synthetic approaches to geometry as follows. The early axiom systems of Pasch, Peano, Pieri, Hilbert, Veblen, Robinson, and others follow the tradition established by Euclid, beginning from a strictly geometric basis. The axiom systems developed by Birkhoff, MacLane, Moise, and the UCSMP assume a rigorous foundation of the real number system and develop geometry by establishing correlations between lengths and angles with the real numbers, an analytic approach to synthetic geometry. The axiom system developed by Tarski is a strictly logical system which is developed have as a model the familiar Euclidean geometry. All are valid approaches to geometry, each with benefits and drawbacks. The choice to follow Euclid’s tradition is made because it is intuitive and does not require a familiarity or mastery of working with the real numbers or abstract logic. The choice to follow Hilbert’s interpretation of Euclid, as opposed to another, is made for the ease with which it can be presented.

131 Tarski pg. 211 130