What is [a] geometry?

John Haas

April 13, 2018

Contents

I Introduction to geometry and logic 3

II Neutral geometry 6

1 Hilbert’s neutral axioms 61.1 Axioms of Incidence . . . . . . . . . . . . . . . . . . . . . . . 61.2 Axioms of Betweenness . . . . . . . . . . . . . . . . . . . . . . 71.3 Axioms of Congruence . . . . . . . . . . . . . . . . . . . . . . 91.4 Axiom of Archimedes . . . . . . . . . . . . . . . . . . . . . . . 10

2 The boring part of neutral geometry 102.1 Right angles and a small dose of rigor . . . . . . . . . . . . . . 112.2 A montage of “boring” neutral facts . . . . . . . . . . . . . . . 13

2.2.1 Bisecting stuff and and another result about perpen-dicular lines . . . . . . . . . . . . . . . . . . . . . . . . 13

2.2.2 Adding, subtracting and comparing stuff . . . . . . . . 142.2.3 Parallel lines and the alternating interior angles theorem 172.2.4 Some basic stuff about triangles . . . . . . . . . . . . . 18

3 Getting weird 193.1 Summing the interior angles of a triangle . . . . . . . . . . . . 193.2 Saccheri quadrilaterals . . . . . . . . . . . . . . . . . . . . . . 20

3.2.1 Definition and basic properties . . . . . . . . . . . . . . 20

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3.2.2 A brief discussion about “models” which really belongsin Part I . . . . . . . . . . . . . . . . . . . . . . . . . . 25

3.2.3 Saccheri’s big theorem . . . . . . . . . . . . . . . . . . 263.3 Two neutral geometries? . . . . . . . . . . . . . . . . . . . . . 27

III Hyperbolic geometry 30

4 The upper half plane model 304.1 Is this new model consistent? . . . . . . . . . . . . . . . . . . 32

4.1.1 Hyperbolic Length/Magnitude . . . . . . . . . . . . . . 32

IV Exams

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Part I

Introduction to geometry andlogicAs we have discussed, the purpose of this course is somewhat philosophicalin nature. We investigate how a small adjustment to our core beliefs canprofoundly alter our interpretation of reality, and we get to do so with math-ematical rigor. Of course, in the context of this course, by “core beliefs” wemean some given axiomatic system and by “interpretation of reality” werefer to a geometric model that fits the theory generated by the axiomaticsystem.

If that last parapraph sounded neat but didn’t really make sense, it may bebecause you are unfamiliar with some of the bold-faced words. Let’s definethem. After rereading this section - if needed, then reread that last paragraph.

In mathematics, an axiom is a statement which is accepted as fundamen-tally true and does not require proof; accordingly, an axiomatic system isa collection of axioms. A theorem is a statement that one can conclude tobe true from a given axiomatic system and a theory is the whole body oftheorems that one can derive. Note that even undiscovered theorems (state-ments that are true but which – to date – have not been proven) are stillconsidered a part of the theory. Thus, a theory is the “entire truth” thatfollows from an axiomatic system.

A few nonmathematical analogies:

• The first amendment from the Bill of Rights is kind of like a singleaxiom, whereas the entire constitution is kind of like an axiomaticsystem. A single law generated by legislature which has been approvedby the judicial branch as consistent with the constitution is kind of likea theorem. The whole body of law in existence today is sort of like atheory, ie “the theory of law”. (Note, this is only a very rough analogy.For one thing, logic and rigor often do not work out well in law.)

• Many religions are structured around some sort of fundamental doctrinethat lays out rules for morality and belief. In this analogy, the doctrines

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of a religion are kind of like an “axiomatic system”, and the wholebody of religion derived from a given doctrine can be thought of as theresulting “theory”.

• Less than a millenium ago, a common belief was that the world is a flatsheet and not infinitely wide. These statements could be thought of astwo axioms. Embracing these as axioms, a resulting theorem was thatif one sailed far enough in a straight line, then one would eventually falloff the edge of the world. Although it seems a funny conclusion withour advanced knowledge today, few mock the ancients for this; afterall, it seems a perfectly reasonable conclusion given the premises.

Of course, our focus is on geometry. A geometry is a theory in whichthe axioms endow meaning to objects that are well-suited to visualization, egpoints, lines, planes, lengths, angles, etc. Some theories are not geometric; forexample, there are other types of axiomatic systems that generate algebraictheories, which are more suited to interpretation via symbolic manipulation.For example, one could study the theory of polynomials.

Our foremost goal is to ponder the meaning and existence of so-callednonEuclidean geometries. Plural. We emphasize one of them, the hy-perbolic geometry, because its axioms are very similar the axioms of Eu-clidean geometry, the familiar “flat” space from elementary school that ismodeled by the xy-plane and in which we expect the angles of a triangle tosum to π. As it turns out, there are several axiomatic systems that generatethis familiar geometry, ranging from the classical axioms from Euclid’s El-ements to more modern systems, as laid out by more recent geometers likeHilbert, Tarski, or Birkhoff, for example. Indeed, it is possible to swap outaxioms from one axiomatic system with theorems produced by a differentaxiomatic system and still end up with a set of axioms which “generates”what we know to be Euclidean geometry. Given this, we could easily ex-haust the whole semester studying the different axiomitizations of Euclideangeometry, but then it could scarcely be said that we ever really studied the“nonEuclidean” geometries.

In order not to dwell too long on something with which we are already fa-miliar, we give Euclid a nod for his foundational efforts and embrace Hilbert’smodern axiomitization. Hilbert’s system is broken down into five sets of ax-ioms:

([I.1] through [I.3]

), the axioms of Incidence,

([B.1] through [B.4]

),

the axioms of Betweenness,([C.1] through [C.6]

), the axioms of Congruence,

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[A], the Archimediam axiom and [P(E)], the axiom of Parallel lines forEuclidean space. As we shall see, axiom [P(E)], which states

“given a line l and a point P not on l, then there exists a uniqueline m containing P such that m and l do not intersect”,

is the sole axiom upon which the content of this course pivots. By simplyreplacing “there exists a unique line” with “there exists at least two lines”,we generate an exotic new geometry (called hyperbolic geometry) wherethe nature of lines, triangles and even infinity take on a new and unfamiliarstructure. Denote this alternative axiom of Parallel lines for Hyperbolicspace as [P(H)]. Now we may define the three geometries that will constituteour primary objects of study.

Definition 0.1.

Neutral geometry is the theory (ie, entire body of theorems) that is gen-erated by axioms [I.1] through [I.3], [B.1] though [B.4], [C.1] through [C.6],and [A], where no assumption is made about the nature of parallel lines.

Euclidean geometry is the theory (ie, entire body of theorems) that isgenerated by axioms [I.1] through [I.3], [B.1] though [B.4], [C.1] through[C.6], [A] and [P(E)].

Hyperbolic geometry is the theory (ie, entire body of theorems) that isgenerated by axioms [I.1] through [I.3], [B.1] though [B.4], [C.1] through[C.6], [A] and [P(H)].

Note that any theorem from neutral geometry will hold in both of theother two geometries, as its proof will not depend on either version of theparallel axiom. Accordingly, we call such a theorem a neutral theorem,and this explains why we use the word “neutral” to describe this geometry.

Finally, before laying out Hilbert’s framework, we conclude with a remarkabout mathematical rigor. Although the ideas of proof and logic underly thefundamental philosophy of this course, I want to avoid that we accidentallyslide into a course on logic itself. After all, our ultimate goal is to see theweird world of hyperbolic geometry. Therefore, we will often skip or merelysketch proofs of lemmas, propositions, theorems and corollaries with the un-derstanding that, despite the omissions of proofs, these statements actuallycan be proven from the axioms.

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Part II

Neutral geometry

1 Hilbert’s neutral axioms

We begin by declaring the existence of two undefined objects: points andlines, which we denote by lower and upper case letters, respectively. (Thereare also planes, but we ignore those because we are restricting ourselves to asingle plane.)

There are also three undefined prepositional terms which relate theseobjects: betweenness, containment and congruence.

1.1 Axioms of Incidence

Although lines and points are never assigned formal meaning, they arefleshed out using the undefined relationship of containment in the followingthree axioms.

[I.1] Any two distinct points A and B are contained by a unique line l. Inthis case, we may also denote l as AB, ie, l = AB.

[I.2] Every line contains at least two points.

[I.3] There exist three points that are not all contained by the same line.

These three basic axioms allow us to define two fundamental notions:intersecting lines and parallel lines.

Definition 1.1. A pair of lines l and m intersect or are intersecting ifthey have at least one point in common; otherwise, they are parallel.

From this, we observe an important fact about pairs of distinct lines.

Proposition 1.2. If l and m are a pair of distinct lines that intersect, thenthey have only one point in common.

Proof. Homework.

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1.2 Axioms of Betweenness

Next, we flesh out the notion of betweenness, a preposition that relatesthree points on a line. We adopt the notation A ∗ B ∗ C to mean “B isbetween A and C”. The following axioms lead to the notions of ‘sideness’and ‘insideness/outsideness’, ie, “a point can be on one side of a line” or “apoint can be inside a triangle”.

[B.1] (Symmetry/linearity of betweenness) If A ∗ B ∗ C, then A, B and Care three distinct points which are collinear, meaning they are on thesame line line, and C ∗B ∗ A.

[B.2] (Extendability of lines) For any two distinct points A and B, thereexists a point C such that A ∗B ∗ C.

[B.3] (Uniqueness/noncyclicity of betweenness) Given three distinct pointson a line, there is one and only one point that is between the othertwo.

[B.4] (Pasch’s axiom) Given three noncollinear points A, B and C, let l bea line not containing A, B or C. If l contains a point between A andB, then it must contain a point between A and C or between B andC, but not both.

With the first two sets of axioms established, we can now formally defineseveral familiar geometric concepts using the undefined but now axiomitizedterms, “points”, “lines”, “contained by” and “between”.

Definition 1.3. Given two distinct points A and B, the line segment ABis the set of all points between A and B, including A and B, which arecalled the endpoints. Given three distinct points not on the same line, thetriangle ∆ABC is the union of the three line sements AB, AC and BC,which are called the sides of the triangle. The points A, B and C are calledthe vertices of the triangle.

An essential concept that follow from these sets of axioms is the “side-ness”, as mentioned before. The ability to divide lines and planes into dis-tinct “sides” is not stated in Hilbert’s axioms, but the ability to do thiscan be proven from the axioms laid out so far. We will skip the proofs forthe following two propositions, but it is important to realize the followingtwo propositions are not axioms but theorems that can be proven from theaxioms.

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Proposition 1.4. (Plane separation) Given a line, l, then the set of allpoints on the plane not contained in l can be divided into two nonempty,nonintersecting subsets (let’s call them halves or sides of the plane) suchthat:

• Two distinct points A and B (neither contained by l) are on the samehalf of the plane if and only if AB does not intersect l.

• Two distinct points A and B (neither contained by l) are in differenthalves of the plane if and only if AB does intersect l.

Sketch of proof. Omitted. Follows from the axioms of betweenness andincidence and the definition of line segments. �

Proposition 1.5. (Line separation). Given a point A on a line l, thenthe set of all points on l not equal to A can be divided into two nonempty,nonintersecting subsets (let’s call them halves or sides of the line) such that:

• Two distinct points B and C in l are on the same side of the line ifand only if BC does not contain A.

• Two distinct points B and C in l are on different sides of the line ifand only if BC does contain A.

Sketch of proof. Omitted. Follows from the axioms of betweenness andincidence, the definition of line segments, and the preceding proposition. �

Thus, when we speak of the two sides or halves of a plane (when dividedby a line) or the two sides or halves of a line (when divided by a point),we mean it in accordance with the last two propositions. In particular, theseallow us to define rays and angles.

Definition 1.6. Given two distinct points A and B, the ray−→AB is the set of

all points of the line AB that are on the same side of A as B along with thepoint A itself, which is called the vertex of the ray. Given three points A,

B and C not on the same line, the angle ∠ABC is the pair of rays−→BA and−−→

BC. The interior of ∠ABC is the set of all points D such that (i) D and Aare together on the same side of the line BC and (ii) D and C are togetheron the same side of the line BA. The interior of the triangle ∆ABC is theintersection of the interiors of the angles ∠ABC, ∠BCA and ∠CAB.

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The following theorem is very similar to Pasch’s axiom, [B.4], except thatthe line passes through one of the triangle’s vertices. It’s so similar that Inearly skipped it...but we can’t do that. It’s pretty important.

Theorem 1.7. (The crossbar theorem) Let ∠BAC be an angle and let D be

a point in its interior, then the ray−−→AD must intersect the segment BC.

Sketch of proof. This follows from Pasch’s axiom and several applicationsof the plane and line separation theorems above. �

Remark 1.8. Unfortunately, notation will almost certainly vary throughoutthe semester, so we proceed with the assumption that we are all familiarwith these variations, as seen in high school geometry and determined bythe nuances of the English language. Also, as an addendum to a notationestablished in class, because we are only studying plane geometry, we reassignlower-cased Greek letters to denote angles (instead of planes, which was theoriginal plan).

1.3 Axioms of Congruence

Now we are in a position to axiomitize the undefined relationship of con-gruence, which allows us to compare line segments and angles. We writeAB ∼= CD to mean “AB is congruent CD” and we write α ∼= β to mean“the angle α is congruent to the angle β”. Ultimately, the axioms allow us tomeasure and order these objects. It is worth noting that these axioms cannoteven be stated without first developing the previous two sets of axioms tothe extent that rays, angles and line segments are well-defined.

[C.1] (Transportation of line segments) Given a line segment AB and a rayr with vertex C, then there exists a unique point D on r such thatAB ∼= CD.

[C.2] (Transitivity/Reflexivity of line segment congruence) If AB ∼= CD andCD ∼= EF , then AB ∼= EF . Also, every line segment is congruent toitself.

[C.3] (Additivity of line segment congruence) Suppose A,B,C,A′, B′ and C ′

are points, where A ∗ B ∗ C and A′ ∗ B′ ∗ C ′. If AB ∼= A′B′ andBC ∼= B′C ′, then AC ∼= A′C ′.

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[C.4] (Transportation of angles) Given an angle ∠ABC and a ray overrightarrowEF ,

then on a given side of the line EF there exists a unique ray−−→ED such

that ∠ABC ∼= ∠DEF .

[C.5] (Transitivity/reflexivity of angle congruence) Given angles α, β, and γ,where α ∼= β and β ∼= γ, then α ∼= γ. Also, every angle is congruent toitself.

[C.6] (SAS rule for triangles) Given triangles ∆ABC and ∆DEF , if AB ∼=DE, BC ∼= EF , and ∠ABC ∼= ∠DEF , then CA ∼= FD, ∠BCA ∼=∠EFD and 〈CAB ∼= ∠FDE. In this case, the two triangles are saidto be congruent.

1.4 Axiom of Archimedes

The following axiom is often called the “axiom of contituity”, as it laysthe foundation for calculus. Effectively, it establishes that there is no linesegment so long or so short that it transcends the notion of measurability;that is, there are no inifinitely long or infinitesimally short line segments.

[A] Given points A, B and A1, A2, A3, A4,... on a line such that

(i) A ∗ A1 ∗B,

(ii) A ∗ A1 ∗ A2, A1 ∗ A2 ∗ A3, A2 ∗ A3 ∗ A4, ..., and

(iii) AA1∼= A1A2

∼= A2A3∼= A3A4

∼= ...,

then there exists a positive integer n such that An ∗B ∗ An+1.

2 The boring part of neutral geometry

Now that we have stated the axioms of neutral geometry, let us derive (or inmany cases, lazily state without proof) some of the more “obvious” neutraltheory that follows. The use of the word “obvious” here is a serious abuse,so to be clear, by “boring” results, we mean the theorems/propositions/etcto which we are already accustomed from twelve years of studying Euclideangeometry from elementary school to high school, which happen to hold truein neutral geometry.

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2.1 Right angles and a small dose of rigor

We’ll start by proving or at least sketching the proofs of a few neutral theo-rems about right angles. This will require some definitions.

Definition 2.1. Given an angle ∠ABC and a point D on the line BC onthe side of B that is opposite to that of C, then ∠ABC and ∠ABD aresupplementary angles.

Note that every angle must have a supplementary angle (easily proved).This allows us to define right angles and verticle angles.

Definition 2.2. An angle which is congruent to its supplementary angle iscalled a right angle.

Definition 2.3. When two lines intersect at a point O, four distinct anglesare formed by considering the four distinct rays emerging from O along theselines. If α and β two of such angles and they are not supplementary, thenthey are verticle angles.

The following sheds light on the invariant nature of supplementary angles.

Proposition 2.4. If (i) α and β are supplementary angles, (ii) α′ and β′

are supplementary angles and (iii) α ∼= α′, then also β ∼= β′.

Sketch of proof. We showed this in class. The proof largely follows fromthree applications of [C.6], the SAS axiom. �

This yields three corollaries about verticle angles and right angles.

Corollary 2.5. Verticle angles are congruent.

Corollary 2.6. If two lines intersect and one of the four angles formed bythis intersection is a right angle, then all four of the angles formed are rightangles.

The following seems obvious, but its truth is only revealed as a corollaryto the preceding proposition about supplementary angles. This will be quiteimportant later.

Corollary 2.7. If α and β are congruent angles and α is a right angle, thenβ is also a right angle.

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We follow by stating another invariant property of supplementary angles.By definition, we know that two supplementary angles are formed by extend-ing a ray away from some point on a line. The following proposition statesthat if two angles are congruent to a pair of supplementary angles, then theymust be supplementary as well.

Proposition 2.8. Let AD be a line and let B and D be points on oppositesides of AD. Furthermore, suppose that α and β are supplementary angles.If α ∼= ∠BAD and β ∼= ∠DAC, then B ∗ A ∗ C and, in particular, ∠BADis supplementary to ∠DAC.

Sketch of proof. The key to the proof is showing that B ∗A∗C, or, in otherwords, that these three points are collinear, because then the two anglesare supplementary by definition. This follows by considering another lineBE such that B ∗ A ∗ E, where E exists by [B.2]. Using the definition of“sidedness”, Proposition 2.4 and [C.5] allows us to conclude that AC andAE are the same line, so that B ∗ A ∗ C. �

At this point, a seemingly silly question needs to be addressed.

Question 2.9. Do right angles exist?

We have defined them, but a definition is hardly an existential statement;otherwise, the world would be overrun with dragons and unicorns. Fortu-nately, right angles do exist, which is pretty important for the developmentof both Euclidean and hyperbolic geometry. Because the following statementis so important, we will call it a theorem and we will prove it.

Theorem 2.10. Given a line l and a point P not on l, there exists a line mpassing through P and intersecting l so that the four angles formed by theirintersection are right angles.

Proof. Using [I.2], let A and B be distinct points on l. Using [C.1] and [C.4],let P ′ be a point on the oposite side of l from P such that ∠BAP ∼= ∠BAP ′

and AP ∼= AP ′. By definition of “being on opposite sides”, there is a pointQ on l which is also on the segment PP ′ such that P ∗Q ∗ P ′.

It is possible that A = Q. If this happens, then ∠BQP and ∠BQP ′ aresupplementary and ∠BQP ∼= ∠BQP ′, so they are right angles by definition.

Now consider the possibility that A 6= Q. THIS IS NOT EUCLIDEANGEOMETRY. If you are looking at a diagram of this situation, then you

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would likely conclude that Q and B must be on the same side of A. This isnot acceptable. We must work around this by considering both cases. Either(i) Q and B are on the same side of A or (ii) they are on opposite sides.

If (i), then ∠BAP and ∠QAP are the same angle and, similarly, ∠BAP ′

and ∠QAP ′ are the same angle. In this case, ∆AQP ∼= ∆AQP ′ by SAS [C.6].In particular, ∠PQA ∼= ∠P ′QA and since these angles are supplementary,they are right angles.

If (ii), then by Proposition 2.4 (concerning supplementary angles), wemay still conclude that ∠QAP ∼= ∠QAP ′. (Why?) Thus, again, we see that∆AQP ∼= ∆AQP ′ by SAS [C.6], so that we again have ∠PQA ∼= ∠P ′QA,which are supplementary and therefore right angles.

We have exhausted all possible situations and found each time that settingm = PP ′ satisfies the theorem’s conclusion.

2.2 A montage of “boring” neutral facts

Now that we’ve had a taste of “rigor” in the development of neutral geometry,let’s skip some details and just state the important stuff. In what follows,we list several important neutral theorems without proof. It’s importantto realize that neutral theorems can be proved without [P(E)] or [P(H)].Since neutral theorems can be proved using only the first four sets of axioms,anything stated here is true in both Euclidean and hyperbolic geometry.

2.2.1 Bisecting stuff and and another result about perpendicularlines

Two distinct, intersecting lines are perpendicular if the angles they formare right angles, and we say that two line segments AB and CD are per-pendicular if the corresponding lines AB and CD are perpendicular. Giventhree distinct points A,B, and C such that A ∗ B ∗ C, then B is the mid-point of the segment AC if AB ∼= BC. A line not equal to the line AC thatpscriptscriptstylees through the midpoint B is said to bisect AC. Given

an angle α = ∠ABC and a point D in the interior of α, then the ray−−→BD

bisects α if ∠ABD ∼= ∠DBC. In this case, the ray−−→BD is the (angular)

bisector of α.Each object defined in this last paragraph exists, and under the appro-

priate conditions, it is unique.

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Theorem 2.11.

(1) Given a line segment AC , then a unique mipoint B exists. Moreover, aline l which bisects AC exists. Furthermore, a perpendicular bisector of ACexists and it is unique.

(2) Given an angle α = ∠ABC, then there exists a point D in the interior

of α such that the ray r =−−→BD bisects α. Moreover, the ray r is unique

(although D is not).

Sketch of proof. The verity of this theorem can be shown by constructing theclaimed objects, using the neutral axioms and the statements we’ve provenso far. �

Corollary 2.12. Given any line l and point P on l, then a unique line mperpendicular to l exists which passes through P .

2.2.2 Adding, subtracting and comparing stuff

Before we can get to the fun part of this course (ie, hyperbolic geometry),we need to know that it is possible to add, subtract and compare line seg-ments with other segments and angles with other angles. Fortunately, theseconcepts can all be defined in neutral geometry, so they will carry over tohyperbolic geometry. In the following, we define and state their importantproperties without proof, stressing again that despite their omissions, theproofs exist and rely on neither [P(E)] nor [P(H)].

Let’s start with line segments.

Definition 2.13. Let AB and CD be two segments and consider the ray

r =−→AB which lies on the line l = AB. Let E be the unique point on r on

the side of B oposite from A such that BE ∼= CD (as given by our segmenttransportation axiom, [C.1]). We say that the segment AE is the sum ofAB and CD and denote this by writing AE = AB + CD. Given distinctpoints A,B and C with A ∗ B ∗ C, then we say that BC is the differencebetween AC and AB, written BC = AC − AB.

Now we state invariance results, namely that sums of congruent segmentsare congruent.

Proposition 2.14. If AB ∼= A′B′ and CD ∼= C ′D′, then AB + CD ∼=A′B′ + C ′D′.

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We have a similar statement for differences.

Proposition 2.15. Given points A,B and C with A∗B∗C and points E anfF on a ray with vertex D, if AB ∼= DE and AC ∼= DF , then (i) D ∗ E ∗ Fand (ii) BC ∼= EF .

Definition 2.16. Given segments AB and CD, we say that AB is greaterthan CD and write AB > CD if there exists a point E with A ∗E ∗B suchthat AE ∼= CD. In this case, we also say CD is less than AB and writeCD > AB.

The “’greater than/less than” properties also remain invariant under con-gruence and they establish an ordering on angles.

Proposition 2.17.

(1) (Invariance of segment inequalities) If AB ∼= A′B′ and CD ∼= C ′D′, thenAB < CD if and only if ′A′B < C ′D′.

(2) (Ordering of segments)

(i) (Transitivity) If AB < CD and CD < EF , then AB < EF .

(ii) (Trichotomy) Given two segments AB and CD, then one and only oneof the following holds: AB < CD, AB ∼= CD or AB > CD.

Now let’s collect similar definitions and results about angles.

Definition 2.18. If ∠BAC is an angle and−−→AD is in its interior, then we say

that ∠BAC is the sum of the angles ∠BAD and ∠DAC, which we denoteby ∠BAC = ∠BAD+∠DAC. In this case, we may also say that ∠DAC isthe difference ∠BAD from ∠BAC, written ∠DAC = ∠BAC − ∠BAD

Remark 2.19. This definition is subtle, as there exist angles α and β forwhich α+β is not defined. (See Exam 1.) For this reason, we must be carefulwhen discussing sums of angles.

Although we have an axiom that states that sums of congruent segmentsare congruent, we do not have an axiom of this form for angles. This isbecause the version for angles can be proven from the axioms.

Proposition 2.20. (Additivity of angles) Suppose that the ray−−→AD is interior

to the ∠BAC and ∠BAC = ∠BAD + ∠DAC. If we have another pair ofangles such that ∠BAD ∼= ∠B′A′D and ∠DAC ∼= ∠D′A′C, where B′ andC ′ are on opposite sides of the line A′D′, then also ∠BAC ∼= ∠B′A′C.

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Now we can “compare” angles. Given angles α = ∠BAC and β =

∠EDF , then α is less than β, written α < β, if there exists a ray−−→DG

in the interior of ∠EDF such that ∠BAC ∼= ∠GDF . Alternatively andequivalently, we may say that β is greater than α, written β > α.

We have an analogous version of Proposition 2.17 for angles.

Proposition 2.21.

(1) (Invariance of angle inequalities) Given angles α, α′, β and β′, if α ∼= α′

and β ∼= β′, then α < β if and only if α′ < β′.

(2) (Ordering of angles)

(i) (Transitivity) Given angles α, β and γ, if α < β and β < γ, then α < γ.

(ii) (Trichotomy) Given two two angles α and β, then one and only oneof the following holds: α < β, α ∼= β or α > β.

Surprisingly, another important and seemingly obvious fact about rightangles could not be proven until we introduced the ability to compare angles.

Proposition 2.22. Any two right angles are congruent.

Sketch of proof. Because the proof of this involves a new type of argument,ie using “¡” and “¿”, I will try to write up the actual proof later. For now,this is omitted so that we can trudge forward. �

Now we may speak of acute and obtuse angles.

Definition 2.23. An angle α is acute if it is less than a right angle. It isobtuse if it is greater than a right angle.

Finally, we observe that the notions presented above can be extended toa well-defined measure on angles that addresses the problems mentioned inProblem 3 of Exam 1.

Theorem 2.24. In neutral geometry, there exists a unique function m whosedomain is the set of all angles and whose range is (0,∞) such that

1. (Positivity) for every angle α, we have 0 < m(α) < 180,

2. (Right angles) if α is a right angle, then m(α) = 90,

3. (Congruence) if α and β are angles, then m(α) = m(β) if and only ifα ∼= beta,

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4. (Ordering) if α and β are angles, then m(α) < m(β) if and only ifα < β, and

5. (Additivity) if α, β and γ are angles with α ∼= β + γ, then m(α) =m(β) +m(γ).

As with angles, we can improve our currently primitive notion of “seg-ment comparison” with a well-defined measure on segments, which we callmagnitude.

Theorem 2.25. Given a neutral geometry, fix distinct two points, O and I,and call the segment OI the unit segment. There exists a unique function,called the magnitude function and denoted | · |, whose domain is the set ofall segments and whose range is the interval (0,∞), such that

1. (Unit length)∣∣OI∣∣ = 1,

2. (Congruence)∣∣AB∣∣ =

∣∣CD∣∣ if and only if AB ∼= CD,

3. (Betweenness) A ∗B ∗ C if and only if∣∣AC∣∣ =

∣∣AB∣∣+∣∣BC∣∣

4. (Ordering)∣∣AB∣∣ < ∣∣CD∣∣ if and only if AB < CD, and

5. (Surjectivity) for every positive number x ∈ (0,∞), there exists a seg-ment AB such that

∣∣AB∣∣ = x.

As a corollary, it is not too difficult to deduce that the celebrated triangleinequality holds as a neutral theorem.

Corollary 2.26. Given distinct points A,B and C, then∣∣AC∣∣ ≤ ∣∣AB∣∣+∣∣BC∣∣ .

2.2.3 Parallel lines and the alternating interior angles theorem

In neutral geometry, it turns out that the elementary “alternating interiorangles theorem” is only valid in one direction. That is, if a pair of alter-nating interior angles are congruent, then it is true (in both hyperbolic andEuclidean geometry) that we end up with a pair of parallel lines. However,the converse is false in neutral geometry. If we have pair of distinct parallellines and a third line that bisects them (called the transversal), then we

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can connot conclude that their alternating interior angles are congruent. Theconverse is only true in Euclidean geometry, which serves to foreshadow thestrangeness of hyperbolic geometry.

Let’s be more precise.

Theorem 2.27. Suppose C and D are points on opposite sides of the lineAB. If ∠CAB ∼= ∠DBA and the lines l = AC and m = BD are distinct,then they are parallel.

Proof. Homework.

As a corollary, we conclude that the perpendicular line, m, from Theo-rem 2.10 must be unique. (Why is this obvious?)

Corollary 2.28. Given a line l and a point P not on l, there exists a uniqueline m passing through P and intersecting l so that the four angles formedby their intersection are right angles.

Surprisingly, this leads to another fundamental question that we havenot been able to answer until now. We have defined them and proven severaltheorems about them, but...

Question 2.29. Do parallel lines exist?

They do.

Theorem 2.30. Given a line l and a point P not on l, then there exists atleast one line m passing through P and parallel to l.

Proof. Homework.

2.2.4 Some basic stuff about triangles

We finish our montage by laying out a few familiar facts about triangleswhich continue to hold in the neutral setting.

Fortunately, the so-called side-side-side (SSS), angle-angle-side (AAS),and angle-side-angle (ASA) theorems still hold.

Theorem 2.31. (SSS) Given triangles ∆ABC and ∆A′B′C ′, if AB ∼= A′B′,BC ∼= B′C ′ and CA ∼= C ′A′, then the triangles are congruent, ie, the remain-ing respective angles and segments are congruent.

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Theorem 2.32. (AAS) Given triangles ∆ABC and ∆A′B′C ′, if ∠ABC ∼=∠A′B′C ′, ∠BCA ∼= ∠B′C ′A′ and AC ∼= A′C ′, then the triangles are congru-ent, ie, the remaining respective angles and segments are congruent.

Theorem 2.33. (ASA) Given triangles ∆ABC and ∆A′B′C ′, if ∠ABC ∼=∠A′B′C ′, ∠BCA ∼= ∠B′C ′A′ and BC ∼= B′C ′, then the triangles are con-gruent, ie, the remaining respective angles and segments are congruent.

Unfortunately, we do not have an angle-side-side triangle congruence the-orem, but - just as in the case of Euclidean geometry - we still have a “weak”version that is true for the special case of right triangles. I’m especially fondof the name of this theorem.

Theorem 2.34. (The Weak ASS Theorem.) Given triangles ∆ABC and∆A′B′C ′ where ∠BAC and B′A′C are both right angles, AB ∼= A′B′ andBC ∼= B′C ′, then the triangles are congruent, ie, the remaining respectiveangles and segments are congruent.

Another familiar property of triangles which upholds in neutral geometryis the “exterior angle theorem”.

Theorem 2.35. Given a triangle ∆ABC and a point D on the line AB suchthat A ∗ B ∗D, then the exterior angle ∠DBC is greater than both of theopposite interior angles, ∠CAB and ∠BCA.

3 Getting weird

Now we have laid out the important “obvious” facts of neutral geometry, bywhich I mean the definitions and theorems from Euclidean geometry withwhich we are comfortable but happen to be provable without [P(E)]. Withthis basic structure in place, we can begin an examination of those thingsthat will defy our intuition without the Euclidean parallel axiom in place.

3.1 Summing the interior angles of a triangle

Before we begin, we state a major fact - without proof - about interior anglesof triangles and quadrilaterals, which are weaker in neutral geometry to thatwith which we are accustomed.

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Theorem 3.1. Given a triangle with interior angles α, β and γ, then

m(α) +m(β) +m(γ) ≤ 180.

With this unfamiliar notion of triangles in mind, we turn our attention tocertain special quadrilaterals, which were the objects of study of an ItalianJesuit priest, Giovanni Girolamo Saccheri, who lived during the late Renais-sance.

3.2 Saccheri quadrilaterals

Let’s take a look at quadrilaterals.

Definition 3.2. Given a set of points A,B,C,and D, such that no threepoints are collinear, and a set of four line segments AB, BC, CD and DA,where AB and CD do not intersect and BC and DA do not intersect, then thequadrilateral 2ABCD is the union of the segments AB, BC, CD and DA.The four segments are called the sides of 2ABCD and the points A,B,Cand D are called its vertices. The four (interior) angles of 2ABCD are∠ABC, ∠BCD, ∠CDA and ∠DAB and the interior of 2ABCD is theintersection of the interiors of these four angles.

The following fact about quadrilaterals in neutral geometry is an obviouscorollary from upper bound for the measure of the interior angles of triangles,as stated in Theorem 3.1.

Corollary 3.3. Given a quadrilateral with interior angles α, β, γ and δ, then

m(α) +m(β) +m(γ) +m(δ) ≤ 360.

Of particular interest, we study a type of quadrilateral which - accordingto our Euclidean biases - ought to be a rectangle. We call these objectsSaccheri quadrilaterals, an acknowledgment of Girolamo Saccheri, whostudied them extensively toward the end of the 18th century.

3.2.1 Definition and basic properties

Definition 3.4. A Saccheri quadrilateral is a quadrilateral 2ABCD suchthat

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(i) AB ∼= CD and

(ii) the lines AB and BC are perpendicular and the lines CD and BC areperpendicular.

In order to clarify a Saccheri quadrilateral’s orientation, we refer to the seg-ment BC as the floor, we refer to DA as the roof, and we refer to ABand CD as the walls. Furthermore, we refer to the right angles ∠ABC and∠BCD as the bottom angles and we refer to ∠CDA and ∠DAB as thetop angles. The segment which joins the midpoint of the roof of 2ABCDwith the floor of 2ABCD is called the midline of 2ABCD.

Recall that a rectangle is a quadrilateral with interiorangles that are all right angles. As depicted in the diagramon the left, a Saccheri quadrilateral is a quadrilateral whichis, roughly speaking, “half of what we expect a rectangleto be.” I have drawn the roof purposefully amorphous toemphasize that we know nothing about it nor the anglesat its vertices. If we were in Euclidean geometry, then we

would conclude from elementary school that the top angles, α and β, shouldbe right angles and that the roof and floor, BC and AD, should be congruent.However, as we have alluded to many times, strange things happen in neutralgeometry, so this peculiar diagram is not unreasonable. Nevertheless, bydeveloping some neutral theorems about this object, we will improve thispicture, as we’ll find that Saccheri quadrilaterals share several propertieswith rectangles which are not apparent yet.

Proposition 3.5. Given a Saccheri quadrilateral 2ABCD, where BC is itsfloor, then its top angles, ∠BAD and ∠ADC, are congruent. Furthermore,midline is perpendicular to the walls AD and BC. In addition, the walls andthe midline are all parallel with each other.

Proof. Let F be the midpoint BC and let l be the perpendicular bisectorof BC, so it contains F by definition. Since the four angles formed by theintersection of l and the line BC are all right angles, it follows from the al-ternating interior angles theorem (Theorem 2.27) that the line AB is parallelwith l and similarly that CD is parallel with l. By definition of “oppositesides” and “parallelism”, this means that A and B are on one side of l whileC and D are on the other side. Again by definition of “oppositie sides”, thereexists some point E in AD with A ∗ E ∗ D such that l passes through E.

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By [C.6], the SAS axiom, it follows that ∆BEF ∼= ∆CEF ; in particular,BE ∼= CE and ∠EBF ∼= ∠ECF . By our theorems on “addition/subtrac-tion of angles”, it follows that ∠EBA ∼= ∠ECD, so again by the SAS axiom,we see that ∆AEB ∼= ∆DEC, so we have

∠BAD = ∠BAE ∼= ∠CDE = ∠CDA,

as claimed. (Note the difference between “=” and ∼=”.) Furthermore, thislast application of SAS showed that AE ∼= ED and ∠AEB ∼= ∠DEC.This shows that E is the midpoint of AD, and using theorems about angleaddition, we conclude that ∠AEF ∼= ∠DEF . Since these are supplementaryangles, they are right angles by definition, which completes the proof.

Thus, we have refined our knowledge about the struc-ture of Saccheri quadrilaterals. Admittedly, the diagram onthe right looks worse than the previous one, but we haveactually nailed down two new things that Saccheri quadri-laterals have in common with rectangles. The top anglesare congruent and the transversal which bisects the roofand floor must be perpedicular to them. Note that the roofin this diagram is presented in a purposefully absurd way to emphasize thatwe still don’t know much about it.

Shortly, it will be convenient to have a name for the segment joining themidpoints of the floor and roof of a Saccheri quadrilateral.

We further resolve theory of Saccheri quadrilaterals by examining howthe relationship between the top angles varies as we tweak the length of oneof the walls.

Proposition 3.6. Given a Saccheri quadrilateral 2ABCD, where BC is its

floor, let E be a point on the ray−−→CD with E 6= C, and consider the newly

formed angles as α = ∠BAE and β = ∠CEA.

1. If CE < CD, then α < β.

2. If CE ∼= CD, then α ∼= β.

3. If CE > CD, then α > β.

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Sketch of proof. If CE ∼= CD, then 2ABCE (ie, D = E, by uniquenessof segment transportation) is a Saccheri quadrilateral, so we are done byProposition 3.5. The proofs for the remaining two cases, (1) AB < CEor (3) AB > CE, both follow by the previous result (Proposition 3.5), an-gle subtraction/addition, and an application of the exterior angle theorem(Theorem 2.35) applied to the triangle ∆ADE. �

The diagram on theleft relects our latestinsight into the natureof Saccheri quadrilater-als, according to Propo-sition 3.5. As the pointE slides along the lineCD, the relationship be-tween two “top” angles

of the new quadrilateral 2ABCE varies with the relationship between thelengths its “left” and “right” sides in way that is intuitive to us from aEuclidean point of view.

Next, we consider the structure of the new quadrilateral formed when we“lift” a perpendicular segment from any point along the floor of a Saccheriquadrilateral to join its roof.

Proposition 3.7. Let 2ABCD be a Saccheri quadrilateral, where BC is itsfloor, let P be any point on its floor that is not a vertex (so B ∗P ∗C) and letQ be the unique point on the roof AD with A ∗Q ∗D such that the lines PQand BC are perpendicular lines.

(Why do we know that Q exists uniquely?

)Let α be either of the top angles of 2ABCD, which we know to be congruentfrom Proposition 3.5.

(a) If PQ < CD, then α is an acute angle.

(b) If PQ ∼= CD, then α is a right angle.

(c) If PQ > CD, then α is an obtuse angle.

Proof. We only prove case (a), because case (b) and case (c) are similar.Write β = ∠AQP and β = ∠DQP for the supplementary angles formed withvertex Q. If PQ < CD, then also PQ < CD by our inequality theorems. Bythe segment transportation axiom, [C.1], and the definition of “less than”,there exists a point R such that secAB ∼= CR and P ∗Q ∗ R. The line PR

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is parallel to both AB and CD by the alternating interior angles theorem(Theorem 2.27), so an application of the crossbar theorem (Theorem 1.7)assures that 2ABPR and 2PCDR both form quadrilaterals; moreover, theyare both Saccheri quadrilaterals by design. Next, apply Proposition 3.6 toeach of these newly formed Saccheri quadrilaterals and conclude that α < βand α < γ. Since β and γ are supplementary, then either (i) they are bothright angles or (ii) one is acute while the other is obtuse (Why?). In eithercase, since α is less than both of them, it follows that α is acute.

Informally speaking,Proposition 3.7 suggeststhat the roof of ev-ery Saccheri quadrilat-eral must be either “sortof concave”, “sort offlat”, or “sort of convex”,in the sense that every

point on the roof must be altogether either “below‘, ”of equal altitude with”or above the “heights” of the walls. This is depicted in the diagram on theleft. The angles β and γ in this image represent the supplementary anglesfrom the proof.

Next, we consider a sort of “complement” of the preceding proposotion.In Proposition 3.7, we examined properties of the “smaller” quadrilateralsformed by “lifting” a perpindicular segment from an arbitrary point on thefloor of a Saccheri quadrilateral to meet its roof. Now, let’s study the “larger”quadrilateral obtained by “lifting” a perpendicular from a point on the linecontaing the floor, but where our “lifted” segment is “outside” the Saccheriquadrilateral this time.

Proposition 3.8. Let 2ABCD be a Saccheri quadrilateral, where BC is itsfloor, and denote its congruent top angles as α. Let Q be a point on the lineAD such that A ∗D ∗Q and let P be the unique point on the line BC suchthat BC and PQ are perpendicular. (Note: Q exists uniquely and B ∗C ∗P .Why?)

(a) If PQ < CD, then α is an obtuse angle.

(b) If PQ ∼= CD, then α is a right angle.

(c) If PQ > CD, then α is an acute angle.

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Sketch of proof. See the solution to Problem 3 of Exam 2. �The new insight gained

from Proposition 3.8 isdepicted in this latestimage on the left. If we“slide” one of the wallsof a Sachheri quadri-lateral “outward”, thenthe “height” of the “new

wall” tells us about the top angles of the Saccheri retangle.Combining the knowledge about Saccheri quadrilaterals that we have col-

lected so far (ie, Propositions 3.5 through Proposition 3.8), we are now ableto state a very profound theorem about the nature of neutral geometry. How-ever, before we do this, an aside about the difference between a geometriesand a models is necessary.

3.2.2 A brief discussion about “models” which really belongs inPart I

We have thrown the word model around a few times now without reallyclarifying its meaning. For example, the definition was implied in Problem 2of Exam 1, where we constructed two distinct models for an incidence geom-etry. Let’s be more explicit, although - I fear - the following definition maynot be perfectly rigorous.

Definition 3.9. A model of a given theory is a mathematical structurewhich satisfies (or “fits”) all of the theories underlying axioms.

Recall: A geometry is a type of theory, and a theory is just a collection ofsentences derived from a fixed set of axioms; thus, a geometry is just a bunchof sentences. Geometric models are what we use to visualize geometries. Ofcourse, we are mainly concerned with models of neutral geometry, whichare models that fit the four sets of neutral axioms.

To make the distinction clear, consider the familiar xy-plane. We thinkof the xy-plane as ‘Euclidean geometry’, but this is a serious misconception.Euclidean geometry is a theory. It is the vast collection of sentences (ie,theorems) that one deduces from the five sets of Euclidean axioms; however,Euclidean geometry is not the xy-plane and the xy-plane is not Euclidean

25

geometry. Instead, the main point here is that the xy-plane is a model forEuclidean geometry.

3.2.3 Saccheri’s big theorem

Now that we have distinguished between the meanings of the terms “geome-try” and “model”, we present Saccheri’s fundamental theorem about neutralgeometry. This result is so impressive that we will call it a theorem. Theproof boils down to two lemmas.

Lemma 3.10. Given two Saccheri quadrilaterals with congruent midlines,then either

(i) the top angles of both Saccheri quadrilaterals are acute,

(ii) the top angles of both Saccheri quadrilaterals are right, or

(ii) the top angles of both Saccheri quadrilaterals are obtuse.

Proof. Homework. Because all three proof are similar, you will only showpart (i).

Lemma 3.11. Another set of three lemmas, rolled into one.

(i) If a Saccheri quadrilateral with top acute top angles exists, then givenany line segment PQ (ie, any “length”), there exists a Saccheri quadri-lateral with acute top angles and with midline congruent to PQ.

(ii) If a Saccheri quadrilateral with top right top angles exists, then givenany line segment PQ (ie, any “length”), there exists a Saccheri quadri-lateral with right top angles and with midline congruent to PQ.

(iii) If a Saccheri quadrilateral with obtuse top angles exists, then given anyline segment PQ (ie, any “length”), there exists a Saccheri quadrilateralwith obtuse top angles and with midline congruent to PQ.

Proof. Homework.

Now we can state the main theorem of Saccheri.

Theorem 3.12. (Saccheri’s Big Theorem)Given a model of neutral geometry, then exactly one of the following state-ments is true about it:

26

(i) If a single Saccheri quadrilateral with acute top angles exists, then thetop angles of every Saccheri quadrilateral are acute.

(ii) If a single Saccheri quadrilateral with right top angles exist in the model,then the top angles of every Saccheri quadrilateral are right.

(iiii) If a single Saccheri quadrilateral with obtuse top angles exist in themodel, then the top angles of every Saccheri quadrilateral are obtuse.

Proof. Yet again, because the proofs for the three cases are so similar, we onlyprove Case (i). Suppose the existence of Saccheri quadrilateral 20 with acutetop angles and let 21 be any other Saccheri quadrilateral. By Lemma 3.11,the existence of 20 implies the existence of Saccheri quadrilateral, 22, whichhas acute top angles and with a midline that is congruent to the midline of21. By Lemma 3.10, it follows that 21 must also have acute top angles,which completes the proof.

3.3 Two neutral geometries?

To conclude our chapter on neutral geometry, we show how Saccheri’s culmi-nating thereom leads to the main topic of this class: hyperbolic geometry.In particular, we examine the three types of neutral models for neutral geom-etry allowed by Theorem 3.12. In fact, only two models are allowable, as itturns out that Saccheri quadrilaterals with obtuse top angles are impossible;that is, every neutral model is either a Euclidean model or a hyperbolicmodel.

Theorem 3.13. (The birth of hyerbolic geometry.)Given a model of neutral geometry, then either

1. it is a Euclidean model, meaning that all Saccheri quadrilaterals arerectangles and [P(E)] holds, or

2. it is a hyperbolic model, meaning that all Saccheri quadrilateralshave acute top angles and [P(H)] holds.

Moreover, both Euclidean and hyperbolic models exist, and no neutral modelsadmitting Saccheri quadrilaterals with obtuse top angles exist.

This is a huge theorem with lots of assertions, so we will prove eachstatement separately. First, we show the last statement, that a neutral modeladmitting Saccheri quadrilaterals with obtuse top angles is impossible.

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Proposition 3.14. Given a model of neutral geometry, then Saccheri quadri-laterals with obtuse top angle do not exist.

Proof. By way of contradiction, suppose a Saccheri quadrilateral with obtusetop angle(s) α exists. Therefore, m(α) = 90 + ε for some number ε > 0, butthen the sum of its interior angles is

90 + 90 + 2m(α) = 360 + 2ε > 360,

which contradicts Corollary 3.3

Thus, in a neutral model, either all Saccheri quadrilaterals have acutetop angles or all Saccheri quadrilaterals are rectangles. Next, we show thata neutral model admits a Saccheri quadrilateral with acute top angles if andonly if [P(H)] holds, meaning that it is a hyperbolic model.

Proposition 3.15. Given a model of neutral geometry, then Saccheri quadri-laterals with acute top angles exist if and only if the hyperbolic axiom forparallel lines, [P(H)], holds, meaning the model is a hyperbolic model.

Proof.( ⇒) Suppose a Saccheri quadrilateral with acute top angles exists. By Sac-cheri’s big theorem, Theorem 3.12, it follows that all Saccheri quadrilateralshave acute top angles. Now we show that [P(H)] holds. Let l be any lineand let A be any point not on l. Let B be the unique point on l such thatthe line w = AB is perpendicular to l, the existence of which is assured byCorollary 2.28 and let C be any other point on l such that B 6= C. Next, letv be the the unique line passing through C and perpendicular to l and, onthe same side of l as A, let D be the unique point on v such that AB ∼= CD.Thus, 2ABCD is a Saccheri quadrilateral with floor BC. By Proposition 3.5,the midline is perpendicular to the floor and the ceiling, so by the alternat-ing interior angles theorem, Theorem 2.27, the line m = AD is parallel to l.Finally, let n be the unique line passing through A that is perpendicular tov. Again, by the alternating intertior angles theorem, it follows that n is alsoparallel to l. It is clear by angle subtraction that n 6= m, so we have shownthat [P(H)] holds.( ⇐) Homework.

Finally, we show that if a neutral model admits rectangles if and only ifit is a Euclidean model.

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Proposition 3.16. Given a model of neutral geometry, then rectangles existif and only if [P(E)] holds, meaning the model is a Euclidean model.

Proof. We know that parallel lines exist in any model, so this follows bythe contrapositive of both directions from the preceding proposition, Propo-sition 3.15.

Now we have shown all of the claims from Theorem 3.13 except for thestatement,

“ both Euclidean and hyperbolic models exist”.

Of course, we already know that a Euclidean model exists - namely, thefamiliar xy-plane that we learned about in elementary school, where thelength of the line segment between two points P (x0, y0) and Q(x1, y1) is givenby ∣∣PQ∣∣ =

√(x0 − x1)2 + (y0 − y1)2,

and given two rays with direction vectors, v and w, the measure of the angle,θ, formed by the rays is given by

cos θ =v ·w|v| |w|

.

Showing the existence of a hyperbolic model is the content of the next chap-ter.

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Part III

Hyperbolic geometry

4 The upper half plane model

As we mentioned in class, there are two equivalent models for hyperbolicgeometry: the disc model and the upper half plane model, both of whichare equivalent up to congruence. For the rest of the course, we focus on thelatter model.

A simple description is as follows. Imagine the usual xy-plane, but discardthe lower half – ie, everything below the x-axis. Moreover, we regard the x-axis as infinity, best depicted with a dashed line. We can “fit” a hyperbolicmodel to this space defining all lines in of the following two forms:

1. Verticle lines, originating from the x-axis (or line of infinity)

2. Semi-circles of any radius, where the centers are on the x-axis (or lineof infinity)

As depicted below, where we naively accept that this model fits ALL ofthe neutral axioms (we’ll deal with this), we see that [P(H)] holds. Thepoint P is not contained by the line l, but there are two distinct lines, m andn, passing through P , both parallel to l.

Indeed, all of Hilbert’s neutral axioms hold in this model, but it dependsprofoundly on how we define our measure(s) on angles and segments.

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The is the upper half-plane model, for hyperbolic geometry, commonlydenoted H2.

31

4.1 Is this new model consistent?

To verify that that H2 is a valid model for neutral geometry, we must checkall of Hilbert’s axioms. However, we are ultimately lazy and – as in class– for the first two sets of axioms, the Incidence and Between axioms, weremark that they are fairly obvious after noting the midpoint/perp.bisectortechnique for determining a semi-circle type line.

For the congruence axioms, we must describe the unique metrics for seg-ment and angle measurements.

However, the verity of the Congruence axioms and the Archimediam ax-iom depend far more on the relationship between the definition of angle ndsegment measure - to be presented next – and their invariance with respectto angle/segmet congruence according to Theorem 2.24 and Theorem 2.25.

4.1.1 Hyperbolic Length/Magnitude

Recall from Cal 3 (for the the Euclidean two independent variable case), acurve/vector fuction (segment) is continuous fuction from an intervalI ⊂ R to R2. Geometrically every such curve,

C : r(t) = 〈x(t), y(t)〉, t ∈ I,

where x(t and y(t are its components,) has a visual representation as we traceout the parameter t through the interval, I. Thus, if we identify the lines inH2 by their parametrizations, we can them define them as follows.

Definition 4.1. Given distinct points P = (x0, y0) and Q = (x1, y1) in theupper half plane, there exists a unique hyperbolic line, l=PQ, passing throughP and Q. In other words, [I.1] holds.

There are two types of lines.

• CASE 1: verticle lines If x0 = x1, the verticle line

l : l(t) = 〈xo, t〉, t ∈ (0,∞)

gets the job done. It is clear that it is unique, by the requirementx(t) = x0 for all t.

• CASE 2: semi-circles If x0 6= x1,, we have shown an elementaryruler/compass type construction that leads to a semi-circle,

l(t) = 〈r cos t+ x0, r sin t〉

of radius r, centered on the x-axis at (x0, 0) passing through both points.

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Of course, not all curves must be lines.

Example 4.2. In R2, the curve r(t) = 〈et cos(2πt), et sin(2πt)〉, t ∈ [0, 2] isa spiral of exponentially growing radius.

Recall the Euclidean length of such a curve.

Definition 4.3. If C : r(t) = 〈x(t),y(t)〉, t ∈ [a,b], then the Euclideanlength of C is

LE(r([a,b])) =

∫ b

a

√(x′(t))2 + (y′(t))2)dt

As in class, we computed the length of the spiral segment in Example 4.2.

Example 4.4. We computed:

LE(r([0, 2])) =

∫ 2

0

√(et sin(2πt)− 2π cos(2πt))2 + (etcos(2πt) + 2πsin(2t))2

After FOIL-expanding both terms and adding under the radical, we find thecross terms cancel (yay!); moreover, the F-terms of FOIL simplify by theidentity cos(2t) + sin(2t) = 1 as do the L-terms. Conveniently, this simplifiesthe integrand greatly, yielding

LE(r([0, 2])) =√

4π2 + 1

∫ 2

0

etdt =

√4π2 + 1(e2 − 1).

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Hyperbolic length for a curve in the upper half-plane is defined asfollows.

Definition 4.5. In H2, the hyperboic length of any curve segment, C :r(t) = 〈x(t),y(t)〉, t ∈ [a,b], y(t) > 0 is

LH(r([a, b])) =

∫ b

a

√(x′(t))2 + (y′(t))2)

y(t)dt.

A motivation inherent to our axiomatization of lines is the phrase: “theshortest path between two points is a line.” A quick example demonstratesthis phenonomenon might be obeyed by our new definition.

Example 4.6. Compute the hyperbolic distance between P=(0, 2) and Q =(√

2,√

2) along a hyperbolic line segment versus the hyperbolic distance alongthe Euclidean line connecting the two points.

1. For the straight line segment, as we learned in Cal 3 and was shown inclass, we can parametrize it as

C : r(t) = 〈√

2t, 2 + t(√

2− 2)〉, t ∈ [0, 1],

so the derivative is

C : r′(t) = 〈√

2, (√

2− 2)〉, t ∈ [0, 1].

Plugging this into our new definition and applying u-substitution gives

LH(r([0, 1])) =

∫ 1

0

2√

2−√

2

2 + (√

2− 2)tdt =

2

2−√

2log(√

2).

≈ 1.18

34

2. On the other hand, this example was carefully setup so that it is fairlyclear that the hyperbolic line determined by P and Q is the semicirclularline of radius 2, centered at the origin, parametrized as

L : l(t) = 〈2 cos t, 2 sin t〉, t ∈ (π

4,π

2)

yielding a derivative of

l′(t) = 〈−2 sin t, 2 cos t〉.

Plugging this into our definition, the expression simplifies to the firstintegral, which – after several integral tricks - leads to the solution

LH(r([0, 1])) =

∫ π2

π4

dt

sin t=

1

2log

(1 +√

2/2

1−√

2/2

)

≈ .88

35

Part IV

ExamsExam 1.Due: Wednesday, February 14, 2018Instructions: Use a separate sheet of paper to answer/solve each of thefollowing questions/problems. Clearly indicate which page(s) correspond(s)to which problem(s). Please write your name on every page. Staple yoursolutions together and submit them on the due date.

1 Prove Proposition 1.2 from the notes using only the axioms of inci-dence.Solution:Suppose distinct lines l and m intersect in two distinct points A andB. By axiom [I.1], it follows that l = AB = m, thereby contradictingthat l and m are distinct.

2 Incidence geometry is the theory that satisfies the axioms of inci-dence, [I.1], [I.2], and [I.3]. Thus, both Euclidean geometry and hyper-bolic geometry (yet to be seen) are incidence geometries. However, thereare other types of incidence geometries. In fact, models of incidencegeometries do not need to have an infinite number of points. Using dia-grams, provide models of two different incidence geometries containingexactly four distinct points A, B, C, and D, where one one satisfies[P(E)] and the other does not. In the two models, use bold-faced dotsto represent the points (label them A,B,C,D) and draw curves betweenthem to represent the lines (label these too, ie l,m, n, etc.) For eachline, list all the points contained by the line.

Solution:

It is easy to verify that both models satisfy the axioms of incidence, andthat only the first one satisfies [P(E)]. In fact, the second model fails[P(E)] terribly, because all lines intersect!

The point of this exercise is to stress how very different models can fitto the same geometry. Note that the first model has six lines while thesecond model has only four. Also, notice that in the second model, theline p contains 3 points while the others contain only two points.

3 Read Remark 2.19. Given an example of two angles α and β so thatα + β is not defined. Explain your answer.

Solution:The sum of two supplementary angles is currently undefined, because

their “sum” would correspond to two collinear, oppositely facing rays.Our definition of “angle” requires that the two rays not lay on the sameline, so this “sum” would not actually be an angle.

4 Using only the neutral axioms and neutral theorems presented beforeTheorem 2.27 in the notes, prove Theorem 2.27.

Solution:Suppose C and D are on opposite sides of the line n = AB and∠CAB ∼= ∠DBA, and that l = AC and m = BD are distinct,as given. With the intention of reaching a contradiction, suppose land m intersect at some point P . By the transportation axiom [C.1],there exist a point Q on the line m on the opposite side of B from Psuch that AP ∼= BQ. Thus, the SAS axiom, [C.6], implies that the∆ABP ∼= ∆AQB. In particular, this shows that ∠BAQ ∼= ∠ABP .

By our result on supplementary angles, Proposition 2.8, it follows that∠BAQ and ∠BAP must be supplementary, and, in particular, P∗A∗Q.However, we also have P ∗B∗Q, which implies that l = m, contradictingthat l and m are distinct.

5 Assume [P(E)], the axiom of parallel lines in Euclidean space, andprove the converse of Theorem 2.27. HINT: It is okay to use the “for-ward direction” of Theorem 2.27, because it is a neutral theorem.

(IMPORTANT: THE CONVERSE IS NOT A “NEUTRAL” THE-OREM BECAUSE ITS PROOF REQUIRES [P(E)]!! IT IS A/AN“EUCLIDEAN” THEOREM.)

Solution:Since we are proving the converse, we start with distinct parallel

lines l = AC, m = BD cut by a transversal n = AB, and our goal isto prove that ∠CAB ∼= ∠DBA. By our axiom on angle transportation,[C.4], there exists a point P such that ∠ABP is congruent to ∠BAC.Call the newly formed line q = BP . By the (forward direction of) thealternating interior angles theorem, Theorem 2.27, it follows that q andl are parallel. Because we are assuming [P(E)], there can be only oneline passing through B that is parallel to l. Therefore, q = BP andm = BD are the same line, so

∠CAB ∼= ∠ABP = ∠ABD,

by design. (Note the difference between ‘=’ and ‘∼=’.)

6 Prove Theorem 2.30, using only the neutral theorems and axioms pre-sented before it.

Solution:Given a line l and a point P not on l, we know by Theorem 2.10 that a

line m exists which is perpendicular to l and passes through P . Let R bethe point where the two lines intersect. By the existence of right anglesand our angle transportation axiom [C.4], there exists a point Q suchthat ∠RPQ is a right angle. If we let n = PQ be the newly formed line,it follows from the alternating interior angle theorem, Theorem 2.27,that l and n are parallel, which completes the proof.

Exam 2. 1 Prove Theorem 2.34, the “weak” ASS theorem.You may only use theorems preceding it to prove it.Solution:

Let P be the unique point on the line AC such that P ∗ A ∗ Cand AP ∼= A′C ′, using the usual axioms of betweenness and segmenttransportation. By the SAS triangle congruence axiom, it follows that∆BAP ∼= B′A′C ′, so ∠APB ∼= A′C ′B′, PB ∼= C ′B′ and by the ax-iom of transitivity for segment congruence, it follows that BP ∼= BC.Now, apply the SAS triangle congruence axiom to show that the trian-gle ∆PBC is congruent to itself; that is, since ∆PBC has two con-gruent sides (ie, it is isosceles), we have ∆PBC ∼= ∆CBP (congru-ent to itself in a different orientation). In particular, this shows that∠BPA ∼= ∠BCA. By the axiom of transitivity for angle congruence,it follows that ∠BCA ∼= ∠B′C ′A′, so by the AAS triangle congruencetheorem, Theorem 2.32, it follows that ∆ABC ∼= ∆A′B′C ′, completingthe proof.

2 Prove part (c) of Proposition 3.8. The proofs for the other parts aresimilar.You may only use preceding theory to prove it.For Case (a), PQ < CD, let G be the unique point such that P ∗Q∗G

and AB ∼= GP , which exists by our rules for line segment addition.Now 2ABPG is a Saccheri quadrilateral with floor BP and 2DCPGis a Saccheri quadrilateral with floor CP , so they both have congruenttop angles by Proposition 3.5. Denote the top angles of 2ABPG by γ,ie γ ∼= ∠BAG ∼= PGA, and denote the top angles of 2DCPG by β, ieβ ∼= ∠CDG ∼= ∠PGD. Let θ = ∠DAG and let δ = ∠GDQ. Applyingthe exterior angle theorem, Theorem 2.27, to the triangle ∆AGD, itfollows that δ > θ ∼= γ − α, where the last congruence follows by anglesubtraction. Finally, observe that the angle φ = ∠CDQ is supplemen-tary to α and it φ is congruent (by angle subtraction) to the differenceφ ∼= β − δ. Thus,

m(α) +m(φ) = m(α) +m(β)−m(δ) = 180.

Combining these results, we obtain

180 = m(α) +m(β)−m(δ) < m(α) +m(γ)−m(δ) < 2m(α),

which impies that α is obtuse.

For Case (b), PQ ∼= CD, we have that 2ABPQ is a Saccheri quadri-lateral with floor BP , so by Proposition 3.7, it follows that α is a rightangle.

The proof of Case (c), where PQ > CD, is very similar to Case (a).As a sketch, we start by letting G be the unique point such that P ∗G∗Qwith BA ∼= PQ, examine the two new Saccheri quadrilaterals along withthe triangles that are formed (as we did in Case (a)), and conclude that

180 = ... > ... > 2m(α),

thereby showing that α is acute.

3 Prove part (i) of Lemma 3.10. Proofs for the other statements aresimilar.You may only use theorems preceding it to prove it.Solution: Because the proofs for the three cases are so similar, wewill only show Case (i). Suppose 2ABCD is a Saccheri quadrilateralwith floor BC such that its top angles (let’s call them both α) are acute.Let E be the midpoint of the floor and let F be the midpoint of the roof.Thus, EF is the midline of 2ABCD and is perpendicular to the floorand the roof of 2ABCD by Proposition 3.5.

Next, suppose that 2WXY Z is any other Saccheri quadrilateral withfloor XY , where S is the midpoint of its floor, T is the midpoint of itsroof and the midlines of both Saccheri quadrilaterals are congruent, ieST ∼= EF . Let β denote top angles of 2WXY Z.

By using our “transportation” axioms, [C.1] and [C.4], several times,applying our rules for addition and substraction of angles several timesand applying the SAS axiom, [C.6], several times, we can construct aSaccheri quadrilatal 2W ′X ′Y ′Z ′, where the midline EF of 2ABCDconcides with the midline of 2W ′X ′Y ′Z ′, where X ′ and Y ′ lie on theline BC, where W ′ and Z ′ lie on the line AD, where the walls andfloor of 2W ′X ′Y ′Z ′ are congruent to those of 2WXY Z and wherethe the top angles of 2W ′X ′Y ′Z ′ are congruent to β (CONVINCEYOURSELF OF THIS.) Informally, we have “superimposed a clone”of 2WXY Z onto 2ABCD.

Concerning the floors of this “superimposed” Saccheri quadrilateral,there are three cases to consider: (a) X ′Y > BC, (b) X ′Y ∼= BC, or(c) X ′Y < BC.

Suppose (a), that X ′Y ′ > BC. Since α is acute, it follows from Propo-sition 3.8 that Y ′Z ′ > CD. Thus, by Proposition 3.7, it follows that βis also acute. Case (b) is trivial, since then 2ABCD and 2W ′X ′Y ′Z ′

are literally the same Saccheri quadrilateral, so β is acute. For case (c),we suppose X ′Y ′ > BC, and then the acuteness of α imples thatY ′Z ′ < CD, by Proposition 3.7, so it follows from Proposition 3.8that β must be acute.

4 Prove part (i) of Lemma 3.11. The proofs of the other two statementsare similar.You may only use theorems preceding it to prove it.Solution: Again, the proofs for the three cases are very similar, so weonly show Case (i). Assume the existence of a Saccheri quadrilateral,20 = 2ABCD (ie, we call it 20 for brevity), with floor BC and withacute top angles. Furthermore, let EF be its midline, where E is themidpoint of its floor and F is the midpoint of the roof.

We use the existence of 20 to construct the desired Saccheri quadrilat-eral.

Denote r = AD, so r is the line containing the roof of 20, let f = BC,so f is the line containing the floor of 20, and let m = EF , so mcontains the midline EF . Thus, F lies on r and E lies on f .

Note, there are three cases for how PQ and EF can relate in terms ofcongruence. If EF ∼= PQ then we are done. If PQ < EF , then is fairlysimple to show that a Saccherri quadrilateral with midline congruent toPQ and acute top angles “sits within” 20, so we omit the proof for thiscase; indeed, the method of proof for this “simpler” case will appear aswe prove the remaining “harder” case: from here on, we assume thatPQ > EF .

Choose one side of the line f relative to E and apply the segment trans-portation axiom, [C.1], to obtain the unique point Y on the chosen sideso that PQ ∼= EY .

Apply Corollary 2.28 to obtain the unique line s that is perpendicular tof and passes through Y . Let X be any point on the same side of the linef as D. Again, apply Corollary 2.28 to obtain the unique point G onm such that the line w = GX is perpendicular to m. By the alternatinginterior angles theorem, Theorem 2.27, f and w are parallel, so G andX are together on the same side of the line f .

Next, “reflect” the quadrilateral 2EGXY across the line f by repeatingthe construction from the previous paragraph to obtain the mirroredquadrilateral 2EG′X ′Y , as in the following diagram.

Note: We have drawn E ∗G ∗ F and E ∗G′ ∗ F ′ in the diagram, withthe assumption that G 6= F and G′ 6= F ′, in which case the alternatinginterior angles theorem would force this betweenness; however, it ispossible that G = F and G′ = F ′, in which case r = w and r′ = w′.Either way, the remainder of the proof will hold.

Also, in the diagram, we have draw the line s “outside” of the originalquadrilateral, but this is only for illustration, as this only depicts oneof three cases. It is possible that E ∗ Y ∗ C, Y = C, or E ∗ C ∗ Y (thecase depicted in the diagram). This is okay, because the following proofholds for all three cases.

By the SAS triangle congruence axiom, [C.6], it follows that ∆EYX ∼=∆EYX ′. Therefore, EX ∼= EX ′ and ∠Y EX ∼= ∠Y EX ′. By anglesubtraction, it then follows that ∠EGX ∼= ∠EG′X ′, so by the AAStriangle congruence theorem, Theorem 2.32, it follows that ∆EGX ∼=∆EG′X ′; in particular, EG ∼= E ′G′ GX ∼= G′X ′, so the quadrilateral2XGG′X ′ is a Saccheri quadrilateral with midline congruent to PQ.

It remains to show that the top angles, β ∼= ∠GXY and β ∼= G′X ′Y , of2XGG′X ′ are acute. Let H be the unque point on the line f such that

EH ∼= EF and E ∗H ∗ Y , which exists by segment transportation andthe assumption EF < PQ. Let t be the unique line passing through Hthat is perpendicular to f , and let I be the point where t intersects w andlet I ′ be the point where t intersects w′ (Why do I and I ′ exist?). By therepeating the last argument (ie, two applications of triangle congruencetheorems and angle subtraction), it follows that 2IGG′I ′ is a Sach-heri quadrilateral with midline congruent to EF ; thus, by Lemma 3.10,it follows that 2IGG′I ′ has accute top angles; that is, the congruentangles ∠GII ′ and ∠GI ′I are accute. Therefore, their correspondingsupplementary angles, γ ∼= ∠XIH and γ ∼= ∠X ′I ′H are obtuse, so wecan write m(γ) = 90 + ε for some real number ε > 0.

Finally, computing the sum of the interior angles of the quadrilateral2XIHY and applying Corollary 3.3 yields

m(γ) + 90 + 90 +m(β) = 270 + ε+m(β) ≤ 360,

which implies that β must be acute, completing the proof.

5 Prove the converse of Proposition 3.15; that is:

“If [P(H)] holds, then all Saccheri quadrilaterals have accute topangles.”

You may only use theorems preceding it to prove it.

Assume [P(H)].

In light of Saccheri’s Big Theorem, it is sufficient to construct a singleSaccheri quadrilateral with acute top angles, since the the theorem thenstates that all such quadrilaterals have acute top angles. I Indeed, byconnecting opposite corners of said quadrilateral, we obtain two trian-gles. With this in mind, it is sufficient to show that a triangle withinterior angle sum strictly less that 180 exists.

Let l be a line and P a point not on l such that two parallel lines to l passthrough P. We can construct one of these parallel lines as perpendicularto the unique line passing through P and perpendicular to l. Let Q bethe point of intersection of the line perpendicular to l through P. Let mbe the perpendicular line bisecting the line PQ passing through P. Thenm and l are parallel, by the weak AIA theorem. Let n be another line

through P which does not intersect l. This line exists by the [P(H)].

Let−−→PX be a ray of n lying between PQ and a ray

−→PY of m.

To complete the proof, we construct a sequence of angles ∠QR1P,∠QR2P, ...,∠QRnPso that ∠QRj+1P < 1/2∠QRjP , to obtain a contradiction.

There is a point R1 on l so that QR1 ≡ PQ. Since ∆QR1P is isosce-les, m(∠QR1P ) < 45. Continuing, there is R2 on l with PR1 ≡ R1R2.Now, ∆PR1R2 is isosceles, meaning ∠R1PR2 ≡ ∠QR2P . By the exte-rior angle theorem, we have ∠R1PR2+∠QR2P ≤ ∠QR1P . Continuingin this fashion, we obtain an integer n

m(∠QRnP ) < 45/2n.

By the Archimedian axiom, for any positive real number x, there is apoint R on the same side of PQ as X and Y , and ∠QRP < ∠XPY.By the crossbar theorem, PX intersects l, a contradiction.

Thus, ∠RPQ < ∠XPQ. Together, this implies that the angle sums of∆RPQ are strictly less than 180 degrees.