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Euclidean Plane and its Relatives

A minimalistic introduction

Anton Petrunin

http://arxiv.org/abs/1302.1630v9

This work is licensed under the Creative CommonsAttribution-ShareAlike 4.0 International License. To view a copy ofthis license, visit http://creativecommons.org/licenses/by-sa/4.0/.

http://www.createspace.com/4122116

ISBN-13: 978-1481918473

ISBN-10: 1481918478

Contents

Introduction 6

Prerequisite. Overview.

1 Preliminaries 9

What is the axiomatic approach? What is a model? Met-ric spaces. Examples. Shortcut for distance. Isometries,motions and lines. Half-lines and segments. Angles. Realsmodulo 2·π. Continuity. Congruent triangles.

Euclidean geometry

2 The Axioms 20

The Axioms. Lines and half-lines. Zero angle. Straightangle. Vertical angles.

3 Half-planes 25

Sign of angle. Intermediate value theorem. Same sign lem-mas. Half-planes. Triangle with the given sides.

4 Congruent triangles 32

Side-angle-side condition. Angle-side-angle condition.Isosceles triangles. Side-side-side condition.

5 Perpendicular lines 36

Right, acute and obtuse angles. Perpendicular bisector.Uniqueness of perpendicular. Reflection. Perpendicular isshortest. Angle bisectors. Circles. Geometric construc-tions.

3

4 CONTENTS

6 Parallel lines and

similar triangles 45

Parallel lines. Similar triangles. Pythagorean theorem.Angles of triangle. Transversal property. Parallelograms.Method of coordinates.

7 Triangle geometry 55

Circumcircle and circumcenter. Altitudes and orthocenter.Medians and centroid. Bisector of triangle. Incenter.

Inversive geometry

8 Inscribed angles 60

Angle between a tangent line and a chord. Inscribed angle.Inscribed quadrilaterals. Arcs.

9 Inversion 67

Cross-ratio. Inversive plane and circlines. Ptolemy’s iden-tity. Perpendicular circles. Angles after inversion.

Non-Euclidean geometry

10 Absolute plane 77

Two angles of triangle. Three angles of triangle. How toprove that something can not be proved? Curvature.

11 Hyperbolic plane 86

Conformal disk model. The plan. Auxiliary statements.Axioms: I, II, III, IV, h-V.

12 Geometry of h-plane 97

Angle of parallelism. Inradius of triangle. Circles, horo-cycles and equidistants. Hyperbolic triangles. Conformalinterpretation.

Incidence geometry

13 Affine geometry 106

Affine transformations. Constructions with parallel tooland ruler. Matrix form. On inversive transformations.

CONTENTS 5

14 Projective geometry 113

Real projective plane. Euclidean space. Perspective pro-jection. Projective transformations. Desargues’ theorem.Duality. Axioms.

Additional topics

15 Spherical geometry 122

Spheres in the space. Pythagorean theorem. Inversion ofthe space. Stereographic projection. Central projection.

16 Projective model 128

Special bijection of h-plane to itself. Projective model.Hyperbolic Pythagorean theorem. Bolyai’s construction.

17 Complex coordinates 136

Complex numbers. Complex coordinates. Conjugationand absolute value. Euler’s formula. Argument and polarcoordinates. Mobius transformations. Elementary trans-formations. Complex cross-ratio. Schwarz–Pick theorem.

18 Geometric constructions 145

Classical problems. Constructable numbers. Construc-tions with set square. More impossible constructions.

19 Area 152

Solid triangles. Polygonal sets. Definition of area. Vanish-ing area and subdivisions. Area of solid rectangles, paral-lelograms and triangles. Area method. Area in absoluteplanes and spheres.

Index 163

Used resources 166

Introduction

The book is meant to be rigorous, conservative, elementary and min-imalistic. At the same time it includes about the maximum whatstudents can absorb in one semester.

Approximately third of the material used to be covered in highschool, not any more.

The present book is based on the courses taught by the author atthe Penn State as an introduction into Foundations of Geometry. Thelectures were oriented to sophomore and senior university students.These students already had a calculus course. In particular they arefamiliar with the real numbers and continuity. It makes possible tocover the material faster and in a more rigorous way than it could bedone in high school.

Prerequisite

The students has to be familiar with the following topics.

⋄ Elementary set theory: ∈, ∪, ∩, \, ⊂, ×.

⋄ Real numbers: intervals, inequalities, algebraic identities.

⋄ Limits, continuous functions and Intermediate value theorem.

⋄ Standard functions: absolute value, natural logarithm, expo-nent. Occasionally, trigonometric functions are used, but theseparts can be ignored.

⋄ Chapter 13 use matrix algebra of 2×2-matrices.

⋄ To read Chapter 15, it is better to have some previous experiencewith scalar product, also known as dot product.

⋄ To read Chapter 17, it is better to have some previous experiencewith complex numbers.

6

CONTENTS 7

Overview

We use so called metric approach introduced by Birkhoff. It meansthat we define Euclidean plane as a metric space which satisfies alist of properties. This way we minimize the tedious parts which areunavoidable in the more classical Hilbert’s approach. At the same timethe students have chance to learn basic geometry of metric spaces.

In the Chapter 1 we give all definitions necessary to formulate theaxioms; it includes metric space, lines, angle measure, continuous mapsand congruent triangles.

Euclidean geometry is discussed in the chapters 2–7. In the Chap-ter 2, we formulate the axioms and prove immediate corollaries. Inthe chapters 3–6 we develop Euclidean geometry to a dissent level. InChapter 7 we give the most classical theorem of triangle geometry;this chapter included mainly as an illustration.

In the chapters 8–9 we discuss geometry of circles on the Euclideanplane. These two chapters will be used in the construction of the modelof hyperbolic plane.

In the chapters 10–12 we discuss non-Euclidean geometry. In Chap-ter 10, we introduce the axioms of absolute geometry. In Chapter 11we describe so called conforlal disc model. This is a construction ofhyperbolic plane, an example of absolute plane which is not Euclidean.In the Chapter 12 we discuss geometry of the constructed hyperbolicplane.

Chapters 13 and 14 discuss so called incidence geometry, it includesaffine and projective geometries.

The last few chapters contain additional topics: Spherical geome-try, Projective model, Complex coordinates, Geometric constructionsand Area. The proofs in these chapters are not completely rigorous.

Disclaimer

I am not doing history. It is impossible to find the original referenceto most of the theorems discussed here, so I do not even try. Mostof the proofs discussed in the book appeared already in the Euclid’sElements and the Elements are not the original source anyway.

Recommended books

⋄ Kiselev’s textbook [5] — a classical book for school students.Should help if you have trouble to follow my book.

⋄ Moise’s book, [10] — should be good for further study.

8 CONTENTS

⋄ Greenberg’s book [4] — a historical tour through the axiomaticsystems of various geometries.

⋄ Prasolov’s book [11] is perfect to master your problem-solvingskills.

⋄ Methodologically my lectures were very close to Sharygin’s text-book [13]. Which I recommend to anyone who can read Russian.

Acknowlegments.

I would like to thank Matthew Chao for thoughtful reading and cor-recting dozens of mistakes.

Chapter 1

Preliminaries

What is the axiomatic approach?

In the axiomatic approach, one defines the plane as anything whichsatisfy a list of properties called axioms. Axiomatic system for thetheory is like rules for the game. Once the axiom system is fixed,a statement considered to be true if it follows from the axioms andnothing else is considered to be true.

The formulations of the first axioms were not rigorous at all. Forexample, Euclid described a line as breadthless length and straight lineas a line which lies evenly with the points on itself. On the other hand,these formulations were clear enough so that one mathematician couldunderstand the other.

The best way to understand an axiomatic system is to make one byyourself. Look around and choose a physical model of the Euclideanplane, say imagine an infinite and perfect surface of chalk board. Nowtry to collect the key observations about this model. Assume for nowthat we have intuitive understanding of such notions as line and point.

(i) We can measure distances between points.

(ii) We can draw unique line which pass though two given points.

(iii) We can measure angles.

(iv) If we rotate or shift we will not see the difference.

(v) If we change scale we will not see the difference.

These observations are good enough to start with. Further we willdevelop the language to reformulate them rigorously.

9

10 CHAPTER 1. PRELIMINARIES

What is a model?

Euclidean plane can be defined rigorously the following way.Define a point in the Euclidean plane is a pair of real numbers

(x, y) and define the distance between two points (x1, y1) and (x2, y2)by the following formula.

√

(x1 − x2)2 + (y1 − y2)2).

That is it! We gave a numerical model of Euclidean plane; it buildsthe Euclidean plane from the real numbers while the later is assumedto be known.

Shortness is the main advantage of the model approach, but it isnot intuitively clear why we define points and the distances this way.

On the other hand, the observations made in the previous sectionare intuitively obvious — this is the main advantage of the axiomaticapproach.

An other advantage lies in the fact that the axiomatic approachis easily adjustable. For example we may remove one axiom from thelist, or exchange it to an other axiom. We will do such modificationsin Chapter 10 and further.

Metric spaces

The notion of metric space provides a rigorous way to say “we canmeasure distances between points”. That is, instead of (i) on page 9,we can say “Euclidean plane is a metric space”.

1.1. Definition. Let X be a nonempty set and d be a function whichreturns a real number d(A,B) for any pair A,B ∈ X . Then d iscalled metric on X if for any A,B,C ∈ X , the following conditionsare satisfied.

(a) Positiveness:

d(A,B) > 0.

(b) A = B if and only if

d(A,B) = 0.

(c) Symmetry:

d(A,B) = d(B,A).

(d) Triangle inequality:

d(A,C) 6 d(A,B) + d(B,C).

11

A metric space is a set with a metric on it. More formally, a metricspace is a pair (X , d) where X is a set and d is a metric on X .

Elements of X are called points of the metric space. Given twopoints A,B ∈ X the value d(A,B) is called distance from A to B.

Examples

⋄ Discrete metric. Let X be an arbitrary set. For any A,B ∈ X ,set d(A,B) = 0 if A = B and d(A,B) = 1 otherwise. The metricd is called discrete metric on X .

⋄ Real line. Set of all real numbers (R) with metric defined as

d(A,B)def== |A−B|.

⋄ Metrics on the plane. Let us denote by R2 the set of all pairs(x, y) of real numbers. Assume A = (xA, yA) and B = (xB , yB)are arbitrary points in R2. One can equip R2 with the followingmetrics.

Euclidean metric, denoted as d2 and defined as

d2(A,B) =√

(xA − xB)2 + (yA − yB)2.

Manhattan metric, denoted as d1 and defined as

d1(A,B) = |xA − xB |+ |yA − yB|.

Maximum metric, denoted as d∞ and defined as

d∞(A,B) = max|xA − xB |, |yA − yB|.

1.2. Exercise. Prove that the following functions are metrics on R2:(a) d1; (b) d2; (c) d∞.

Shortcut for distance

Most of the time we study only one metric on the space. For exampleR will always refer to the real line. Thus we will not need to name themetric function each time.

Given a metric space X , the distance between points A and B willbe further denoted as

AB or dX (A,B);


the later is used only if we need to emphasize that A and B are pointsof the metric space X .

For example, the triangle inequality can be written as

AC 6 AB +BC.

For the multiplication we will always use “·”, so AB should not beconfused with A·B.

Isometries, motions and lines

In this section we define lines in a metric space. Once it is donethe sentence “We can draw unique line which pass though two givenpoints.” becomes rigorous; see (ii) on page 9.

Recall that a map f : X → Y is a bijection if it gives an exactpairing of the elements of two sets. Equivalently, f : X → Y is abijection if it has an inverse; that is, a map g : Y → X such thatg(f(A)) = A for any A ∈ X and f(g(B)) = B for any B ∈ Y.1.3. Definition. Let X and Y be two metric spaces and dX , dY betheir metrics. A map

f : X → Yis called distance-preserving if

dY(f(A), f(B)) = dX (A,B)

for any A,B ∈ X .A bijective distance-preserving map is called an isometry.Two metric spaces are called isometric if there exists an isometry

from one to the other.The isometry from a metric space to itself is also called motion of

the space.

1.4. Exercise. Show that any distance preserving map is injective;that is, if f : X → Y is a distance preserving map then f(A) 6= f(B)for any pair of distinct points A,B ∈ X .

1.5. Exercise. Show that if f : R → R is a motion of the real linethen either

f(x) = f(0) + x for any x ∈ R

orf(x) = f(0)− x for any x ∈ R.

13

1.6. Exercise. Prove that (R2, d1) is isometric to (R2, d∞).

1.7. Advanced exercise. Describe all the motions of the Manhattanplane.

If X is a metric space and Y is a subset of X , then a metric on Ycan be obtained by restricting the metric from X . In other words, thedistance between points of Y is defined to be the distance between thesame points in X . This way any subset of a metric space can be alsoconsidered as a metric space.

1.8. Definition. A subset ℓ of metric space is called line if it isisometric to the real line.

Note that a space with discrete metric has no lines. The followingpicture shows examples of lines on the Manhattan plane (R, d1).

1.9. Exercise. Consider graph y = |x| in R2. In which of the fol-lowing spaces (a) (R2, d1), (b) (R2, d2) (c) (R2, d∞) it forms a line?Why?

1.10. Exercise. How many points M on the line (AB) for which wehave

1. AM = MB ?2. AM = 2·MB ?

Half-lines and segments

Assume there is a line ℓ passing through two distinct points P and Q.In this case we might denote ℓ as (PQ). There might be more thanone line through P and Q, but if we write (PQ) we assume that wemade a choice of such line.


Let us denote by [PQ) the half-line which starts at P and containsQ. Formally speaking, [PQ) is a subset of (PQ) which correspondsto [0,∞) under an isometry f : (PQ) → R such that f(P ) = 0 andf(Q) > 0.

1.11. Exercise. Show that if X ∈ [PQ) then QX = |PX − PQ|.The subset of line (PQ) between P andQ is called segment between

P and Q and denoted as [PQ]. Formally, segment can defined as theintersection of two half-lines: [PQ] = [PQ) ∩ [QP ).

Angles

Our next goal is to introduce angles and angle measures ; after thatthe statement “we can measure angles” will become rigorous; see (iii)on page 9.

O

B

Aα

An ordered pair of half-lines whichstart at the same point is called angle.An angle formed by two half-lines [OA)and [OB) will be denoted as ∠AOB. Inthis case the point O is called vertex ofthe angle.

Intuitively, the angle measure tellshow much one has to rotate the first half-line counterclockwise so it gets the position of the second half-line ofthe angle. The full turn is assumed to be 2·π; it corresponds to theangle measure in radians.

The angle measure of ∠AOB is denoted as ∡AOB; it is a realnumber in the interval (−π, π]. The notations ∠AOB and ∡AOBlook similar, they also have close but different meanings, which betternot to be confused. For example, the equality

∠AOB = ∠A′O′B′

means that [OA) = [O′A′) and [OB) = [O′B′), in particular O = O′.On the other hand the equality

∡AOB = ∡A′O′B′

means only equality of two real numbers; in this case O may be distinctfrom O′.

Here is the first property of angle measure which will become apart of the axiom.

Given a half-line [OA) and α ∈ (−π, π] there is unique half-line[OB) such that ∡AOB = α.

15

Reals modulo 2·π

Consider three half-lines starting from the same point, [OA), [OB)and [OC). They make three angles ∠AOB, ∠BOC and ∠AOC, so theangle measure ∡AOC should coincide with the sum ∡AOB +∡BOCup to full rotation. This property will be expressed by a formula

∡AOB + ∡BOC ≡ ∡AOC,

where “≡” is a new notation which we are about to introduce. Thelast identity will become a part of the axiom.

We will write

α ≡ β or α ≡ β (mod 2·π)

if α = β + 2·π·n for some integer n. In this case we say

“α is equal to β modulo 2·π”.

For example

−π ≡ π ≡ 3·π and 12 ·π ≡ − 3

2 ·π.

The introduced relation “≡” behaves roughly as equality, but theangle measures which differ by full turn

. . . , α− 2·π, α, α+ 2·π, α+ 4·π, . . .

are considered to be the same. We can do addition subtraction andmultiplication by integer number without getting into trouble. Thatis, if

α ≡ β and α′ ≡ β′

then

α+ α′ ≡ β + β′, α− α′ ≡ β − β′ and n·α ≡ n·β

for any integer n. But “≡” does not in general respect multiplicationby non-integer numbers; for example

π ≡ −π but 12 ·π 6≡ − 1

2 ·π.

1.12. Exercise. Show that 2·α ≡ 0 if and only if α ≡ 0 or α ≡ π.


Continuity

The angle measure is also assumed to be continuous. Namely, thefollowing property of angle measure which will become a part of theaxiom.

The function∡ : (A,O,B) 7→ ∡AOB

is continuous at any triple of points (A,O,B) such that O 6= A andO 6= B and ∡AOB 6= π.

To explain this property we need to extend the notion of continuityto the functions between metric spaces. The definition is a straight-forward generalization of the standard definition for the real-to-realfunctions.

Further X and Y be two metric spaces and dX , dY be their metrics.A map f : X → Y is called continuous at point A ∈ X if for any

ε > 0 there is δ > 0 such that if dX (A,A′) < δ then

dY(f(A), f(A′)) < ε.

The same way one may define a continuous map of several variables.Say, assume f(A,B,C) is a function which returns a point in the spaceY for a triple of points (A,B,C) in the space X . The map f might bedefined only for some triples in X .

Assume f(A,B,C) is defined. Then we say that f continuous atthe triple (A,B,C) if for any ε > 0 there is δ > 0 such that

dY(f(A,B,C), f(A′, B′, C′)) < ε.

if dX (A,A′) < δ, dX (B,B′) < δ and dX (C,C′) < δ.

1.13. Exercise. Let X be a metric space.(a) Let A ∈ X be a fixed point. Show that the function

f(B)def== dX (A,B)

is continuous at any point B.(b) Show that dX (A,B) is a continuous at any pair A,B ∈ X .

1.14. Exercise. Let X , Y and Z be a metric spaces. Assume thatthe functions f : X → Y and g : Y → Z are continuous at any pointand h = g f is its composition; that is, h(A) = g(f(A)) for anyA ∈ X . Show that h : X → Z is continuous at any point.

1.15. Exercise. Show that any distance preserving map is continu-ous.

17

Congruent triangles

Our next goal is to give a rigorous meaning for (iv) on page 9. Todo this, we introduce the notion of congruent triangles so instead of“if we rotate or shift we will not see the difference” we say that fortriangles side-angle-side congruence holds; that is, if two triangles arecongruent if they have two pairs of equal sides and the same anglemeasure between these sides.

An ordered triple of distinct points in a metric space X , say A,B,Cis called triangle and denoted as ABC. So the triangles ABC andACB are considered as different.

Two triangles A′B′C′ and ABC are called congruent (brieflyA′B′C′ ∼= ABC) if there is a motion f : X → X such that

A′ = f(A), B′ = f(B) and C′ = f(C).

Let X be a metric space and f, g : X → X be two motions. Notethat the inverse f−1 : X → X , as well as the composition fg : X → Xare also motions.

It follows that “∼=” is an equivalence relation; that is, the followingtwo conditions hold.

⋄ If A′B′C′ ∼= ABC then ABC ∼= A′B′C′.⋄ If A′′B′′C′′ ∼= A′B′C′ and A′B′C′ ∼= ABC then

A′′B′′C′′ ∼= ABC.

Note that if A′B′C′ ∼= ABC then AB = A′B′, BC = B′C′

and CA = C′A′.For discrete metric, as well some other metric spaces the converse

also holds. The following example shows that it does not hold in theManhattan plane.

Example. Consider three points A = (0, 1), B = (1, 0) and C == (−1, 0) on the Manhattan plane (R2, d1). Note that

d1(A,B) = d1(A,C) = d1(B,C) = 2.

A

BC

On one hand

ABC ∼= ACB.

Indeed, it is easy to see that themap (x, y) 7→ (−x, y) is a motion of(R2, d1) which sends A 7→ A, B 7→ Cand C 7→ B.


On the other hand

ABC ≇ BCA.

Indeed, assume there is a motion f of (R2, d1) which sends A 7→ Band B 7→ C. Note that a point M is a midpoint1 of A and B if andonly if f(M) is a midpoint of B and C. The set of midpoints for Aand B is infinite, it contains all points (t, t) for t ∈ [0, 1] (it is the darkgray segment on the picture). On the other hand the midpoint for Band C is unique (it is the black point on the picture). Thus the mapf can not be bijective, a contradiction.

1M is a midpoint of A and B if d1(A,M) = d1(B,M) = 1

2·d1(A,B).

19

Chapter 2

The Axioms

The Axioms

Summarizing the above discussion, let us give an axiomatic system ofthe Euclidean plane.

I. The Euclidean plane is a metric space with at least two points.

II. There is one and only one line, that contains any two givendistinct points P and Q in the Euclidean plane.

III. Any angle ∠AOB in the Euclidean plane defines a real numberin the interval (−π, π]. This number is called angle measure of∠AOB and denoted by ∡AOB. It satisfies the following condi-tions:

(a) Given a half-line [OA) and α ∈ (−π, π] there is unique half-line [OB) such that ∡AOB = α

(b) For any points A, B and C distinct from O we have

∡AOB + ∡BOC ≡ ∡AOC.

(c) The function

∡ : (A,O,B) 7→ ∡AOB

is continuous at any triple of points (A,O,B) such thatO 6= A and O 6= B and ∡AOB 6= π.

IV. In the Euclidean plane, we have ABC ∼= A′B′C′ if and onlyif

A′B′ = AB, A′C′ = AC, and ∡C′A′B′ = ±∡CAB.

20

21

V. If for two triangles ABC, AB′C′ in the Euclidean plane andk > 0 we have

B′ ∈ [AB), C′ ∈ [AC)

AB′ = k·AB, AC′ = k·AC

then

B′C′ = k·BC, ∡ABC = ∡AB′C′ and ∡ACB = ∡AC′B′.

This set of axioms is very close to the one given by Birkhoff in [3].

From now on, we can use no information about Euclidean planewhich does not follow from the five axioms above.

Lines and half-lines

2.1. Proposition. Any two distinct lines intersect at most at onepoint.

Proof. Assume two lines ℓ and m intersect at two distinct points Pand Q. Applying Axiom II, we get ℓ = m.

2.2. Exercise. Suppose A′ ∈ [OA) and A′ 6= O show that

[OA) = [OA′).

2.3. Proposition. Given r > 0 and a half-line [OA) there is uniqueA′ ∈ [OA) such that OA′ = r.

Proof. According to definition of half-line, there is an isometry

f : [OA) → [0,∞),

such that f(O) = 0. By the definition of isometry, OA′ = f(A′) forany A′ ∈ [OA). Thus, OA′ = r if and only if f(A′) = r.

Since isometry has to be bijective, the statement follows.

22 CHAPTER 2. THE AXIOMS

Zero angle

2.4. Proposition. ∡AOA = 0 for any A 6= O.

Proof. According to Axiom IIIb,

∡AOA + ∡AOA ≡ ∡AOA.

Subtract ∡AOA from both sides, we get ∡AOA ≡ 0. Hence ∡AOA == 0.

2.5. Exercise. Assume ∡AOB = 0. Show that [OA) = [OB).

2.6. Proposition. For any A and B distinct from O, we have

∡AOB ≡ −∡BOA.

Proof. According to Axiom IIIb,

∡AOB + ∡BOA ≡ ∡AOA

By Proposition 2.4, ∡AOA = 0. Hence the result follows.

Straight angle

If ∡AOB = π, we say that ∠AOB is a straight angle. Note that byProposition 2.6, if ∠AOB is a straight angle then so is ∠BOA.

We say that point O lies between points A and B if O 6= A, O 6= Band O ∈ [AB].

2.7. Theorem. The angle ∠AOB is straight if and only if O liesbetween A and B.

B O A

Proof. By Proposition 2.3, we may as-sume that OA = OB = 1.

(⇐). Assume O lies between A and B.Let α = ∡AOB.

Applying Axiom IIIa, we get a half-line [OA′) such that α == ∡BOA′. By Proposition 2.3, we can assume that OA′ = 1. Ac-cording to Axiom IV,

AOB ∼= BOA′.

23

Denote by h the corresponding motion of the plane; that is, h is amotion such that h(A) = B, h(O) = O and h(B) = A′.

Then (A′B) = h(AB) ∋ h(O) = O. Therefore both lines (AB) and(A′B), contain B and O. By Axiom II, (AB) = (A′B).

By the definition of the line, (AB) contains exactly two points Aand B on distance 1 from O. Since OA′ = 1 and A′ 6= B, we getA = A′.

By Axiom IIIb and Proposition 2.4, we get

2·α ≡ ∡AOB + ∡BOA′ ≡≡ ∡AOB + ∡BOA ≡≡ ∡AOA ≡≡ 0

Therefore, by Exercise 1.12, α is either 0 or π.Since [OA) 6= [OB), we have α 6= 0, see Exercise 2.5. Therefore

α = π.

(⇒). Suppose that ∡AOB ≡ π. Consider line (OA) and choose pointB′ on (OA) so that O lies between A and B′.

From above, we have ∡AOB′ = π. Applying Axiom IIIa, we get[OB) = [OB′). In particular, O lies between A and B.

A triangle ABC is called degenerate if A, B and C lie on oneline. The following corollary is just a reformulation of Theorem 2.7.

2.8. Corollary. A triangle is degenerate if and only if one of itsangles is equal to π or 0.

2.9. Exercise. Show that three distinct points A, O and B lie on oneline if and only if

2·∡AOB ≡ 0.

2.10. Exercise. Let A, B and C be three points distinct from O.Show that B, O and C lie on one line if and only if

2·∡AOB ≡ 2·∡AOC.

Vertical angles

A pair of angles ∠AOB and ∠A′OB′ is called vertical if the point Olies between A and A′ and between B and B′ at the same time.

2.11. Proposition. The vertical angles have equal measures.

24 CHAPTER 2. THE AXIOMS

Proof. Assume that the angles ∠AOB and ∠A′OB′ are vertical.Note that the angles ∠AOA′ and ∠BOB′ are straight. Therefore

∡AOA′ = ∡BOB′ = π.

A

A′

O

B

B′

It follows that

0 = ∡AOA′ − ∡BOB′ ≡≡ ∡AOB + ∡BOA′ − ∡BOA′ − ∡A′OB′ ≡≡ ∡AOB − ∡A′OB′.

Hence the result follows.

2.12. Exercise. Assume O is the midpoint for both segments [AB]and [CD]. Prove that AC = BD.

Chapter 3

Half-planes

This chapter contains long proofs of intuitively evident statements.It is okay to skip it, but make sure you know definitions of posi-tive/negative angles and that your intuition agrees with 3.8, 3.10, 3.11and 3.16.

Sign of angle

⋄ The angle ∠AOB is called positive if 0 < ∡AOB < π;

⋄ The angle ∠AOB is called negative if ∡AOB < 0.

Note that according to the above definitions the straight angle aswell as zero angle are neither positive nor negative.

3.1. Exercise. Show that ∠AOB is positive if and only if ∠BOA isnegative.

3.2. Exercise. Let ∠AOB be a straight angle. Show that ∠AOX ispositive if and only if ∠BOX is negative.

3.3. Exercise. Assume that the angles ∠AOB and ∠BOC are posi-tive. Show that

∡AOB + ∡BOC + ∡COA = 2·π.

if ∠COA is positive and

∡AOB + ∡BOC + ∡COA = 0.

if ∠COA is negative.

25

26 CHAPTER 3. HALF-PLANES

Intermediate value theorem

3.4. Intermediate value theorem. Let f : [a, b] → R be a contin-uous function. Assume f(a) and f(b) have the opposite signs thenf(t0) = 0 for some t0 ∈ [a, b].

f(b)

f(a)

t0 b

a

The Intermediate value theorem is as-sumed to be known; it should be covered inany calculus course. We will use the follow-ing corollary.

3.5. Corollary. Assume that for any t ∈∈ [0, 1] we have three points in the plane Ot,At and Bt such that

(a) Each function t 7→ Ot, t 7→ At and t 7→ Bt is continuous.(b) For for any t ∈ [0, 1], the points Ot, At and Bt do not lie on one

line.Then the angles ∠A0O0B0 and ∠A1O1B1 have the same sign.

Proof. Consider the function f(t) = ∡AtOtBt.Since the points Ot, At and Bt do not lie on one line, Theorem 2.7

implies that f(t) = ∡AtOtBt 6= 0 nor π for any t ∈ [0, 1].Therefore by Axiom IIIc and Exercise 1.14, f is a continuous func-

tion.Further, by Intermediate value theorem, f(0) and f(1) have the

same sign; hence the result follows.

Same sign lemmas

3.6. Lemma. Assume Q′ ∈ [PQ) and Q′ 6= P . Then for any X /∈/∈ (PQ) the angles ∠PQX and ∠PQ′X have the same sign.

PQ′Q

X Proof. By Proposition 2.3, for any t ∈ [0, 1]there is unique point Qt ∈ [PQ) such that

PQt = (1− t)·PQ+ t·PQ′.

Note that the map t 7→ Qt is continuous,

Q0 = Q, Q1 = Q′

and for any t ∈ [0, 1], we have P 6= Qt.Applying Corollary 3.5, for Pt = P , Qt and Xt = X , we get that

∠PQX has the same sign as ∠PQ′X .

27

3.7. Lemma. Assume [XY ] does not intersect (PQ) then the angles∠PQX and ∠PQY have the same sign.

PQ

X

Y

The proof is nearly identical to the oneabove.

Proof. According to Proposition 2.3, for anyt ∈ [0, 1] there is a point Xt ∈ [XY ] such that

XXt = t·XY.

Note that the map t 7→ Xt is continuous,X0 = X and X1 = Y and for any t ∈ [0, 1],we have Yt /∈ (QP ).

Applying Corollary 3.5, for Pt = P , Qt = Q and Xt, we get that∠PQX has the same sign as ∠PQY .

Half-planes

3.8. Proposition. The complement of a line (PQ) in the plane canbe presented in the unique way as a union of two disjoint subsets calledhalf-planes such that(a) Two points X,Y /∈ (PQ) lie in the same half-plane if and only

if the angles ∠PQX and ∠PQY have the same sign.(b) Two points X,Y /∈ (PQ) lie in the same half-plane if and only

if [XY ] does not intersect (PQ).

Further we say that X and Y lie on one side from (PQ) if they liein one of the half-planes of (PQ) and we say that P and Q lie on theopposite sides from ℓ if they lie in the different half-planes of ℓ.

PQ

H+

H−

Proof. Let us denote by H+ (correspondinglyH−) the set of points X in the plane suchthat ∠PQX is positive (correspondingly neg-ative).

According to Theorem 2.7, X /∈ (PQ) ifand only if ∡PQX 6= 0 nor π. Therefore H+

and H− give the unique subdivision of thecomplement of (PQ) which satisfies property(a).

Now let us prove that the this subdivision depends only on theline (PQ); that is, if (P ′Q′) = (PQ) and X,Y /∈ (PQ) then theangles ∠PQX and ∠PQY have the same sign if and only if the angles∠P ′Q′X and ∠P ′Q′Y have the same sign.


Applying Exercise 3.2, we can assume that P = P ′ and Q′ ∈ [PQ).It remains to apply Lemma 3.6.

(b). Assume [XY ] intersects (PQ). Since the subdivision dependsonly on the line (PQ), we can assume that Q ∈ [XY ]. In this case, byExercise 3.2, the angles ∠PQX and ∠PQY have opposite signs.

O

A

B

A′B′

Now assume [XY ] does not intersect (PQ). Inthis case, by Lemma 3.7, ∠PQX and ∠PQY havethe same sign.

3.9. Exercise. Assume that the angles ∠AOB and∠A′OB′ are vertical. Show that the line (AB) doesnot intersect the segment [A′B′].

Consider triangle ABC. The segments [AB], [BC] and [CA] arecalled sides of the triangle.

The following theorem is a corollary of Proposition 3.8.

3.10. Pasch’s theorem. Assume line ℓ does not pass through anyvertex a triangle. Then it intersects either two or zero sides of thetriangle.

3.11. Signs of angles of triangle. In any nondegenerate triangleABC the angles ∠ABC, ∠BCA and ∠CAB have the same sign.

ZAB

C Proof. Choose a point Z ∈ (AB) so that Alies between B and Z.

According to Lemma 3.6, the angles∠ZBC and ∠ZAC have the same sign.

Note that ∡ABC = ∡ZBC and

∡ZAC + ∡CAB ≡ π.

Therefore ∠CAB has the same sign as ∠ZACwhich in turn has the same sign as ∡ABC = ∡ZBC.

Repeating the same argument for ∠BCA and ∠CAB, we get theresult.

PQ

X

Y

BA

A′B′

C

29

3.12. Exercise. Show that two points X,Y /∈ (PQ) lie on the sameside from (PQ) if and only if the angles ∠PXQ and ∠PY Q have thesame sign.

3.13. Exercise. Let ABC be a nondegenerate triangle, A′ ∈ [BC]and B′ ∈ [AC]. Show that the segments [AA′] and [BB′] intersect.

3.14. Exercise. Assume that the points X and Y lie on the oppo-site sides from the line (PQ). Show that the half-line [PX) does notinterests [QY ).

3.15. Advanced exercise. Note that the following quantity

∡ABC =

[

π if ∡ABC = π

−∡ABC if ∡ABC < π

can serve as the angle measure; that is, the axioms hold if one changeseverywhere ∡ to ∡.

Show that ∡ and ∡ are the only possible angle measures on theplane.

Show that without Axiom IIIc, this is not longer true.

Triangle with the given sides

Consider a triangle ABC. Set

a = BC, b = CA, c = AB.

Without loss of generality we may assume that

a 6 b 6 c.

Then all three triangle inequalities for ABC hold if and only if

c 6 a+ b.

The following theorem states that this is the only restriction on a, band c.

3.16. Theorem. Assume that 0 < a 6 b 6 c 6 a+ b. Then there isa triangle ABC such that a = BC, b = CA and c = AB.

The proof requires some preparation.Assume r > 0 and π > β > 0. Consider triangle ABC such that

AB = BC = r and ∡ABC = β. The existence of such triangle followfrom Axiom IIIa and Proposition 2.3.


A C

B

s(β, r)

r r

β

Note that according to Axiom IV, the valuesβ and r define the triangle up to congruence. Inparticular the distance AC depends only on β andr. Set

s(β, r)def== AC.

3.17. Proposition. Given r > 0 and ε > 0 thereis δ > 0 such that if 0 < β < δ then s(r, β) < ε.

Proof. Fix two point A and B such that AB = r.Choose a point X such that ∡ABX is positive. Let Y ∈ [AX) be

the point such that AY = ε8 ; it exists by Proposition 2.3.

Note that X and Y lie on the same side from (AB); therefore∠ABY is positive. Set δ = ∡ABY .

Assume 0 < β < δ, ∡ABC = β and BC = r.Applying Axiom IIIa, we can choose a half-line [BZ) such that

∡ABZ = 12 ·β. Note that A and Y lie on the opposite sides from

(BZ). Therefore (BZ) intersects [AY ]; denote by D the point ofintersection.

Since D ∈ (BZ), we get ∡ABD = β2 or β

2 − π. The later isimpossible since D and Y lie on the same side from (AB). Therefore

∡ABD = ∡DBC = β2 .

AB

C

D Z

Y

X

r

r

By Axiom IV, ABD ∼= CBD. In par-ticular

AC 6 AD +DC =

= 2·AD 6

6 2·AY =

= ε4 .


3.18. Corollary. Fix a real number r > 0 and two distinct points Aand B. Then for any real number β ∈ [0, π], there is unique point Cβ

such that BCβ = r and ∡ABCβ = β. Moreover, the map β 7→ Cβ isa continuous map from [0, π] to the plane.

Proof. The existence and uniqueness of Cβ follows from Axiom IIIaand Proposition 2.3.

Note that if β1 6= β2 then

Cβ1Cβ2

= s(r, |β1 − β2|).

31

Therefore Proposition 3.17 implies that the map β 7→ Cβ is con-tinuous.

Proof of Theorem 3.16. Fix points A and B such that AB = c. Givenβ ∈ [0, π], denote by Cβ the point in the plane such that BCβ = a and∡ABC = β.

According to Corollary 3.18, the map β 7→ Cβ is a continuous.Therefore function b(β) = ACβ is continuous (formally it follows fromExercise 1.13 and Exercise 1.14).

Note that b(0) = c− a and b(π) = c+ a. Since c− a 6 b 6 c+ a,by Intermediate value theorem (3.4) there is β0 ∈ [0, π] such thatb(β0) = b. Hence the result follows.

Chapter 4

Congruent triangles

Side-angle-side condition

Our next goal is to give conditions which guarantee congruence of twotriangles. One of such conditions is Axiom IV, it is also called side-angle-side congruence condition or briefly SAS congruence condition.

Angle-side-angle condition

4.1. ASA condition. Assume that

AB = A′B′, ∡ABC ≡ ±∡A′B′C′, ∡CAB ≡ ±∡C′A′B′

and A′B′C′ is nondegenerate. Then

ABC ∼= A′B′C′.

Note that for degenerate triangles the statement does not hold, sayconsider one triangle with sides 1, 4, 5 and the other with sides 2, 3, 5.

A′

B′

C′ C′′

Proof. According to Theorem 3.11, either

➊∡ABC ≡ ∡A′B′C′,

∡CAB ≡ ∡C′A′B′

or

➋∡ABC ≡ −∡A′B′C′,

∡CAB ≡ −∡C′A′B′.

32

33

Further we assume that ➊ holds; the case ➋ is analogous.Let C′′ be the point on the half-line [A′C′) such that that A′C′′ =

= AC.By Axiom IV, A′B′C′′ ∼= ABC. Applying Axiom IV again, we

get∡A′B′C′′ ≡ ∡ABC ≡ ∡A′B′C′.

By Axiom IIIa, [B′C′) = [BC′′). Whence C′′ lies on (B′C′) as well ason (A′C′).

Since A′B′C′ is not degenerate, (A′C′) is distinct from (B′C′).Applying Axiom II, we get C′′ = C′.

Therefore A′B′C′ = A′B′C′′ ∼= ABC.

Isosceles triangles

A triangle with two equal sides is called isosceles ; the remaining sideis called base of isosceles triangle.

4.2. Theorem. Assume ABC is an isosceles triangle with the base[AB]. Then

∡ABC ≡ −∡BAC.

Moreover, the converse holds if ABC is nondegenerate.

A B

CThe following proof is due to Pappus ofAlexandria.

Proof. Note that

CA = CB, CB = CA, ∡ACB ≡ −∡BCA.

Therefore by Axiom IV,

CAB ∼= CBA.

Applying the theorem on the signs of angles of triangles (3.11) andAxiom IV again, we get

∡CAB ≡ −∡CBA.

To prove the converse, we assume ∡CAB ≡ −∡CBA. By ASAcondition 4.1, CAB ∼= CBA. Therefore CA = CB.

A triangle with three equal sides is calledequilateral.

4.3. Exercise. Let ABC be an equilateral triangle. Show that

∡ABC = ∡BCA = ∡CAB.

34 CHAPTER 4. CONGRUENT TRIANGLES

Side-side-side condition

4.4. SSS condition. ABC ∼= A′B′C′ if

A′B′ = AB, B′C′ = BC and C′A′ = CA.

Proof. Choose C′′ so that A′C′′ = A′C′ and ∡B′A′C′′ ≡ ∡BAC.According to Axiom IV,

A′B′C′′ ∼= ABC.

A′ B′

C′

C′′

It will suffice to prove that

➌ A′B′C′ ∼= A′B′C′′.

The condition ➌ trivially holds if C′′ == C′. Thus it remains to consider thecase C′′ 6= C′.

Clearly, the corresponding sides ofA′B′C′ and A′B′C′′ are equal.Whence the triangles C′A′C′′ andC′B′C′′ are isosceles. By Theorem 4.2,we have

∡A′C′′C′ ≡ −∡A′C′C′′,

∡C′C′′B′ ≡ −∡C′′C′B′.

By addition∡A′C′B′ ≡ −∡A′C′′B′.

Applying Axiom IV again, we get ➌.

BA

A′B′

C4.5. Advanced exercise. Let M be themidpoint of side [AB] of a triangle ABCand M ′ be the midpoint of side [A′B′] ofa triangle A′B′C′. Assume C′A′ = CA,C′B′ = CB and C′M ′ = CM . Prove thatA′B′C′ ∼= ABC.

4.6. Exercise. Let ABC be an isoscelestriangle with the base [AB] and the points A′ ∈

[BC] and B′ ∈ [AC] be such that CA′ = CB′. Show that

(a) AA′C ∼= BB′C;(b) ABB′ ∼= BAA′.

35

4.7. Exercise. Show that if AB +BC = AC then B ∈ [AC].

4.8. Exercise. Let ABC be a nondegenerate triangle and let f bea motion of the plane such that

f(A) = A, f(B) = B and f(C) = C.

Show that f is the identity; that is, f(X) = X for any point X onthe plane.

Chapter 5

Perpendicular lines

Right, acute and obtuse angles

⋄ If |∡AOB| = π2 , we say that the angle ∠AOB is right ;

⋄ If |∡AOB| < π2 , we say that the angle ∠AOB is acute;

⋄ If |∡AOB| > π2 , we say that the angle ∠AOB is obtuse.

On the diagrams, the right angles will bemarked with a little square, as shown.

If ∠AOB is right, we say also that [OA)is perpendicular to [OB); it will be written as[OA) ⊥ [OB).

From Theorem 2.7, it follows that two lines(OA) and (OB) are appropriately called perpen-dicular, if [OA) ⊥ [OB). In this case we alsowrite (OA) ⊥ (OB).

5.1. Exercise. Assume point O lies between A and B. Show thatfor any point X the angle ∠XOA is acute if and only if ∠XOB isobtuse.

Perpendicular bisector

Assume M is the midpoint of the segment [AB]; that is, M ∈ (AB)and AM = MB.

The line ℓ passing through M and perpendicular to (AB) is calledperpendicular bisector to the segment [AB].

5.2. Theorem. Given distinct points A and B, all points equidistantfrom A and B and no others lie on the perpendicular bisector to [AB].

36

37

A BM

P

Proof. Let M be the midpoint of [AB].Assume PA = PB and P 6= M . Ac-

cording to SSS (4.4), AMP ∼= BMP .Whence

∡AMP ≡ ±∡BMP.

Since A 6= B, we have “−” in the aboveformula. Further,

π ≡ ∡AMB ≡≡ ∡AMP + ∡PMB ≡≡ 2·∡AMP.

That is, ∡AMP ≡ ±π2 and therefore P lies on the perpendicular

bisector.To prove converse, suppose P 6= M is any point in the perpendic-

ular bisector to [AB]. Then ∡AMP ≡ ±π2 , ∡BMP ≡ ±π

2 and AM == BM . Therefore AMP ∼= BMP ; in particular AP = BP .

5.3. Exercise. Let ℓ be the perpendicular bisector to the segment[AB] and X be an arbitrary point on the plane.

Show that AX < BX if and only if X and A lie on the same sidefrom ℓ.

5.4. Exercise. Let ABC be nondegenerate. Show that AB > BCif and only if |∡BCA| > |∡ABC|.

Uniqueness of perpendicular

5.5. Theorem. There is one and only one line which pass through agiven point P and perpendicular to a given line ℓ.

A B

ℓ

P

P ′

According to the above theorem, thereis unique point Q ∈ ℓ such that (QP ) ⊥ ℓ.This point Q is called foot point of P on ℓ.

Proof. If P ∈ ℓ then both, existence anduniqueness, follow from Axiom III.

Existence for P 6∈ ℓ. Let A, B be two dis-tinct points of ℓ. Choose P ′ so that AP ′ == AP and ∡P ′AB ≡ −∡PAB. Accordingto Axiom IV, AP ′B ∼= APB. ThereforeAP = AP ′ and BP = BP ′.

38 CHAPTER 5. PERPENDICULAR LINES

According to Theorem 5.2, A and B lie on perpendicular bisectorto [PP ′]. In particular (PP ′) ⊥ (AB) = ℓ.

Q Q′

P

P ′

ℓ

m

Uniqueness for P 6∈ ℓ. From above we canchoose a point P ′ in such a way that ℓ formsthe perpendicular bisector to [PP ′].

Assume m ⊥ ℓ and m ∋ P . Then ma perpendicular bisector to some segment[QQ′] of ℓ; in particular, PQ = PQ′.

Since ℓ is perpendicular bisector to[PP ′], we get PQ = P ′Q and PQ′ = P ′Q′. Therefore

P ′Q = PQ = PQ′ = P ′Q′.

By Theorem 5.2, P ′ lies on the perpendicular bisector to [QQ′] whichis m. By Axiom II, m = (PP ′).

Reflection

Assume a point P and a line (AB) are given. To find the reflectionP ′ of P in (AB), one drops a perpendicular from P onto (AB), andcontinues it to the same distance on the other side.

According to Theorem 5.5, P ′ is uniquely determined by P .

Note that P = P ′ if and only if P ∈ (AB).

5.6. Proposition. Assume P ′ is a reflection of the point P in theline (AB). Then AP ′ = AP and if A 6= P then ∡BAP ′ ≡ −∡BAP .

Proof. Note that if P ∈ (AB) then P = P ′ and by Corollary 2.8∡BAP = 0 or π. Hence the statement follows.

A B

P

P ′

If P /∈ (AB), then P ′ 6= P . Byconstruction, the line (AB) is per-pendicular bisector of [PP ′]. There-fore, according to Theorem 5.2, AP ′ =AP and BP ′ = BP . In particu-lar, ABP ′ ∼= ABP . Therefore∡BAP ′ ≡ ±∡BAP .

Since P ′ 6= P and AP ′ = AP , weget ∡BAP ′ 6= ∡BAP . That is, we areleft with the case

∡BAP ′ ≡ −∡BAP.

39

5.7. Corollary. Reflection through the line is a motion of the plane.More over if P ′Q′R′ is the reflection of PQR then

∡Q′P ′R′ ≡ −∡QPR.

Proof. From the construction it follows that the composition of tworeflections through the same line, say (AB), is the identity map. Inparticular reflection is a bijection.

Assume P ′, Q′ and R′ denote the reflections of the points P , Qand R through (AB). Let us first show that

➊ P ′Q′ = PQ and ∡AP ′Q′ ≡ −∡APQ.

Without loss of generality we may assume that the points P andQ are distinct from A and B. By Proposition 5.6,

∡BAP ′ ≡ −∡BAP, ∡BAQ′ ≡ −∡BAQ,

AP ′ = AP, AQ′ = AQ.

It follows that ∡P ′AQ′ ≡ −∡PAQ. Therefore P ′AQ′ ∼= PAQand ➊ follows.

Repeating the same argument for P and R, we get

∡AP ′R′ ≡ −∡APR.

Subtracting the second identity in ➊, we get

∡Q′P ′R′ ≡ −∡QPR.

5.8. Exercise. Show that any motion of the plane can be presentedas a composition of at most three reflections.

Applying the exercise above and Corollary 5.7, we can divide themotions of the plane in two types, direct and indirect motions. Themotion m is direct if

∡Q′P ′R′ = ∡QPR

for any PQR and P ′ = m(P ), Q′ = m(Q) and R′ = m(R); if insteadwe have

∡Q′P ′R′ ≡ −∡QPR

for any PQR then the motion m is called indirect.

5.9. Exercise. Let X and Y be the reflections of P through the lines(AB) and (BC) correspondingly. Show that

∡XBY ≡ 2·∡ABC.


Perpendicular is shortest

5.10. Lemma. Assume Q is the foot point of P on line ℓ. Then

PX > PQ

for any point X on ℓ distinct from Q.

Xℓ

P

Q

P ′

Proof. If P ∈ ℓ then the result follows since PQ = 0.Further we assume that P /∈ ℓ.

Let P ′ be the reflection of P in ℓ. Note that Q isthe midpoint of [PP ′] and ℓ is perpendicular bisectorof [PP ′]. Therefore

PX = P ′X and PQ = P ′Q = 12 ·PP ′

Note that ℓmeets [PP ′] at the point Q only. There-fore by the triangle inequality and Exercise 4.7,

PX + P ′X > PP ′.


5.11. Exercise. Assume ∠ABC is right or obtuse. Show that

BC > AB.

Angle bisectors

If ∡ABX ≡ −∡CBX then we say that line (BX) bisects angle ∠ABC,or line (BX) is the bisector of ∠ABC. If ∡ABX ≡ π − ∡CBX thenthe line (BX) is called external bisector of ∠ABC.

Note that bisector and external bisector are uniquely defined bythe angle.

A

B

C

bisector

extern

al

bisector

Note that if ∡ABA′ = π; that is, ifB lies between A and A′, then bisector of∠ABC is the external bisector of ∠A′BCand the other way around.

5.12. Exercise. Show that for any an-gle, its bisector and external bisector areperpendicular.

5.13. Lemma. Assume ∡ABC 6≡ π nor 0. Given angle ∠ABC anda point X, consider foot points Y and Z of X on (AB) and (BC).

41

Then XY = XZ if and only if X lies on the bisector or externalbisector of ∠ABC.

Proof. Let Y ′ and Z ′ be the reflections of X through (AB) and (BC)correspondingly. By Proposition 5.6, XB = Y ′B = Z ′B.

A

B

C

ZX

Y

Y ′

Z ′

Note that

XY ′ = 2·XY and XZ ′ = 2·XZ.

Applying SSS and then SAS congruence condi-tions, we get

➋

XY = XZ

mXY ′ = XZ ′

mBXY ′ ∼= BXZ ′

m∡XBY ′ ≡ ±∡BXZ ′.

According to Proposition 5.6,

∡XBA ≡ −∡Y ′BA,

∡XBC ≡ −∡Z ′BC.

Therefore

2·∡XBA ≡ ∡XBY ′ and 2·∡XBC ≡ −XBZ ′.

That is, we can continue the chain of equivalence conditions ➋ thefollowing way

∡XBY ′ ≡ ±∡BXZ ′ ⇐⇒ 2·∡XBA ≡ ±2·∡XBC.

Since (AB) 6= (BC), we have

2·∡XBA 6≡ 2·∡XBC

(compare to Exercise 2.10). Therefore

XY = XZ ⇐⇒ 2·∡XBA ≡ −2·∡XBC.

The last identity means either

∡XBA+ ∡XBC ≡ 0

or

∡XBA+ ∡XBC ≡ π.



Circles

Given a positive real number r and a point O, the set Γ of all pointson distant r from O is called circle with radius r and center O.

We say that a point P lies inside Γ if OP < r and if OP > r, wesay that P lies outside Γ.

5.14. Exercise. Let Γ be a circle and P /∈ Γ. Assume a line ℓ ispassing through the point P intersects Γ at two distinct points X andY . Show that P is inside Γ if and only if P lies between X and Y .

A segment between two points on Γ is called chord of Γ. A chordpassing through the center is called diameter.

5.15. Exercise. Assume two distinct circles Γ and Γ′ have a commonchord [AB]. Show that the line between centers of Γ and Γ′ forms aperpendicular bisector to [AB].

5.16. Lemma. A line and a circle can have at most two points ofintersection.

A B C

ℓ

m n

Proof. Assume A, B and C are distinct points which lie on a line ℓand a circle Γ with center O.

Then OA = OB = OC; in particular O lies on the perpendicularbisectors m and n to [AB] and [BC] correspondingly.

Note that the midpoints of [AB] and [BC] are distinct. There-fore m and n are distinct. The later contradicts the uniqueness ofperpendicular (Theorem 5.5).

5.17. Exercise. Show that two distinct circles can have at most twopoints of intersection.

In consequence of the above lemma, a line ℓ and a circle Γ mighthave 2, 1 or 0 points of intersections. In the first case the line is calledsecant line, in the second case it is tangent line; if P is the only pointof intersection of ℓ and Γ, we say that ℓ is tangent to Γ at P .

Similarly, according Exercise 5.17, two circles might have 2, 1 or 0points of intersections. If P is the only point of intersection of circlesΓ and Γ′, we say that Γ is tangent to Γ at P .

43

5.18. Lemma. Let ℓ be a line and Γ be a circle with center O. As-sume P is a common point of ℓ and Γ. Then ℓ is tangent to Γ at P ifand only if (PO) ⊥ ℓ.

Proof. Let Q be the foot point of O on ℓ.Assume P 6= Q. Denote by P ′ the reflection of P through (OQ).Note that P ′ ∈ ℓ and (OQ) is perpendicular bisector of [PP ′].

Therefore OP = OP ′. Hence P, P ′ ∈ Γ ∩ ℓ; that is, ℓ is secant to Γ.If P = Q then according to Lemma 5.10, OP < OX for any point

X ∈ ℓ distinct from P . Hence P is the only point in the intersectionΓ ∩ ℓ; that is, ℓ is tangent to Γ at P .

5.19. Exercise. Let Γ and Γ′ be two distinct circles with centers at Oand O′ correspondingly. Assume Γ and Γ′ intersect at point P . Showthat Γ is tangent to Γ′ if and only if O, O′ and P lie on one line.

5.20. Exercise. Let Γ and Γ′ be two distinct circles with centers atO and O′ and radii r and r′.(a) Show that Γ is tangent to Γ′ if and only if

OO′ = r + r′ or OO′ = |r − r′|.

(b) Show that Γ intersects Γ′ if and only if

|r − r′| 6 OO′ 6 r + r′.

5.21. Exercise. Assume three circles intersect at two points. Showthat the centers of these circles lie on one line.

Geometric constructions

The ruler-and-compass constructions in the plane is the constructionof points, lines, and circles using only an idealized ruler and compass.These construction problems provide a valuable source of exercises ingeometry which we will use further in the book. In addition, Chap-ter 18 is devoted completely to the subject.

The idealized ruler can be used only to draw a line through giventwo points. The idealized compass can be used only to draw a circlewith given center and radius. That is, given three points A, B andO we can draw the set of all points on distant AB from O. We mayalso mark new points in the plane as well as on the constructed lines,circles and their intersections (assuming that such points exist).


We can also look at the different set of instruments. For example,we may only use the ruler or we may invent a new instrument, say aninstrument which produce midpoint for given two points.

As an example, let us consider the following problem:

5.22. Construction of midpoint. Construct the midpoint of thegiven segment [AB].

A

B

P

Q

M

Construction.1. Construct the circle with center at A

which is passing through B.2. Construct the circle with center at B

which is passing through A.3. Mark both points of intersection of these

circles, label them by P and Q.4. Draw the line (PQ).5. Mark the point of intersection of (PQ)

and [AB]; this is the midpoint.

Typically, you need to proof that the con-struction produce what was expected. Here is

a proof for the example above.

Proof. According to Theorem 5.2, (PQ) is the perpendicular bisectorto [AB]. Therefore M = (AB) ∩ (PQ) is the midpoint of [AB].

5.23. Exercise. Make a ruler-and-compass construction of the linethough the given point which is perpendicular to the given line.

5.24. Exercise. Make a ruler-and-compass construction of the centerof given circle.

5.25. Exercise. Make a ruler-and-compass construction of the linestangent to the given circle which pass through given point.

5.26. Exercise. Given two circles Γ1 and Γ2 and a segment [AB]make a ruler-and-compass construction of a circle of radius AB whichis tangent to each circle Γ1 and Γ2.

Chapter 6

Parallel lines and

similar triangles

Parallel lines

In consequence of Axiom II, any two distinct lines ℓ and m have eitherone point in common or none. In the first case they are intersecting;in the second case, ℓ and m are said to be parallel (briefly ℓ ‖ m); inaddition, a line is always regarded as parallel to itself.

6.1. Proposition. Let ℓ, m and n be the lines in the plane. Assumethat n ⊥ m and m ⊥ ℓ. Then ℓ ‖ n.

Proof. Assume contrary; that is, ℓ ∦ n. Then there is a point, say Z,of intersection of ℓ and n. Then by Theorem 5.5, ℓ = n. In particularℓ ‖ n, a contradiction.

6.2. Theorem. Given a point P and line ℓ in the Euclidean planethere is unique line m which pass though P and parallel to ℓ.

The above theorem has two parts, existence and uniqueness. Inthe proof of uniqueness we will use Axiom V for the first time in thisbook.

Proof; existence. Apply Theorem 5.5 two times, first to construct linem through P which is perpendicular to ℓ and second to construct linen through P which is perpendicular to m. Then apply Proposition 6.1.

Uniqueness. If P ∈ ℓ then m = ℓ by the definition of parallel lines.Further we assume P /∈ ℓ.

45

46 CHAPTER 6. PARALLEL LINES AND SIMILAR TRIANGLES

Let us construct lines n ∋ P andm ∋ P as in the proof of existence,so m ‖ ℓ.

Assume there is yet an other line s ∋ P which is distinct from mand parallel to ℓ. Choose a point Q ∈ s which lies with ℓ on the sameside from m. Let R be the foot point of Q on n.

Let D be the point of intersection of n and ℓ. According to Propo-sition 6.1 (QR) ‖ m. Therefore Q, R and ℓ lie on the same side fromm. In particular, R ∈ [PD).

P

R

D

Q

Zℓ

m

s

n

Choose Z ∈ [PQ) such that

PZ

PQ=

PD

PR.

Then by Axiom V, (ZD) ⊥ (PD); that is, Z ∈ ℓ ∩ s, a contradiction.

6.3. Corollary. Assume ℓ, m and n are lines in the Euclidean planesuch that ℓ ‖ m and m ‖ n. Then ℓ ‖ n.

Proof. Assume contrary; that is, ℓ ∦ n. Then there is a point P ∈ ℓ∩n.By Theorem 6.2, n = ℓ, a contradiction.

Note that from the definition, we have ℓ ‖ m if and only if m ‖‖ ℓ. Therefore according to the above corollary “‖” is an equivalencerelation.

6.4. Exercise. Let k, ℓ, m and n be the lines in Euclidean plane.Assume that k ⊥ ℓ and m ⊥ n. Show that if k ‖ m then ℓ ‖ n.

6.5. Exercise. Make a ruler-and-compass construction of the linethough the given point which is parallel to the given line.

Similar triangles

Two triangles A′B′C′ and ABC are similar (briefly A′B′C′ ∼∼ ABC) if their sides are proportional, that is,

➊ A′B′ = k·AB, B′C′ = k·BC and C′A′ = k·CA

47

for some k > 0 and

➋

∡A′B′C′ = ±∡ABC,

∡B′C′A′ = ±∡BCA,

∡C′A′B′ = ±∡CAB.

Remarks.

⋄ According to 3.11, in the above three equalities the signs can beassumed to me the same.

⋄ If A′B′C′ ∼ ABC with k = 1 in ➊, then A′B′C′ ∼=∼= ABC.

⋄ Note that “∼” is an equivalence relation. That is,

(i) ABC ∼ ABC for any ABC.

(ii) If A′B′C′ ∼ ABC then

ABC ∼ A′B′C′

(iii) If A′′B′′C′′ ∼ A′B′C′ and A′B′C′ ∼ ABC then

A′′B′′C′′ ∼ ABC.

Using “∼”, the Axiom V can be formulated the following way.

6.6. Reformulation of Axiom V. If for two triangles ABC,AB′C′ and k > 0 we have B′ ∈ [AB), C′ ∈ [AC), AB′ = k·AB andAC′ = k·AC then ABC ∼ AB′C′.

In other words, the Axiom V provides a condition which guaran-tee that two triangles are similar. Let us formulate yet three suchconditions.

6.7. Similarity conditions. Two triangles ABC and A′B′C′

in the Euclidean plane are similar if one of the following conditionshold.

(SAS) For some constant k > 0 we have

AB = k·A′B′, AC = k·A′C′

and ∡BAC = ±∡B′A′C′.

(AA) The triangle A′B′C′ is nondegenerate and

∡ABC = ±∡A′B′C′, ∡BAC = ±∡B′A′C′.

(SSS) For some constant k > 0 we have

AB = k·A′B′, AC = k·A′C′, CB = k·C′B′.


Each of these conditions is proved by applying the Axiom V withSAS, ASA and SSS congruence conditions correspondingly (see Ax-iom IV and the conditions 4.1, 4.4).

Proof. Set k = ABA′B′

. Choose points B′′ ∈ [A′B′) and C′′ ∈ [A′C′) sothat A′B′′ = k·A′B′ and A′C′′ = k·A′C′. By Axiom V, A′B′C′ ∼∼ A′B′′C′′.

Applying SAS, ASA or SSS congruence condition, depending onthe case, we get A′B′′C′′ ∼= ABC. Hence the result follows.

A triangle with all acute angles is called acute.

A B

C

A′B′

6.8. Exercise. Let ABC be an acute trianglein the Euclidean plane. Denote by A′ the foot pointof A on (BC) and by B′ the foot point of B on(AC). Prove that A′B′C ∼ ABC.

A bijection from plane to itself is called anglepreserving transformationangle preserving if

∡ABC = ∡A′B′C′

for any triangle ABC and its image A′B′C′.

6.9. Exercise. Show that any angle-preserving transformation of Eu-clidean plane multiplies all the distance by a fixed constant.

Pythagorean theorem

A triangle is called right if one of its angles is right. The side oppositethe right angle is called the hypotenuse. The sides adjacent to theright angle are called legs.

6.10. Theorem. Assume ABC is a right triangle in the Euclideanplane with right angle at C. Then

AC2 +BC2 = AB2.

Proof. Let D be the foot point of C on (AB).

A B

C

D

According to Lemma 5.10,

AD < AC < AB

and

BD < BC < AB.

49

Therefore D lies between A and B; in particular,

➌ AD +BD = AB.

Note that by AA similarity condition, we have

ADC ∼ ACB ∼ CDB.

In particular

➍AD

AC=

AC

ABand

BD

BC=

BC

BA.

Let us rewrite identities ➍ on an other way:

AC2 = AB ·AD and BC2 = AB ·BD.

Summing up above two identities and applying ➌, we get

AC2 +BC2 = AB ·(AD +BD) = AB2.

6.11. Exercise. Assume A, B, C and D are as in the proof above.Show that

CD2 = AD·BD.

The following exercise is the converse to Pythagorean theorem.

6.12. Exercise. Assume ABC is a triangle in the Euclidean planesuch that

AC2 +BC2 = AB2.

Prove that the angle at C is right.

Angles of triangle

6.13. Theorem. In any triangle ABC in the Euclidean plane, wehave

∡ABC + ∡BCA+ ∡CAB ≡ π.

Proof. First note that if ABC is degenerate then the equality followsfrom Lemma 2.7. Further we assume that ABC is nondegenerate.


A B

C

α α β

γ

β

±γ

M

KL

Set

α = ∡CAB,

β = ∡ABC,

γ = ∡BCA.

We need to prove that

➎ α+ β + γ ≡ π.

Let K, L, M be the midpoints of the sides [BC], [CA], [AB] re-spectively. By Axiom V,

AML ∼ ABC, MBK ∼ ABC, LKC ∼ ABC

and

LM = 12 ·BC, MK = 1

2 ·CA, KL = 12 ·AB.

According to SSS condition (6.7), KLM ∼ ABC. Thus,

➏ ∡MKL = ±α, ∡KLM = ±β, ∡LMK = ±γ.

According to 3.11, the “+” or “−” sign is to be the same throughout➏.

If in ➏ we have “+” then ➎ follows since

β + γ + α ≡ ∡AML+ ∡LMK + ∡KMB ≡ ∡AMB ≡ π

It remains to show that we can not have “−” in ➏. In this case thesame argument as above gives

α+ β − γ ≡ π.

The same way we get

α− β + γ ≡ π

Adding last two identities we get

2·α ≡ 0.

Equivalently α ≡ π or 0; that is, ABC is degenerate, a contradiction.

51

A

B CD

6.14. Exercise. Let ABC be a non-degenerate triangle. Assume there is apoint D ∈ [BC] such that (AD) bisects∠BAC and BA = AD = DC. Find theangles of ABC.

6.15. Exercise. Show that

|∡ABC| + |∡BCA|+ |∡CAB| = π.

for any ABC in the Euclidean plane.

A

B

C

D

6.16. Exercise. Let ABC be an isoscelesnondegenerate triangle with base [AC]. Con-sider the point D on the extension of the side[AB] such that AB = BD. Show that ∠ACD isright.

6.17. Exercise. Let ABC be an isosceles nondegenerate trianglewith base [AC]. Assume that circle Γ pass through A, centered ata point on [AB] and tangent to (BC) at the point X. Show that∡CAX = ±π

4 .

Transversal property

If a line t intersects each line ℓ and m at one point then we say thatt is a transversal to ℓ and m. On the diagram below, line (CB) is atransversal to (AB) and (CD).

6.18. Transversal property. In the Euclidean plane, (AB) ‖ (CD)if and only if

➐ 2·(∡ABC + ∡BCD) ≡ 0.

Equivalently

∡ABC + ∡BCD ≡ 0 or ∡ABC + ∡BCD ≡ π;

in the first case A and D lie on the opposite sides of (BC), in thesecond case A and D lie on the same sides of (BC).


AB

C

D

Proof. If (AB) ∦ (CD) then there is Z ∈ (AB)∩(CD)and BCZ is nondegenerate.

According to Theorem 6.13,

∡ZBC + ∡BCZ ≡ π − ∡CZB 6≡6≡ 0 nor π.

Note that 2·∡ZBC ≡ 2·∡ABC and 2·∡BCZ ≡ 2·∡BCD. Therefore

2·(∡ABC + ∡BCD) ≡ 2·∡ZBC + 2·∡BCZ 6≡ 0;

that is, ➐ does not hold.Note that if the points A, B and C are fixed, the identity ➐

uniquely defines the line (CD). Therefore by Theorem 6.2, if (AB) ‖(CD) then equality ➐ holds.

Applying Proposition 3.8, we get the last part of Transversal prop-erty.

6.19. Exercise. Let ABC be a nondegenerate triangle. Assume B′

and C′ are points on sides [AB] and [AC] such that (B′C′) ‖ (BC).Show that ABC ∼ AB′C′.

6.20. Exercise. Trisect given segment with ruler and compass.

Parallelograms

A quadrilateral is an ordered quadruple of distinct points in the plane.A quadrilateral formed by quadruple (A,B,C,D) will be denoted asABCD.

Given a quadrilateral ABCD, the four segments [AB], [BC],[CD] and [DA] are called sides of ABCD; the remaining two seg-ments [AC] and [BD] are called diagonals of ABCD.

6.21. Exercise. Show for any quadrilateral ABCD in the Eu-clidean plane we have

∡ABC + ∡BCD + ∡CDA+ ∡DAB ≡ 0.

A quadrilateral ABCD in the Euclidean plane is called nonde-generate if any three points from A,B,C,D do not lie on one line.

The nondegenerate quadrilateral ABCD is called parallelogramif (AB) ‖ (CD) and (BC) ‖ (DA).

6.22. Lemma. If ABCD is a parallelogram then

53

(a) ∡DAB = ∡BCD;(b) AB = CD.

Proof. Since (AB) ‖ (CD), the points C and D lie on the same sidefrom (AB). Whence ∠ABD and ∠ABC have the same sign.

AB

C D

Analogously, ∠CBD and ∠CBA have thesame sign.

Since ∡ABC ≡ −∡CBA, we get that theangles ∠DBA and ∠DBC have opposite signs;that is, A and C lie on the opposite sides of(BD).

According to Transversal property (6.18),

∡BDC ≡ −∡DBA and ∡DBC ≡ −∡BDA.

By ASA condition ABD ∼= CDB. Which implies both statementsin the lemma.

6.23. Exercise. Assume ABCD is a quadrilateral such that

AB = CD = BC = DA.

Show that ABCD is a parallelogram.

A quadrilateral as in the exercise above is called rhombus.

6.24. Exercise. Show that diagonals of parallelogram intersect eachother at the midpoint.

A quadraliteral ABCD is called rectangle if the angles ∠ABC,∠BCD, ∠CDA and∠DAB are right. Note that according to Transver-sal property (6.18), any rectangle is a parallelogram.

If in addition AB = BC = CD = DA then the rectangle ABCDis called square. In other words a square is a rectangle which is also arhombus.

6.25. Exercise. Show that parallelogram ABCD is a rectangle ifand only if AC = BD.

6.26. Exercise. Show that parallelogram ABCD is a rhombus ifand only if (AC) ⊥ (BD).

Method of coordinates

The following exercise is important; it shows that our axiomatic defi-nition agrees with the model definition described on page 11.


P

Q

Pℓ Qℓ

Pm

Qm

ℓ

m 6.27. Exercise. Let ℓ and m be perpendic-ular lines in the Euclidean plane. Given apoints P denote by Pℓ and Pm the foot pointsof P on ℓ and m correspondingly.(a) Show that for any X ∈ ℓ and Y ∈ m

there is unique point P such that Pℓ = Xand Pm = Y .

(b) Show that PQ2 = PℓQ2ℓ+PmQ2

m for anypair of points P and Q.

(c) Conclude that Euclidean plane is isometric to (R2, d2) definedon page 11.

Once this exercise solved, we can apply the method of coordinatesto solve any problem in Euclidean plane geometry. This method ispowerful, but it is often considered as a bad style.

6.28. Exercise. Use the Exercise 6.27, to give an alternative proofof Theorem 3.16 in the Euclidean plane.

That is, prove that given real numbers a, b and c such that

0 < a 6 b 6 c 6 a+ c,

there is a triangle ABC such that a = BC, b = CA and c = AB.

6.29. Exercise. Let (xA, yA) and (xB , yB) be the coordinates of dis-tinct points A and B in the Euclidean plane. Show that the line (AB)is formed by points with coordinates (x, y) which satisfy the followingequation

(x− xA)·(yB − yA) = (y − yA)·(xB − xA).

Chapter 7

Triangle geometry

Triangle geometry studies of the properties of triangles, including as-sociated centers and circles.

In this chapter we discuss the most basic results in triangle ge-ometry, mostly to show that we develop enough machinery to provethings.

Circumcircle and circumcenter

7.1. Theorem. Perpendicular bisectors to the sides of any nonde-generate triangle in the Euclidean plane intersect at one point.

The point of the intersection of the perpendicular bisectors is calledcircumcenter. It is the center of the circumcircle of the triangle; thatis, the circle which pass through all three vertices of the triangle. Thecircumcenter of the triangle is usually denoted by O.

B

A

C

O

ℓm

Proof. Let ABC be nondegenerate. Let ℓ and mbe perpendicular bisectors to sides [AB] and [AC]correspondingly.

Assume ℓ and m intersect, let O = ℓ ∩ n.Let us Theorem 5.2. Since O ∈ ℓ, we have OA =

OB and since O ∈ m, we have OA = OC. It followsthat OB = OC; that is, O lies on the perpendicularbisector to [BC].

It remains to show that ℓ ∦ m; assume contrary.Since ℓ ⊥ (AB) and m ⊥ (AC), we get (AC) ‖ (AB) (see Exer-cise 6.4). Therefore by Theorem 5.5, (AC) = (AB); that is, ABC isdegenerate, a contradiction.

55

56 CHAPTER 7. TRIANGLE GEOMETRY

7.2. Exercise. There is unique circle which pass through vertices ofa given nondegenerate triangle in the Euclidean plane.

Altitudes and orthocenter

An altitude of a triangle is a line through a vertex and perpendicularto the line containing the opposite side. The term altitude maybealso used for the distance from the vertex to its foot point on the linecontaining opposite side.

7.3. Theorem. The three altitudes of any nondegenerate triangle inthe Euclidean plane intersect in a single point.

The point of intersection of altitudes is called orthocenter ; it isusually denoted as H .

B′

A

A′

B

C

C′

Proof. Let ABC be nondegenerate.Consider three lines ℓ ‖ (BC) through A, m ‖

‖ (CA) through B and n ‖ (AB) through C.Since ABC is nondegenerate, no pair of thelines ℓ, m and n is parallel. Set A′ = m ∩ n,B′ = n ∩ ℓ and C′ = ℓ ∩m.

Note that ABA′C, BCB′A and CBC′Aare parallelograms. Applying Lemma 6.22 we get

that ABC is the median triangle of A′B′C′; that is, A, B and Care the midpoints of [B′C′], [C′A′] and [A′B′] correspondingly.

By Exercise 6.4, (B′C′) ‖ (BC), the altitudes from A is perpen-dicular to [B′C′] and from above it bisects [B′C′].

Whence the altitudes of ABC are also perpendicular bisectors ofthe triangle A′B′C′. Applying Theorem 7.1, we get that altitudesof ABC intersect at one point.

7.4. Exercise. Assume H is the orthocenter of an acute triangleABC in the Euclidean plane. Show that A is orthocenter of HBC.

Medians and centroid

A median of a triangle is a segment joining a vertex to the midpointof the opposing side.

7.5. Theorem. The three medians of any nondegenerate triangle inthe Euclidean plane intersect in a single point. Moreover the point ofintersection divides each median in ratio 2:1.

57

The point of intersection of medians is called centroid ; it is usuallydenoted by M .

Proof. Consider a nondegenerate triangleABC. Let [AA′] and [BB′]be its medians.

According to Exercise 3.13, [AA′] and [BB′] are intersecting. Letus denote by M the point of intersection.

By SAS, B′A′C ∼ ABC and A′B′ = 12 ·AB. In particular

∡ABC ≡ ∡B′A′C.

Since A′ lies between B and C, we get ∡BA′B′ + ∡B′A′C = π.Therefore

∡B′A′B + ∡A′BA = π.

By Transversal property (6.18), (AB) ‖ (A′B′).

A

A′

B

B′

C

M

Note that A′ and A lie on the opposite sidesfrom (BB′). Therefore by Transversal property(6.18) we get

∡B′A′M = ∡BAM.

The same way we get,

∡A′B′M = ∡ABM.

By AA condition, ABM ∼ A′B′M .

Since A′B′ = 12 ·AB, we have

A′M

AM=

B′M

BM=

1

2.

In particular M divides medians [AA′] and [BB′] in ratio 2:1.

Note that M is unique point on [BB′] such that

B′M

BM=

1

2.

Repeating the same argument for vertices B and C we get that allmedians [CC′] and [BB′] intersect in M .

7.6. Exercise. Let ABCD be a nondegenerate quadrilateral andX, Y , V , W be the midpoints of its sides [AB], [BC], [CD] and [DA].Show that XY VW is a parallelogram.

58 CHAPTER 7. TRIANGLE GEOMETRY

Bisector of triangle

7.7. Lemma. Let ABC be a nondegenerate triangle in the Eu-clidean plane. Assume that the bisector of ∠BAC intersects [BC] atthe point D. Then

➊AB

AC=

DB

DC.

A

BC D

Eℓ

Proof. Let ℓ be the line through C parallel to(AB). Note that ℓ ∦ (AD); set E = ℓ ∩ (AD).

Note that B and C lie on the opposite sides of(AD). Therefore by Transversal property (6.18),

➋ ∡BAD = ∡CED.

Further, note that the angles ∠ADB and∠EDC are vertical; in particular, by 2.11

∡ADB = ∡EDC.

By AA-similarity condition, ABD ∼ ECD. In particular,

➌AB

EC=

DB

DC.

Since (AD) bisects ∠BAC, we get ∡BAD = ∡DAC. Togetherwith ➋, it implies that ∡CEA = ∡EAC. By Theorem 4.2, ACE isisosceles; that is,

EC = AC.

Together with ➌, it implies ➊.

7.8. Exercise. Prove an analog of Lemma 7.7 for the external bisec-tor.

Incenter

7.9. Theorem. The angle bisectors of any nondegenerate triangleintersect at one point.

The point of intersection of bisectors is called incenter ; it is usuallydenoted as I. The point I lies on the same distance from each side, itis the center of a circle tangent to each side of triangle. This circle iscalled incircle and its radius is called inradius of the triangle.

59

Proof. Let ABC be a nondegenerate triangle.Note that points B and C lie on the opposite sides from the bisector

of ∠BAC. Hence this bisector intersects [BC] at a point, say A′.Analogously, there is B′ ∈ [AC] such the (BB′) bisects ∠ABC.

A B

C

I

Z

A′

XY

B′

Applying Pasch’s theorem (3.10), twice forthe triangles AA′C and BB′C, we getthat [AA′] and [BB′] intersect. Let us denoteby I the point of intersection.

Let X , Y and Z be the foot points of I on(BC), (CA) and (AB) correspondingly. Ap-plying Lemma 5.13, we get

IY = IZ = IX.

From the same lemma we get that I lies on abisector or exterior bisector of ∠BCA.

The line (CI) intersects [BB′], the pointsB and B′ lie on opposite sides of (CI). There-fore the angles ∠ICB′ and ∠ICB have oppo-site signs. Note that ∠ICA = ∠ICB′. Therefore (CI) can not beexterior bisector of ∠BCA. Hence the result follows.

More exercises

7.10. Exercise. Assume that bisector at one vertex of a nondegener-ate triangle bisects the opposite side. Show that the triangle is isosce-les.

7.11. Exercise. Assume that at one vertex of a nondegenerate tri-angle bisector coincides with the altitude. Show that the triangle isisosceles.

A B

C

XY

Z

7.12. Exercise. Assume sides [BC], [CA] and[AB] of ABC are tangent to incircle at X, Yand Z correspondingly. Show that

AY = AZ = 12 ·(AB +AC −BC).

By the definition, the orthic triangle is formed by the base pointsof its altitudes of the given triangle.

7.13. Exercise. Prove that orthocenter of an acute triangle coincideswith incenter of its orthic triangle.

What should be an analog of this statement for an obtuse triangle?

Chapter 8

Inscribed angles

Angle between a tangent line and a chord

8.1. Theorem. Let Γ be a circle with center O in the Euclideanplane. Assume line (XQ) is tangent to Γ at X and [XY ] is a chordof Γ. Then

➊ 2·∡QXY ≡ ∡XOY.

Equivalently,

∡QXY ≡ 12 ·∡XOY or ∡QXY ≡ 1

2 ·∡XOY + π.

Q

X

Y

O

Proof. Note that XOY is isosceles.Therefore ∡Y XO = ∡OY X .

Let us applying Theorem 6.13 toXOY . We get

π ≡ ∡Y XO + ∡OY X + ∡XOY ≡≡ 2·∡Y XO + ∡XOY.

By Lemma 5.18, (OX) ⊥ (XQ).Therefore

∡QXY + ∡Y XO ≡ ±π2 .

Therefore

2·∡QXY ≡ π − 2·∡Y XO ≡ ∡XOY.

60

61

Inscribed angleP

X

Y

O

We say that triangle is inscribed in thecircle Γ if all its vertices lie on Γ.

8.2. Theorem. Let Γ be a circle withcenter O in the Euclidean plane, andX,Y be two distinct points on Γ. ThenXPY is inscribed in Γ if and only if

➋ 2·∡XPY ≡ ∡XOY.

Equivalently, if and only if

∡XPY ≡ 12 ·∡XOY or ∡XPY ≡ 1

2 ·∡XOY + π.

Proof. Choose a point Q such that (PQ) ⊥ (OP ). By Lemma 5.18,(PQ) is tangent to Γ.

According to Theorem 8.1,

2·∡QPX ≡ ∡POX,

2·∡QPY ≡ ∡POY.

Subtracting one identity from the other we get ➋.

X

Y

O

O′

P

Γ′Γ

Let us prove the converse. Assumethat ➋ holds for some P /∈ Γ. Note that∡XOY 6= 0 and therefore ∡XPY 6= 0nor π; that is, PXY is nondegenerate.

Let Γ′ be the circumcircle of PXYand O′ be its circumcenter; they exist byExercise 7.2.

Note that O′ 6= O. From above, wehave

∡XOY ≡ 2·∡XPY ≡ ∡XO′Y.

Note that OX = OY and O′X = O′Y . By Theorem 5.2, (OO′) isthe perpendicular bisector to [XY ]; equivalently X is the reflection ofY in (OO′). Applying Proposition 5.6, we get

∡XOO′ ≡ −∡Y OO′, ∡XO′O ≡ −∡Y O′O.

Therefore

2·∡XOO′ ≡ ∡XOO′ + ∡O′OY ≡≡ ∡XOY ≡ ∡XO′Y ≡≡ ∡XO′O + ∡OO′Y

≡ 2·∡XO′O.

62 CHAPTER 8. INSCRIBED ANGLES

By Transversal property 6.18, (XO) ‖ (XO′), a contradiction.

Y ′

Y

P

X

X ′

O

8.3. Exercise. Let [XX ′] and [Y Y ′]be two chords of circle Γ with centerO and radius r in the Euclidean plane.Assume (XX ′) and (Y Y ′) intersect atpoint P . Show that(a) 2·∡XPY = ∡XOY + ∡X ′OY ′;(b) PXY ∼ PY ′X ′;(c) PX ·PX ′ = |OP 2 − r2|.

8.4. Exercise. Assume that the chords[XX ′], [Y Y ′] and [ZZ ′] of the circle Γin the Euclidean plane intersect at onepoint. Show that

XY ′ ·ZX ′ ·Y Z ′ = X ′Y ·Z ′X ·Y ′Z.

C

A

A′

B

B′

8.5. Exercise. Let Γ be a circumcircle ofABC. A′ 6= A and B′ 6= B be the pointsof intersection of altitudes from A and B withΓ. Show that A′B′C is isosceles.

Recall that diameter of circle is its chordwhich pass through the center. Note that if[XY ] is diameter of circle with center O then∡XOY = π. Whence Theorem 8.2 implies thefollowing.

8.6. Corollary. Let Γ be a circle with diameter [XY ]. Assume thatpoint P is distinct from X and Y . Then P ∈ Γ if and only if ∠XPYis right.

8.7. Exercise. Given four points A, B, A′ and B′ construct a pointZ such that both angles ∠AZB and ∠A′ZB′ are right.

L

M

N

O

X8.8. Exercise. Assume three lines ℓ,m and n in-tersect at point O and form six equal angles at O.Let X be a point distinct from O, denote by L, Mand N be the footpoints of X on ℓ,m and n corre-spondingly. Show that LMN is equilateral.

8.9. Exercise. Let ABC be a nondegenerate triangle in the Eu-clidean plane, A′ and B′ be foot points of altitudes from A and B.Show that A, B, A′ and B′ lie on one circle.

63

What is the center of this circle?

8.10. Exercise. Assume a line ℓ and a circle with center on ℓ aregiven. Make a ruler-only construction of the perpendicular to ℓ fromthe given point.

Inscribed quadrilaterals

A quadrilateral ABCD is called inscribed if all the points A, B, Cand D lie on a circle or a line.

8.11. Theorem. A quadrilateral ABCD in the Euclidean plane isinscribed if and only if

➌ 2·∡ABC + 2·∡CDA ≡ 0.

Equivalently, if and only if

∡ABC + ∡CDA ≡ π or ∡ABC ≡ −∡CDA.

A

BC

D

Proof of Theorem 8.11. AssumeABC is degenerate. By Corol-lary 2.8,

2·∡ABC ≡ 0;

From the same corollary, we get

2·∡CDA ≡ 0

if and only if D ∈ (AB); hence theresult follows.

It remains to consider the case ifABC is nondegenerate.

Denote by Γ the circumcircle of ABC and let O be the center ofΓ. According to Theorem 8.2,

➍ 2·∡ABC ≡ ∡AOC.

From the same theorem, D ∈ Γ if and only if

➎ 2·∡CDA ≡ ∡COA.

Adding ➍ and ➎, we get the result.


A

B

Y

X

X ′

Y ′

Γ

Γ′8.12. Exercise. Let Γ and Γ′ be twocircles which intersect at two distinctpoints A and B. Assume [XY ] and[X ′Y ′] be the chords of Γ and Γ′ corre-spondingly such that A lies between Xand X ′ and B lies between Y and Y ′.Show that (XY ) ‖ (X ′Y ′).

8.13. Exercise. Let [XY ] and [X ′Y ′]be two parallel chords of a circle. Showthat XX ′ = Y Y ′.

Arcs

A subset of a circle bounded by two points is called a circle arc.More precisely, let Γ be a circle and A, B, C be distinct points on

Γ. The subset which includes the points A, C as well as all the pointson Γ which lie with B on the same side from (AC) is called circle arcABC.

A

B

C

X

Γ

For the circle arc ABC, the points Aand C are called endpoints. Note thatthere are two circle arcs of Γ with thegiven endpoints.

A half-line [AX) is called tangent toarc ABC at A if the line (AX) is tangentto Γ and the points X and B lie on thesame side from the line (AC).

If B lies on the line (AC), the arcABC degenerates to one of two following a subsets of line (AC).

⋄ If B lies between A and C then we define the arc ABC as thesegment [AC]. In this case the half-line [AC) is tangent to thearc ABC at A.

⋄ If B ∈ (AC)\[AC] then we define the arc ABC as the line (AC)without all the points between A and C. If we choose points Xand Y ∈ (AC) such that the points X , A, C and Y appear inthe same order on the line then the arc ABC is formed by twohalf-lines in [AX) and [CY ). The half-line [AX) is tangent tothe arc ABC at A.

⋄ In addition, any half-line [AB) will be regarded as an arc. Thisdegenerate arc has only one end point A and it assumed to betangent to itself at A.

The circle arcs together with the degenerate arcs will be calledarcs.

65

8.14. Proposition. In the Euclidean plane, a point D lies on thearc ABC if and only if

∡ADC = ∡ABC

or D coincides with A or C.

A

B

C D

Proof. Note that if A, B and C lie on one line thenthe statement is evident.

Assume Γ be the circle passing through A, B andC.

Assume D is distinct from A and C. Accordingto Theorem 8.11, D ∈ Γ if and only if

∡ADC = ∡ABC or ∡ADC ≡ ∡ABC + π.

By Exercise 3.12, the first identity holds then B and D lie on oneside of (AC); that is, D belongs to the arc ABC. If the second identityholds then the points B and D lie on the opposite sides from (AC), inthis case D does not belong to the arc ABC.

8.15. Proposition. In the Euclidean plane, a half-lines [AX) is tan-gent to the arc ABC if and only if

∡ABC + ∡CAX ≡ π.

Proof. For a degenerate arc ABC the statement is evident. Furtherwe assume the arc ABC is nondegenerate.

Applying theorems 8.1 and 8.2, we get

2·∡ABC + 2·∡CAX ≡ 0.

Therefore either

∡ABC + ∡CAX ≡ π or ∡ABC + ∡CAX ≡ 0.

A

B

C

X

Since [AX) is the tangent half-line to thearc ABC, X and B lie on the same side from(AC). Therefore the angles ∠CAX , ∠CAB and∠ABC have the same sign. In particular ∡ABC+∡CAX 6≡ 0; that is, we are left with the case

∡ABC + ∡CAX ≡ π.


8.16. Exercise. Assume that in the Euclidean plane, the half-lines[AX) and [AY ) are tangent to the arcs ABC and ACB correspond-ingly. Show that ∠XAY is straight.

8.17. Exercise. Show that in the Euclidean plane, there is uniquearc with endpoints at the given points A and C which is tangent at Ato the given half line [AX).

A

B1

B2

C

Y1

Y2

X1

X2

8.18. Exercise. Given two arcsAB1C and AB2C in the Euclideanplane, let [AX1) and [AX2) be thehalf-lines tangent to arcs AB1Cand AB2C at A and [CY1) and[CY2) be the half-lines tangent toarcs AB1C and AB2C at C. Showthat

∡X1AX2 ≡ −∡Y1CY2.

8.19. Exercise. Given an acutetriangle ABC make a compass-and-ruler construction of the pointZ such that

∡AZB ≡ ∡BZC ≡ ∡CZA ≡ ± 23 ·π

Chapter 9

Inversion

Let Ω be the circle with center O and radius r. The inversion of apoint P with respect to Ω is the point P ′ ∈ [OP ) such that

OP ·OP ′ = r2.

In this case the circle will be called the circle of inversion and itscenter is called center of inversion.

Ω

O P

P ′

TThe inversion of O is undefined.Note that if P is inside Ω then P ′ is out-

side and the other way around. Further,P = P ′ if and only if P ∈ Ω.

Note that the inversion takes P ′ back toP .

9.1. Exercise. Let P be a point inside of acircle Ω centered at O. Let T be a point wherethe perpendicular to (OP ) from P intersectsΩ. Let P ′ be the point where the tangent toΩ at T intersects (OP ).

Show that P ′ is the inversion of P in the circle Ω.

9.2. Lemma. Let Γ be a circle with center O in the Euclidean plane.Assume A′ and B′ are inversions of A and B in Γ. Then

OAB ∼ OB′A′.

Moreover,

➊

∡AOB ≡ −∡B′OA′,

∡OBA ≡ −∡OA′B′,

∡BAO ≡ −∡A′B′O.

67

68 CHAPTER 9. INVERSION

AA′

B

B′

O

Proof. Let r be the radius of the circle of theinversion.

From the definition of inversion, we get

OA·OA′ = OB ·OB′ = r2.

ThereforeOA

OB′ =OB

OA′ .

Clearly

➋ ∡AOB = ∡A′OB′ ≡ −∡B′OA′.

From SAS, we getOAB ∼ OB′A′.

Applying Theorem 3.11 and ➋, we get ➊.

9.3. Exercise. Let P ′ be the inversion of P in the circle Γ. Assumethat P 6= P ′. Show that the value PX

P ′Xis the same for all X ∈ Γ.

The converse to the above exercise also holds. Namely, given pos-itive real number k 6= 1 and two distinct points P and P ′ in theEuclidean plane the locus of points X such that PX

P ′X= k forms a

circle which is called circle of Apollonius. In this case P ′ is inverse ofP in the circle of Apollonius.

9.4. Exercise. Let A′, B′, C′ be the images of A, B, C under in-version in the incircle of ABC. Show that the incenter of ABC isthe orthocenter of A′B′C′.

9.5. Exercise. Make a ruler-and-compass construction of the inver-sion of given point in the given circle.

Cross-ratio

The following theorem gives some quantities expressed in distances orangles which do not change after inversion.

9.6. Theorem. Let ABCD and A′B′C′D′ be two quadrilateralsin the Euclidean plane such that the points A′, B′, C′ and D′ areinversions of A, B, C, and D correspondingly.

Then(a)

AB ·CD

BC ·DA=

A′B′ ·C′D′

B′C′ ·D′A′ .

69

(b)∡ABC + ∡CDA ≡ −(∡A′B′C′ + ∡C′D′A′).

(c) If the quadrilateral ABCD is inscribed then so is A′B′C′D′.

Proof; (a). Let O be the center of inversion. According to Lemma 9.2,AOB ∼ B′OA′. Therefore

AB

A′B′ =OA

OB′ .

Analogously,

BC

B′C′ =OC

OB′ ,CD

C′D′ =OC

OD′ ,DA

D′A′ =OA

OD′ .

Therefore

AB

A′B′ ·B′C′

BC· CD

C′D′ ·D′A′

DA=

OA

OB′ ·OB′

OC· OC

OD′ ·OD′

OA= 1.

Hence (a) follows.

(b). According to Lemma 9.2,

∡ABO ≡ −∡B′A′O, ∡OBC ≡ −∡OA′B′,

∡CDO ≡ −∡D′C′O, ∡ODA ≡ −∡OA′D′.

Summing these four identities we get

∡ABC + ∡CDA ≡ −(∡D′C′B′ + ∡B′A′D′).

Applying Axiom IIIb and Exercise 6.21, we get

∡A′B′C′ + ∡C′D′A′ ≡ −(∡B′C′D′ + ∡D′A′B′) ≡≡ ∡D′C′B′ + ∡B′A′D′.

Hence (b) follows.

(c). Follows from (b) and Theorem 8.11.

Inversive plane and circlines

Let Ω be a circle with center O and radius r. Consider the inversionin Ω.

Recall that inversion of O is undefined. To deal with this problemit is useful to add to the plane an extra point; it will be called the


point at infinity and we will denote it as ∞. We can assume that ∞is inversion of O and the other way around.

The Euclidean plane with added a point at infinity is called inver-sive plane.

We will always assume that any line and half-line contains ∞.It will be convenient to use notion of circline, which means circle

or line; for instance we may say if circline contains ∞ then it is a lineor circline which does not contain ∞ is a circle.

Note that according to Theorem 7.1, for anyABC there is uniquecircline which pass through A, B and C.

9.7. Theorem. In the inversive plane, inversion of a circline is acircline.

Proof. Denote by O the center of inverse.Let Γ be a circline. Choose three distinct points A, B and C on

Γ. (If ABC is nondegenerate then Γ is the circumcircle of ABC;if ABC is degenerate then Γ is the line passing through A, B andC.)

Denote by A′, B′ and C′ the inversions of A, B and C corre-spondingly. Let Γ′ be the circline which pass though A′, B′ and C′.According to 7.1, Γ′ is well defined.

Assume D is a point of inversive plane which is distinct from A,C, O and ∞. Denote by D′ the inversion of D.

By Theorem 9.6(c), D′ ∈ Γ′ if and only if D ∈ Γ. Hence the resultfollows.

It remains to prove that O ∈ Γ ⇔ ∞ ∈ Γ′ and ∞ ∈ Γ ⇔ O ∈ Γ′.Since Γ is inversion of Γ′ it is sufficient to prove only

∞ ∈ Γ ⇐⇒ O ∈ Γ′.

Since ∞ ∈ Γ we get that Γ is a line. Therefore for any ε > 0, theline Γ contains point P with OP > r2/ε. For the inversion P ′ ∈ Γ′ ofP , we have OP ′ = r2/OP < ε. That is, the circline Γ′ contains pointsarbitrary close to O. It follows that O ∈ Γ′.

Q′ Q

9.8. Exercise. Assume that if circle Γ′ is the in-version of circle Γ in the Euclidean plane. Denoteby Q the center of Γ and by Q′ the inversion of Q.

Show that Q′ is not the center of Γ′.

Assume circumtool is a geometric construction tool which producea circline passing through the given three points.

9.9. Exercise. Show that with circumtool only, it is impossible toconstruct the center of given circle.

71

9.10. Exercise. Show that for any pair of tangent circles in the inver-sive plane there is an inversion which sends them to a pair of parallellines.

9.11. Theorem. Consider inversion with respect to circle Ω withcenter O in the inversive plane. Then(a) Line passing through O is inverted into itself.(b) Line not passing through O is inverted into a circle which pass

through O, and the other way around.(c) A circle not passing through O is inverted into a circle not pass-

ing through O.

Proof. In the proof we use Theorem 9.7 without mentioning.

(a). Note that if line passing through O it contains both ∞ and O.Therefore its inversion also contains ∞ and O. In particular image isa line passing through O.

(b). Since any line ℓ pass through ∞, its image ℓ′ has to contain O.If the line did not contain O then ℓ′ 6∋ ∞; that is ℓ′ is not a line.Therefore ℓ′ is a circle which pass through O.

(c). If circle Γ does not contain O then its image Γ′ does not contain∞. Therefore Γ′ is a circle. Since Γ 6∋ ∞ we get Γ′ 6∋ O. Hence theresult follows.

Ptolemy’s identity

Here is one application of inversion, which we include as an illustrationonly; we will not use it further in the book.

9.12. Theorem. Let ABCD be an inscribed quadrilateral in theEuclidean plane. Assume that the points A, B, C and D appear onthe circline in the same order. Then

AB ·CD +BC ·DA = AC ·BD.

Proof. Assume the points A,B,C,D lie on one line in this order.

A B C Dx y z

Set x = AB, y = BC, z = CD. Notethat

x·z + y ·(x+ y + z) = (x + y)·(y + z).

Since AC = x + y, BD = y + z and DA = x + y + z, it proves theidentity.


A

B CD

A′ B′ C′ D′ It remains to consider the case whenquadrilateral ABCD is inscribed in a cir-cle, say Γ.

The identity can be rewritten as

AB ·DC

BD·CA+

BC ·ADCA·DB

= 1.

On the left hand side we have two cross-ratios. According to Theorem 9.6(a), theleft hand side does not change if we apply

an inversion to each point.Consider an inversion in a circle centered at a point O which lie

on Γ between A and D. By Theorem 9.11, this inversion maps Γ toa line. This reduces the problem to the case when A, B, C and D lieon one line, which was already considered.

A

B

In the proof above we noted that werewrite Ptolemy identity in a form whichis invariant with respect to inversion andthen apply an inversion which makes it evi-dent. The solution of the following exerciseis based on the same idea; one has to applyan inversion with center at A.

9.13. Exercise. Assume that three circlestangent to each other and to two parallel lines as shown on the picture.Show that the line passing through the point of tangency A and B onthe diagram is also tangent to the two circles at A.

Perpendicular circles

Assume two circles Γ and Ω intersect at two points say X and Y .Let ℓ and m be the tangent lines at X to Γ and Ω correspondingly.Analogously, ℓ′ and m′ be the tangent lines at Y to Γ and Ω.

From Exercise 8.18, we get that ℓ ⊥ m if and only if ℓ′ ⊥ m′.We say that circle Γ is perpendicular to circle Ω (briefly Γ ⊥ Ω)

if they intersect and the lines tangent to the circle at one point (andtherefore both points) of intersection are perpendicular.

Similarly, we say that circle Γ is perpendicular to a line ℓ (brieflyΓ ⊥ ℓ) if Γ ∩ ℓ 6= ∅ and ℓ perpendicular to the tangent lines to Γ atone point (and therefore both points) of intersection. According toLemma 5.18, it happens only if the line ℓ pass through the center ofΓ.

Now we can talk about perpendicular circlines.

73

9.14. Theorem. Assume Γ and Ω are distinct circles in the Eu-clidean plane. Then Ω ⊥ Γ if and only if the circle Γ coincides withits inversion in Ω.

X

YQ

O

Proof. Denote by Γ′ the inversion of Γ.

(⇒) Let O be the center of Ω and Q be the centerof Γ. Denote by X and Y the points of intersec-tions of Γ and Ω. According to Lemma 5.18,Γ ⊥ Ω if and only if (OX) and (OY ) are tangentto Γ.

Note that Γ′ is also tangent to (OX) and(OY ) at X and Y correspondingly. It follows that X and Y are thefoot points of the center of Γ′ on (OX) and (OY ). Therefore bothΓ′ and Γ have the center Q. Finally, Γ′ = Γ, since both circles passthrough X .

(⇐) Assume Γ = Γ′.Since Γ 6= Ω, there is a point P which lies on Γ, but not on Ω.

Let P ′ be the inversion of P in Ω. Since Γ = Γ′, we have P ′ ∈ Γ, inparticular the half-line [OP ) intersects Γ at two points. By Exercise5.14, O lies outside of Γ.

As Γ has points inside and outside Ω, the circles Γ and Ω intersect.The later follows from Exercise 5.20(b).

Let X be a point of their intersection. We need to show (OX) istangent to Γ; that is X is the only intersection point of (OX) and Γ.

Assume Z is an other point of intersection. Since O is outside ofΓ, the point Z lies on the half-line [OX).

Denote by Z ′ the inversion of Z in Ω. Clearly the three pointsZ,Z ′, X lie on Γ and (OX). The later contradicts Lemma 5.16.

It is convenient to define inversion in the line as the reflectionthrough this line. This way we can talk about inversion in arbitrarycircline.

9.15. Corollary. Let Ω and Γ be distinct circlines in the inversiveplane. Then the inversion in Ω sends Γ to itself if and only if Ω ⊥ Γ.

Proof. By Theorem 9.14, it is sufficient to consider the case when Ωor Γ is a line.

Assume Ω is a line, so the inversion in Ω is reflection. In this casethe statement follows from Corollary 5.7.

If Γ is a line then the statement follows from Theorem 9.11.

9.16. Corollary. Let P and P ′ be two distinct points in the Euclideanplane such that P ′ is the inversion of P in the circle Ω. Assume thata circline Γ pass through P and P ′. Then Γ ⊥ Ω.


Proof. Without loss of generality we may assume that P is inside andP ′ is outside Ω. By Theorem 3.16, Γ intersects Ω. Denote by A apoint of intersection.

Denote by Γ′ the inversion of Γ. Since A is inversion of itself, thepoints A, P and P ′ lie on Γ′. By Exercise 7.2, Γ′ = Γ and by Theorem9.14, Γ ⊥ Ω.

9.17. Corollary. Let P and Q be two distinct points inside the circleΩ in the Euclidean plane. Then there is unique circline Γ perpendicularto Ω which pass through P and Q.

Proof. Let P ′ be the inversion of point P in a circle Ω. According toCorollary 9.16, the circline passing through P and Q is perpendicularto Ω if and only if it pass though P ′.

Note that P ′ lies outside of Ω. Therefore the points P , P ′ and Qare distinct.

According to Exercise 7.2, there is unique circline passing throughP , Q and P ′. Hence the result follows.

9.18. Exercise. Let Ω1 and Ω2 be two distinct circles in the Eu-clidean plane. Assume that the point P does not lie on Ω1 nor onΩ2. Show that there is unique circline passing through P which isperpendicular Ω1 and Ω2.

9.19. Exercise. Let P , Q, P ′ and Q′ be points in the Euclideanplane. Assume P ′ and Q′ are inversions of P and Q correspondingly.Show that the quadrilateral PQP ′Q′ is inscribed.

9.20. Exercise. Let Ω1 and Ω2 be two perpendicular circles withcenters at O1 and O2 correspondingly. Show that the inversion of O1

in Ω2 coincides with the inversion of O2 in Ω1.

9.21. Exercise. Three distinct circles Ω1, Ω2 and Ω3 intersect attwo points A and B. Assume that a circle Γ ⊥ Ω1 and Γ ⊥ Ω2, Showthat Γ ⊥ Ω3.

9.22. Exercise. Assume you have two tools, first produce a circlinewhich pass through given three points and the second produces inversionof given point in the given circle.

Assume that a point P does not lie on two circles Ω1, Ω2. Usingonly the two tools above, construct a circline Γ which pass through Pand perpendicular to both Ω1 and Ω2.

75

Angles after inversion

9.23. Proposition. In the inversive plane, the inversion of an arcis an arc.

Proof. Consider four distinct points A, B, C and D; let A′, B′, C′ andD′ be their inverses. We need to show that D lies on the arc ABC ifand only if D′ lies on the arc A′B′C′. According to Proposition 8.14,the later is equivalent to the following

∡ADC = ∡ABC ⇐⇒ ∡A′D′C′ = ∡A′B′C′

and it follows from Theorem 9.6(b).

The following theorem roughly says that the angle between arcschanges sign after the inversion.

A

A′

B1

B′1

C1

C′1

X1

Y1

B2C2

B′2

C′2 X2

Y2

9.24. Theorem. Let AB1C1, AB2C2 be two arcs in the inversiveplane and A′B′

1C′1, A

′B′2C

′2 be their inversions. Let [AX1) and [AX2)

be the half-lines tangent to AB1C1 and AB2C2 at A and [A′Y1) and[A′Y2) be the half-lines tangent to A′B′

1C′1 and A′B′

2C′2 at A′. Then

∡X1AX2 ≡ −∡Y1A′Y2.

Proof. Applying to Proposition 8.15,

∡X1AX2 ≡ ∡X1AC1 + ∡C1AC2 + ∡C2AX2 ≡≡ (π − ∡C1B1A) + ∡C1AC2 + (π − ∡AB2C2) ≡≡ −(∡C1B1A+ ∡AB2C2 + ∡C2AC1) ≡≡ −(∡C1B1A+ ∡AB2C1)− (∡C1B2C2 + ∡C2AC1).


The same way we get

∡Y1A′Y2 ≡ −(∡C′

1B′1A

′ + ∡A′B′2C

′1)− (∡C′

1B′2C

′2 + ∡C′

2A′C′

1).

By Theorem 9.6(b),

∡C1B1A+ ∡AB2C1 ≡ −(∡C′1B

′1A

′ + ∡A′B′2C

′1),

∡C1B2C2 + ∡C2AC1 ≡ −(∡C′1B

′2C

′2 + ∡C′

2A′C′

1).


9.25. Corollary. Let P ′, Q′ and Γ′ be the inversions of the pointsP , Q and the circle Γ in the circle Ω of the Euclidean plane. AssumeP is inversion of Q in Γ then P ′ is inversion of Q′ in Γ′.

Proof. If P = Q then P ′ = Q′ ∈ Γ′ therefore P ′ is inversion of Q′ inΓ′.

It remains to consider the case P 6= Q. Let ∆1 and ∆2 be two dis-tinct circles which intersect at P and Q. According to Corollary 9.16,∆1 ⊥ Γ and ∆2 ⊥ Γ.

Denote by ∆′1 and ∆′

2 the inversions of ∆1 and ∆2 in Ω. Clearly∆′

1 and ∆′2 intersect at P ′ and Q′.

From Theorem 9.24, the later is equivalent to ∆′1 ⊥ Γ′ and ∆′

2 ⊥⊥ Γ′. By Corollary 9.15 the later implies P ′ is inversion of Q′ inΓ′.

Chapter 10

Absolute plane

Let us remove Axiom V from our axiomatic system, see pages 20–21.This way we define a new object called absolute plane or neutral plane.(In the absolute plane, the Axiom V may or may not hold.)

Clearly any theorem in absolute geometry holds in Euclidean ge-ometry. In other words, Euclidean plane is an example of absoluteplane. In the next chapter we will show that besides the Euclideanplane there are other examples of absolute plane.

Many theorems in Euclidean geometry hold in absolute geometry.In this book, the Axiom V was used for the first time in the proof of

uniqueness of parallel line in Theorem 6.2. Therefore all the statementsbefore Theorem 6.2 also hold in absolute geometry.

It makes all the discussed results about half-planes, signs of angles,congruence conditions, perpendicular lines and reflections true in ab-solute plane. If in the formulation of a statement above you do notsee words “Euclidean plane” or “inversive plane”, it means that thestatement holds in absolute plane and the same proof works.

Let us give an example of theorem in absolute geometry, whichadmits a simpler proof in Euclidean geometry.

10.1. Theorem. Assume that triangles ABC and A′B′C′ haveright angles at C and C′ correspondingly, AB = A′B′ and AC = A′C′.Then ABC ∼= A′B′C′.

Euclidean proof. By Pythagorean theorem BC = B′C′. Then thestatement follows from SSS congruence condition.

Note that the proof of Pythagorean theorem used properties ofsimilar triangles, which in turn used Axiom V. Whence the aboveproof is not working in absolute plane.

77

78 CHAPTER 10. ABSOLUTE PLANE

A

B

CD

Absolute proof. Denote by D the reflection ofA through (BC) and by D′ the reflection of A′

through (B′C′). Note that

AD = 2·AC = 2·A′C′ = A′D′,

BD = BA = B′A′ = B′D′.

By SSS, we get ABD ∼= A′B′D′.The theorem follows since C is the midpoint of [AD] and C′ is the

midpoint of [A′D′].

10.2. Exercise. Give a proof of Exercise 7.10 which works in theabsolute plane.

10.3. Exercise. Let ABCD be an inscribed quadrilateral in theabsolute plane. Show that

∡ABC + ∡CDA ≡ ∡BCD + ∡DAB.

Note that the Theorem 8.11 can not be applied in the above exer-cise; it use Theorems 8.1 and 8.2; which in turns use Theorem 6.13.

Two angles of triangle

In this section we will prove a weaker form of Theorem 6.13 whichholds in absolute plane.

10.4. Proposition. Let ABC be nondegenerate triangle in the ab-solute plane. Then

|∡CAB| + |∡ABC| < π.

Note according to 3.11, the angles ∠ABC, ∠BCA and ∠CABhave the same sign. Therefore in Euclidean plane the theorem followsimmediately from Theorem 6.13. In absolute geometry we need towork more.

Proof. By 3.11, we may assume that ∠CAB and ∠ABC are positive.Let M be the midpoint of [AB]. Chose C′ ∈ (CM) distinct from

C so that C′M = CM .Note that the angles ∠AMC and∠BMC′ are vertical; in particular

∡AMC = ∡BMC′.

By constructionAM = BM and CM = C′M . ThereforeAMC ∼=BMC′; in particular

∡CAB = ±∡C′BA.

79

B

A

C

C′

M

According to 3.11, the angles ∠CABand ∠C′BA have the same sign as ∠AMCand ∠BMC′. Therefore

∡CAB = ∡C′BA.

In particular,

∡C′BC ≡ ∡C′BA+ ∡ABC ≡≡ ∡CAB + ∡ABC.

Finally note that C′ and A lie on the same side from (CB). There-fore the angles ∠CAB, ∠ABC and ∠C′BC are positive. By Exer-cise 3.3, the result follows.

10.5. Exercise. Assume A, B, C and D be points in absolute planesuch that

2·∡ABC + 2·∡BCD ≡ 0.

Show that (AB) ‖ (CD).

Note that one can not extract the solution of the above exercisefrom the proof of Transversal property (6.18)

10.6. Exercise. Prove side-angle-angle congruence condition in ab-solute plane.

In other words, let ABC and A′B′C′ be two triangles in abso-lute plane. Show that ABC ∼= A′B′C′ if

AB = A′B′, ∡ABC = ±∡A′B′C′ and ∡BCA = ±∡B′C′A′.

Note that in the Euclidean plane, the above exercise follows fromASA and the theorem on sum of angles of triangle (6.13). However,Theorem 6.13 can not be used here since its proof use Axiom V. Later,in theorem Theorem 12.5, we will show that Theorem 6.13 does nothold in absolute plane.

10.7. Exercise. Assume that point D lies between the vertices A andB of triangle ABC in the absolute plane. Show that

CD < CA or CD < CB.

Three angles of triangle

10.8. Proposition. Let ABC and A′B′C′ be two triangles inthe absolute plane such that AC = A′C′ and BC = B′C′. Then

AB < A′B′ if and only if |∡ACB| < |∡A′C′B′|.


AC

BB′ X

Proof. Without loss of generality, wemay assume that A = A′ and C = C′

and ∡ACB,∡ACB′ > 0. In this case weneed to show that

AB < AB′ ⇐⇒ ∡ACB < ∡ACB′.

Choose a point X so that

∡ACX = 12 ·(∡ACB + ∡ACB′).

Note that⋄ (CX) bisects ∠BCB′

⋄ (CX) is the perpendicular bisector of [BB′].⋄ A and B lie on the same side from (CX) if and only if

∡ACB < ∡ACB′.

From Exercise 5.3, A and B lie on the same side from (CX) if andonly if AB < AB′. Hence the result follows.

10.9. Theorem. Let ABC be a triangle in the absolute plane.Then

|∡ABC| + |∡BCA|+ |∡CAB| 6 π.

The following proof is due to Legendre [7], earlier proofs were dueto Saccheri [12] and Lambert [6].

Proof. Let ABC be the given triangle. Set

a = BC, b = CA, c = AB,

α = ∡CAB, β = ∡ABC, γ = ∡BCA.

Without loss of generality, we may assume that α, β, γ > 0.

A0 A1 A2. . . An

C1 C2 . . . Cn

c c c c

a b

d

a b

d

a b

d

a b

αβ

γ

δαβ

γ

Fix a positive integer n. Consider points A0, A1, . . . , An on thehalf-line [BA) so that BAi = i·c for each i. (In particular, A0 = Band A1 = A.) Let us construct the points C1, C2, . . . , Cn, so that∡AiAi−1Ci = β and Ai−1Ci = a for each i.

81

This way we construct n congruent triangles

ABC = A1A0C1∼=

∼= A2A1C2∼=

. . .∼= AnAn−1Cn.

Set d = C1C2 and δ = ∡C2A1C1. Note that

➊ α+ β + δ = π.

By Proposition 10.4, δ > 0.By construction

A1C1C2∼= A2C2C3

∼= . . . ∼= An−1Cn−1Cn.

In particular, CiCi+1 = d for each i.

By repeated application of the triangle inequality, we get that

n·c = A0An 6

6 A0C1 + C1C2 + · · ·+ Cn−1Cn + CnAn =

= a+ (n− 1)·d+ b.

In particular,

c 6 d+ 1n·(a+ b− d).

Since n is arbitrary positive integer, the later implies

c 6 d.

From Proposition 10.8 and SAS, the later is equivalent to

γ 6 δ.

From ➊, the theorem follows.

The defect of triangle ABC is defined as

defect(ABC)def== π − |∡ABC| + |∡BCA|+ |∡CAB|.

Note that Theorem 10.9 sates that, defect of any triangle in ab-solute plane has to be nonnegative. According to Theorem 6.13, anytriangle in Euclidean plane has zero defect.


A

C

BD

10.10. Exercise. Let ABC be nondegenerate trian-gle in the absolute plane. Assume D lies between A andB. Show that

defect(ABC) = defect(ADC) + defect(DBC).

How to prove that something

can not be proved?

Many attempts were made to prove that any theorem in Euclideangeometry holds in absolute geometry. The later is equivalent to thestatement that Axiom V is a theorem in absolute geometry.

Many these attempts being accepted as proofs for long periods oftime until the mistake was found.

There is a number of statements in the geometry of absolute planewhich are equivalent to the Axiom V. It means that if we exchange theAxiom V to any of these statements then we will obtain an equivalentaxiomatic system.

Here we give a short list of such statements. We are not going toprove the equivalence in the book.

10.11. Theorem. An absolute plane is Euclidean if and only if oneof the following equivalent conditions hold.(a) There is a line ℓ and a point P /∈ ℓ such that there is only one

line passing through P and parallel to ℓ.(b) Every nondegenerate triangle can be circumscribed.(c) There exists a pair of distinct lines which lie on a bounded dis-

tance from each other.(d) There is a triangle with arbitrary large inradius.

(e) There is a nondegenerate triangle with zero defect.(f) There exists a quadrilateral in which all angles are right.

It is hard to imagine an absolute plane, which does not satisfy someof the properties above. That is partly the reason why for the largenumber of false proofs; each used one of such statements by accident.

Let us formulate the negation of (a) above.

h-V. For any line ℓ and any point P /∈ ℓ there are at least two lineswhich pass through P and parallel to ℓ.

By Theorem 6.2, an absolute plane which satisfies Axiom h-V isnot Euclidean. Moreover, according to the Theorem 10.11 (which wedo not prove) any non-Euclidean absolute plane Axiom h-V holds.

83

It opens a way to look for a proof by contradiction. Simply ex-change Axiom V to Axiom h-V and start to prove theorems in theobtained axiomatic system. In the case if we arrive to a contradiction,we prove the Axiom V in absolute plane.

This idea was growing since 5th century; the most notable resultwere obtained by Saccheri in [12]. The more this new geometry wasdeveloped, it became more and more believable that there will be nocontradiction; that is, the system of axioms I–IV and h-V is consistent.

This new type of geometry is now called hyperbolic or Lobachevskiangeometry. In fact the following theorem holds.

10.12. Theorem. The hyperbolic geometry is consistent if and onlyif so is the Euclidean geometry.

The statement that hyperbolic geometry has no contradiction ap-pears first in private letters of Bolyai, Gauss, Schweikart and Tauri-nus1. They all seem to be afraid to state it in public. Say, in 1818Gauss writes to Gerling

. . . I am happy that you have the courage to expressyourself as if you recognized the possibility that our paral-lels theory along with our entire geometry could be false.But the wasps whose nest you disturb will fly around yourhead.. . .

Lobachevsky came to the same conclusion independently. Unlikethe others he had courage to state it in public and in print (see [8]).That cost him serious troubles.

It seems that Lobachevsky was also the first who had a proof ofTheorem 10.12. Later Beltrami gave a cleaner proof of “if” part ofthe theorem. It was done by modeling points, lines, distances andangle measures of one geometry using some other objects in the othergeometry. The same idea was used originally by Lobachevsky in theproof of the “only if” part of the theorem, see [9, 34].

The proof of Beltrami is the subject of the next chapter.

Arguably, the discovery of hyperbolic geometry was the secondmain discoveries of 19th century, trailing only the Mendel’s laws.

Curvature

In a letter from 1824 Gauss writes:The assumption that the sum of the three angles is less

than π leads to a curious geometry, quite different from

1The oldest surviving letters were the Gauss letter to Gerling 1816 and yet moreconvincing letter dated by 1818 of Schweikart sent to Gauss via Gerling.


ours but thoroughly consistent, which I have developed tomy entire satisfaction, so that I can solve every problemin it with the exception of a determination of a constant,which cannot be designated a priori. The greater one takesthis constant, the nearer one comes to Euclidean geometry,and when it is chosen indefinitely large the two coincide.The theorems of this geometry appear to be paradoxical and,to the uninitiated, absurd; but calm, steady reflection re-veals that they contain nothing at all impossible. For ex-ample, the three angles of a triangle become as small asone wishes, if only the sides are taken large enough; yetthe area of the triangle can never exceed a definite limit,regardless how great the sides are taken, nor indeed can itever reach it.

In the modern terminology the constant which Gauss mentions,can be expressed as 1/

√−k, where k denotes so called curvature of

the absolute plane which we are about to introduce.The identity in the Exercise 10.10 suggests that defect of triangle

should be proportional to its area.2

In fact for any absolute plane there is a nonpositive real number ksuch that

k· area(ABC) + defect(ABC) = 0

for any triangle ABC. This number k is called curvature of theplane.

For example, by Theorem 6.13, the Euclidean plane has zero curva-ture. By Theorem 10.9, curvature of any absolute plane is nonpositive.

It turns out that up to isometry, the absolute plane is characterizedby its curvature; that is, two absolute planes are isometric if and onlyif they have the same curvature.

In the next chapter we will construct hyperbolic plane, this is anexample of absolute plane with curvature k = −1.

Any absolute planes, distinct from Euclidean, can be obtained byrescaling metric on the hyperbolic plane. Indeed, if we rescale themetric by a positive factor c, the area changes by factor c2, whiledefect stays the same. Therefore taking c =

√−k, we can get the

absolute plane given curvature k < 0. In other words, all the non-Euclidean absolute planes become identical if we use r = 1/

√−k as

the unit of length.

In the Chapter 15, we briefly discuss the geometry of the unitsphere. Although spheres are not absolute planes, the spherical geom-

2The area in absolute plane discussed briefly in the end of Chapter 19, butinstead the reader could also refer to intuitive understanding of area.

85

etry is a close relative of Euclidean and hyperbolic geometries.The nondegenerate spherical triangles have negative defect. More-

over if R is the radius of the sphere then

1R2 · area(ABC) + defect(ABC) = 0

for any spherical triangle ABC. In other words, the sphere of radiusR has positive curvature k = 1

R2 .

Chapter 11

Hyperbolic plane

In this chapter we use inversive geometry to construct the modelof hyperbolic plane — an example of absolute plane which is not Eu-clidean.

Namely, we construct so called conformal disk model of hyperbolicplane. This model was discovered by Beltrami in [2] and often calledPoincare disc model.

86

87

The figure above shows the conformal disk model of hyperbolicplane which is cut into congruent triangles with angles π

3 ,π3 and π

4 .

Conformal disk model

In this section we give new names for some objects in the Euclideanplane which will represent lines, angle measures, distances in the hy-perbolic plane.

Hyperbolic plane. Let us fix a circle on the Euclidean plane andcall it absolute. The set of points inside the absolute will be calledhyperbolic plane (or h-plane).

Note that the points on absolute do not belong to the h-plane. Thepoints in the h-plane will be also called h-points.

Often we will assume that the absolute is a unit circle.

Hyperbolic lines. The intersections of h-plane with circlines per-pendicular to the absolute are called hyperbolic lines or h-lines.

P

Q

A

B

Γh-plane

By Corollary 9.17, thereis unique h-line which passthrough given two distinct h-points P and Q. This h-linewill be denoted as (PQ)h.

The arcs of hyperboliclines will be called hyperbolicsegments or h-segments. Anh-segment with endpoints Pand Q will be denoted as[PQ]h.

The subset of h-line onone side from a point will be called hyperbolic half-line (or h-half-line).More precisely, h-half-line is an intersection of h-plane with arc whichperpendicular to the absolute with only one endpoint in the h-plane.An h-half-line from P passing through Q will be denoted as [PQ)h.

If Γ is the circle containing the h-line (PQ)h then the points ofintersection of Γ with absolute are called ideal points of (PQ)h. (Notethat the ideal points of h-line do not belong to the h-line.)

Similarly a triangle in h-plane; that is, an ordered triple of h-points,say (P,Q,R), will be denoted by PQR.

So far (PQ)h is just a subset of h-plane; below we will introduce h-distance and later we will show that (PQ)h is a line for the h-distancein the sense of the Definition 1.8.

11.1. Exercise. Show that the h-line is uniquely determined by itsideal points.

88 CHAPTER 11. HYPERBOLIC PLANE

11.2. Exercise. Show that the h-line is uniquely determined by oneof its ideal points and one h-point on it.

11.3. Exercise. Show that h-segment [PQ]h coincides with the Eu-clidean segment [PQ] if and only if the line (PQ) pass through thecenter of the absolute.

Hyperbolic distance. Let P and Q be distinct h-points. Denote byA and B be the ideal points of (PQ)h. Without loss of generality, wemay assume that on the Euclidean circle containing the h-line (PQ)h,the points A,P,Q,B appear in the same order.

Consider function

δ(P,Q)def==

AQ·BP

QB ·PA.

Note that right hand side is the cross-ratio, which appeared in Theo-rem 9.6. Set δ(P, P ) = 1 for any h-point P . Set

PQhdef== ln[δ(P,Q)].

The proof that PQh is a metric on h-plane will be given below.For now it is just a function which returns a real value PQh for anypair of h-points P and Q.

11.4. Exercise. Let O be the center of the absolute and the h-pointsO, X and Y lie on the one h-line in the same order. Assume OX == XY . Prove that OXh < XYh.

Hyperbolic angles. Consider three h-points P , Q and R such thatP 6= Q and R 6= Q. The hyperbolic angle ∠hPQR is ordered pair ofh-half-lines [QP )h and [QR)h.

Let [QX) and [QY ) be (Euclidean) half-lines which are tangent to[QP ]h and [QR]h at Q. Then the hyperbolic angle measure (or h-anglemeasure) ∡hPQR is defined as ∡XQY .

11.5. Exercise. Let ℓ be an h-line and P be an h-point which doesnot lie on ℓ. Show that there is unique h-line passing through P andperpendicular to ℓ.

The plan

We defined all the h-notions needed in the formulation of the axiomsI–IV and h-V. It remains to show that all these axioms hold; this willbe done by the end of this chapter.

89

Once we are done with the proofs, we get that the model providesan example of absolute plane; in particular Exercise 11.5 can be provedthe same way as Theorem 5.5.

Most importantly we will prove the “if”-part of Theorem 10.12.Indeed, any statement in hyperbolic geometry can be restated in

the Euclidean plane using the introduced h-notions. Therefore if thesystem of axioms I–IV and h-V leads to a contradiction the so doesthe system axioms I–V.

Auxiliary statements

11.6. Lemma. Consider h-plane with unit circle as the absolute. LetO be the center of absolute and P be an other h-point. Denote by P ′

the inversion of P in the absolute.Then the circle Γ with center P ′ and radius 1/

√1−OP 2 is per-

pendicular to the absolute. Moreover O is the inversion of P in Γ.

Γ

O P P ′

TProof. Follows from Exercise 9.20.

Assume Γ is a circline which is per-pendicular to the absolute. Considerthe inversion X 7→ X ′ in Γ, or if Γ isa line, set X 7→ X ′ to be the reflectionin Γ.

The following observation roughlysays that the map X 7→ X ′ respects allthe notions introduced in the previous section. Together with thelemma above, it implies that in any problem which formulated en-tirely in h-terms we can assume that a given h-point lies in the centerof absolute.

11.7. Main observation. The map X 7→ X ′ described above is abijection of h-plane to itself. Moreover for any h-points P , Q, R suchthat P 6= Q and Q 6= R the following conditions hold(a) The h-lines (PQ)h, [PQ)h and [PQ]h are mapped to the h-lines

(P ′Q′)h, [P ′Q′)h and [P ′Q′]h correspondingly.(b) δ(P ′, Q′) = δ(P,Q) and P ′Q′

h = PQh.(c) ∡hP

′Q′R′ ≡ −∡hPQR.

Proof. According to Theorem 9.14 the map sends the absolute to itself.Note that the points on Γ do not move, it follows that points inside ofabsolute remain inside after the mapping and the other way around.

Part (a) follows from 9.7 and 9.24.


Part (b) follows from Theorem 9.6.Part (c) follows from Theorem 9.24.

11.8. Exercise. Let Γ be a circle which is perpendicular to the abso-lute and Q be an h-point lying on Γ. Assume P is an h-point and P ′

is its inversion in Γ. Show that PQh = P ′Qh.

11.9. Exercise. Consider the function

ϑ(P,Q) = 2·PQ·P ′Q′

PP ′ ·QQ′ ,

where P,Q are points in the h-plane and P ′, Q′ are their inversions inthe absolute. Show that

2 + 2·ϑ(P,Q) = δ(P,Q) +1

δ(P,Q).

Conclude that

PQh = ln[

1 + ϑ(P,Q) +√

ϑ(P,Q)2 + 2·ϑ(P,Q)]

.

11.10. Lemma. Assume that the absolute is a unit circle centered atO. Given a h-point P , set x = OP and y = OPh. Then

y = ln1 + x

1− xand x =

ey − 1

ey + 1.

A O P B

Proof. Note that h-line (OP )h lies in adiameter of absolute. Therefore if A andB are the ideal points as in the definitionof h-distance. Therefore

OA = OB = 1,

PA = 1 + x,

PB = 1− x.

In particular,

y = lnAP ·BO

PB ·OA= ln

1 + x

1− x.

Taking exponent of left and right hand side and applying obviousalgebra manipulations we get

x =ey − 1

ey + 1.

91

11.11. Lemma. Assume points P , Q and R appear on one h-line inthe same order. Then

PQh +QRh = PRh.

Proof. Note thatPQh +QRh = PRh

is equivalent to

➊ δ(P,Q)·δ(Q,R) = δ(P,R).

Let A and B be the ideal points of (PQ)h. Without loss of gener-ality we can assume that the points A,P,Q,R,B appear in the sameorder on the circline containing (PQ)h. Then

δ(P,Q)·δ(Q,R) =AQ·BP

QB ·PA·AR·BQ

RB ·QA=

=AR·BP

RB ·PA=

= δ(P,R)

Hence ➊ follows.

Let P be an h-point and ρ > 0. The set of all h-points Q such thatPQh = ρ is called h-circle with center P and h-radius ρ.

11.12. Lemma. Any h-circle is formed by a Euclidean circle whichlies completely in h-plane.

More precisely for any h-point P and ρ > 0 there is a ρ > 0 and apoint P such that

PQh = ρ ⇐⇒ PQ = ρ.

Moreover, if O is the center of absolute then1. O = O for any ρ and2. P ∈ (OP ) for any P 6= O.

Proof. According to Lemma 11.10, OQh = ρ if and only if

OQ = ρ =eρ − 1

eρ + 1.

Therefore the locus of h-points Q such that OQh = ρ is formed by theEuclidean circle, denote it by ∆ρ.


If P 6= O, applying Lemma 11.6 and the Main observation (11.7)we get a circle Γ perpendicular to the absolute such that P is theinversion of O in Γ.

Let ∆′ρ be the inversion of ∆ρ in Γ. Since the inversion in Γ pre-

serves the h-distance, PQh = ρ if and only if Q ∈ ∆′ρ.

O

Q

P

P

∆′ρ

According to Theorem 9.7, ∆′ρ is a

Euclidean circle. Denote by P the Eu-clidean center and by ρ the Euclideanradius of ∆′

ρ.Finally note that ∆′

ρ reflects to it-

self in (OP ); that is, the center P lieson (OP ).

11.13. Exercise. Assume P , P andO be as in the Lemma 11.12 and P 6=6= O. Show that P ∈ [OP ].

Axiom I

Evidently h-plane contains at least two points. Therefore to showthat Axiom I holds in h-plane we need to show that h-distance definedon page 88 is a metric on h-plane; that is, the conditions (a)–(d) inDefinition 1.1 hold for h-distance.

The following claim says that h-distance meets the conditions (a)and (b).

11.14. Claim. Given h-points P and Q, we have PQh > 0 andPQh = 0 if and only if P = Q.

Proof. According to Lemma 11.6 and Main Observation (11.7), wemay assume that Q is the center of absolute. In this case

δ(Q,P ) =1 +QP

1−QP> 1

and the equality holds only if P = Q.Therefore

QPh = ln[δ(Q,P )] > 0.

and the equality holds if and only if P = Q.

By the following claim says that h-distance meets the conditions1.1(c).

11.15. Claim. For any h-points P and Q, we have PQh = QPh.

93

Proof. Let A and B be ideal points of (PQ)h and A,P,Q,B appearon the circline containing (PQ)h in the same order.

P

Q

A

B

Then

PQh = lnAQ·BP

QB ·PA=

= lnBP ·AQPA·QB

=

= QPh.

The following claim shows in particular that tri-angle inequality (which is condition 1.1(d)) holdsfor h-distance.

11.16. Claim. Given a triple of h-points P , Q and R, we have

PQh +QRh > PRh.

Moreover, the equality holds if and only if P , Q and R lie on one h-linein the same order.

Proof. Without loss of generality, we may assume that RPh > PQh >0 and R is the center of absolute.

Denote by ∆ the h-circle with center P and h-radius PQh. Let Sand T be the points of intersection of (RP ) and ∆.

Since PQh 6 RPh, by Lemma 11.11 we can assume that the pointsR, S, P and T appear on the h-line in the same order.

Set ρ = PQh. According to Lemma 11.12, ∆ is a Euclidean circle;denote by P its Euclidean center. Note that P is the (Euclidean)midpoint of [ST ].

R P

P

∆

S T

Q

By the Euclidean triangle inequality

➋ RT = RP + PQ > RQ

and the equality holds if and only if T = Q.By Lemma 11.10,

RTh = ln1 +RT

1−RT

RQh = ln1 +RQ

1−RQ.

Since the function f(x) = ln 1+x1−x

is increasing for x ∈ [0, 1), inequality➋ implies

RTh > RQh


and the equality holds if and only if only if T = Q.Finally applying Lemma 11.11 again, we get

RTh = RPh + PQh.

Hence the claim follows.

Axiom II

Note that once the following claim is proved, Axiom II follows fromCorollary 9.17.

11.17. Claim. A subset of h-plane is an h-line if and only if it formsa line for h-distance in the sense of Definition 1.8.

Proof. Let ℓ be an h-line. Applying the main observation we canassume that ℓ contains the center of absolute. In this case ℓ is formedby intersection of diameter of absolute and the h-plane. Let A and Bbe the endpoints of the diameter.

Consider the map ι : ℓ → R defined as

ι(X) = lnAX

XB.

Note that ι : ℓ → R is a bijection.Further, if X,Y ∈ ℓ and the points A, X , Y and B appear on [AB]

in the same order then

ι(Y )− ι(X) = lnAY

Y B− ln

AX

XB= ln

AY ·BX

YB ·XB= XYh;

that is, any h-line is a line for h-metric.We proved that any h-line is a line for h-distance. The converse

follows from Claim 11.16.

Axiom III

Note that the first part of Axiom III follows directly from the definitionof h-angle measure defined on page 88. It remains to show that ∡h

satisfies the conditions IIIa, IIIb and IIIc on page 20.The following two claims say that ∡h satisfies IIIa and IIIb.

11.18. Claim. Given an h-half-line [OP ) and α ∈ (−π, π] there isunique h-half-line [OQ) such that ∡POQ = α.

95

11.19. Claim. For any h-points P , Q and R distinct from h-point Owe have

∡POQ+ ∡QOR ≡ ∡POR.

Proof of 11.18 and 11.19. Applying the main observation, we mayassume that O is the center of absolute. In this case, for any h-point P 6= O, [OP )h is the intersection of [OP ) with h-plane. Hencethe claims 11.18 and 11.19 follow from the corresponding axioms ofEuclidean plane.

The following claim says that ∡h satisfies IIIc.

11.20. Claim. The function

∡h : (P,Q,R) 7→ ∡PQR

is continuous at any triple of points (P,Q,R) such that Q 6= P andQ 6= R and ∡hPQR 6= π.

Proof. Denote by O the center of absolute. We can assume that Q isdistinct from O.

Denote by Z the inversion of Q in the absolute and by Γ the circleperpendicular to the absolute which is centered at Q′. According toLemma 11.6, the point O is the inversion of Q in Γ.

Denote by P ′ and R′ the inversions in Γ of the points P and Rcorrespondingly. Note that the point P ′ is completely determined bythe points Q and P ; moreover the map (Q,P ) 7→ P ′ is continuous atany pair of points (Q,P ) such that Q 6= O. The same is true for themap (Q,R) 7→ R′

According to the Main Observation

∡hPQR ≡ −∡hP′OR′.

Since ∡hP′OR′ = ∡P ′OR′ and the maps (Q,P ) 7→ P ′, (Q,R) 7→ R′

are continuous, the claim follows from the corresponding axiom ofEuclidean plane.

Axiom IV

The following claim says that Axiom IV holds holds in the h-plane.

11.21. Claim. In the h-plane, we have hPQR ∼= hP′Q′R′ if and

only if

Q′P ′h = QPh, Q′R′

h = QRh, and ∡hP′Q′R′ = ±∡PQR.


Proof. Applying the main observation, we can assume that both Qand Q′ coincide with the center of absolute. In this case

∡P ′QR′ = ∡hP′QR′ = ±∡hPQR = ±∡PQR.

SinceQPh = QP ′

h and QRh = QR′h,

Lemma 11.10 implies that the same holds for the Euclidean distances;that is,

QP = QP ′ and QR = QR′.

By SAS, there is a motion of Euclidean plane which sends Q to itself,P to P ′ and R to R′

Note that the center of absolute is fixed by the corresponding mo-tion. It follows that this motion gives also a motion of h-plane; inparticular the h-triangleshPQR and hP

′QR′ are h-congruent.

Axiom h-V

Finally we need to check that the Axiom h-V on page 82 holds; thatis, we need to prove the following claim.

11.22. Claim. For any h-line ℓ and any h-point P /∈ ℓ there are atleast two h-lines which pass through P and have no points of intersec-tion with ℓ.

A

B

P

m

n

ℓ

Instead o proof. Applying the main ob-servation we can assume that P is thecenter of absolute.

The remaining part of proof can beguessed from the picture

11.23. Exercise. Show that in the h-plane there are 3 mutually parallel h-lines such that any pair of these threelines lies on one side from the remain-ing h-line.

Chapter 12

Geometry of h-plane

In this chapter we study the geometry of the plane described by con-formal disc model. For briefness, this plane will be called h-plane.Note that we can work with this model directly from inside of Eu-clidean plane but we may also use the axioms of absolute geometrysince according to the previous chapter they all hold in the h-plane.

Angle of parallelism

Let P be a point off an h-line ℓ. Drop a perpendicular (PQ)h from Pto ℓ with foot point Q. Let ϕ be the least angle such that the h-line(PZ)h with |∡hQPZ| = ϕ does not intersect ℓ.

The angle ϕ is called angle of parallelism of P to ℓ. Clearly ϕdepends only on the h-distance s = PQh. Further ϕ(s) → π/2 ass → 0, and ϕ(s) → 0 as s → ∞. (In the Euclidean geometry the angleof parallelism is identically equal to π/2.)

P ℓ

If ℓ, P and Z as above then theh-line m = (PZ)h is called asymptot-ically parallel to ℓ. In other words,two h-lines are asymptotically parallelif they share one ideal point. (In hyper-bolic geometry the term parallel linesis often used for asymptotically paral-lel lines ; we do not follow this conven-tion.)

Given P 6∈ ℓ there are exactly twoasymptotically parallel lines through Pto ℓ; the remaining parallel lines to ℓ through P are called ultra parallel.

97

98 CHAPTER 12. GEOMETRY OF H-PLANE

On the diagram, the two solid h-lines passing through P are asymp-totically parallel to ℓ; the dotted h-line is ultra parallel to ℓ.

12.1. Proposition. Let Q be the foot point of P on h-line ℓ. Denoteby ϕ the angle of parallelism of P to ℓ. Then

PQh = 12 · ln

1+cosϕ1−cosϕ .

A

B

P X ZQ

ϕ

Proof. Applying a mo-tion of h-plane if neces-sary, we may assume Pis the center of absolute.Then the h-lines throughP are formed by the inter-sections of Euclidean lineswith the h-plane.

Let us denote by A andB the ideal points of ℓ.Without loss of generality

we may assume that ∠APB is positive. In this case

ϕ = ∡QPB = ∡APQ = 12 ·∡APB.

Let Z be the center of the circle Γ containing the h-line ℓ. Set X tobe the point of intersection of the Euclidean segment [AB] and (PQ).

Note that, OX = cosϕ therefore by Lemma 11.10,

OXh = ln 1+cosϕ1−cosϕ .

Note that both angles ∠PBZ and ∠BXZ are right. Therefore,sine the ∠PZB is shared, we get ZBX ∼ ZPB. In particular

ZX ·ZP = ZB2;

that is, X is the inversion of P in Γ.

The inversion in Γ is the reflection of h-plane through ℓ. Therefore

PQh = QXh =

= 12 ·OXh =

= 12 · ln

1+cosϕ1−cosϕ .

99

Inradius of h-triangle

12.2. Theorem. Inradius of any h-triangle is less than 12 · ln 3.

Proof. First note that any triangle in h-plane lies in an ideal triangle;that is, a region bounded by three pairwise asymptotically parallellines.

XY

Z

A

B

C

A proof can be seen in the picture.Consider arbitrary h-triangle hXY Z.Denote by A, B and C the ideal pointsof the h-half-lines [XY )h, [Y Z)h and[ZX)h.

It should be clear that inradius ofthe ideal triangle ABC is bigger thaninradius of hXY Z.

Applying an inverse if necessary, wecan assume that h-incenter (O) of theideal triangle is the center of absolute.Therefore, without loss of generality, we may assume

∡AOB = ∡BOC = ∡COA = 23 ·π.

It remains to find the inradius. Denote by Q the foot point of O on(AB)h. Then OQh is the inradius. Note that the angle of parallelismof (AB)h at O is equal to π

3 .

A

B

C O Q

By Proposition 12.1,

OQh = 12 · ln

1 + cos π3

1− cos π3

=

= 12 · ln

1 + 12

1− 12

=

= 12 · ln 3.

12.3. Exercise. Let hABCD be aquadrilateral in the h-plane such thatthe h-angles at A, B and C are right and ABh = BCh. Find theoptimal upper bound for ABh.

Circles, horocycles and equidistants

Note that according to Lemma 11.12, any h-circle is formed by a Eu-clidean circle which lies completely in the h-plane. Further any h-line


is an intersection of the h-plane with the circle perpendicular to theabsolute.

In this section we will describe the h-geometric meaning of theintersections of the other circles with the h-plane.

You will see that all these intersections formed by a perfectly roundshape in the h-plane.

One may think of these curves as about trajectories of a car whichdrives in the plane with fixed position of the wheel. In the Euclideanplane, this way you either run along a circles or along a line.

In hyperbolic plane the picture is different. If you turn wheel farright, you will run along a circle. If you turn it less, at certain positionof wheel, you will never come back, the path will be different from theline. If you turn the wheel further a bit, you start to run along a pathwhich stays on the same distant from an h-line.

m

g

A

B

Equidistants of h-lines. Consider h-plane with absolute Ω. Assume a circleΓ intersects Ω in two distinct points Aand B. Denote by g the intersectionof Γ with the h-plane. Let us draw anh-line m with the ideal points A andB. According to Exercise 11.1, m isuniquely determined by its ideal pointsA and B.

Consider any h-line ℓ perpendicularto m; let ∆ be the circle containing ℓ.

Note that ∆ ⊥ Γ. Indeed, according to Corollary 9.15, m and Ωinverted to themselves in ∆. It follows that A is the inversion of B in∆. Finally, by Corollary 9.16, we get that ∆ ⊥ Γ.

Therefore inversion in ∆ sends both m and g to themselves. So ifP ′, P ∈ g are inversions of each other in ∆ then they lie on the sameh-distance from m. Clearly we have plenty of choice for ℓ, which canbe used to move points along g arbitrary keeping the distance to m.

It follows that g is formed by the set of points which lie on fixedh-distance and the same side from m.

Such curve g is called equidistant to h-line m. In Euclidean ge-ometry the equidistant from a line is a line; apparently in hyperbolicgeometry the picture is different.

Horocycles. If the circle Γ touches the absolute from inside at onepoint A then the complement h = Γ\A lies in the h-plane. Thisset is called horocycle. It also has perfectly round shape in the sensedescribed above.

101

Γ

A

Horocycles are the boarder case be-tween circles and equidistants to h-lines. A horocycle might be consideredas a limit of circles which pass throughfixed point which the centers running toinfinity along a line. The same horocy-cle is a limit of equidistants which passthrough fixed point to the h-lines run-ning to infinity.

12.4. Exercise. Find the leg of isosce-les right h-triangle inscribed in a horocycle.

Hyperbolic triangles

12.5. Theorem. Any nondegenerate hyperbolic triangle has positivedefect.

AC

B

Proof. Consider h-triangle hABC. Ac-cording to Theorem 10.9,

➊ defect(hABC) > 0.

It remains to show that in the case of equalitythe triangle hABC degenerates.

Without loss of generality, we may as-sume that A is the center of absolute; in this case ∡hCAB = ∡CAB.Yet we may assume that

∡hCAB, ∡hABC, ∡hBCA, ∡ABC, ∡BCA > 0.

Let D be an arbitrary point in [CB]h distinct from B and C. FromProposition 8.15

∡ABC − ∡hABC ≡ π − ∡CDB ≡ ∡BCA− ∡hBCA.

From Exercise 6.15, we get

defect(hABC) = 2·(π − ∡CDB).

Therefore if we have equality in ➊ then ∡CDB = π. In particular theh-segment [BC]h coincides with Euclidean segment [BC]. By Exer-cise 11.3, the later can happen only if the h-line passes through thecenter of absolute; that is, if hABC degenerates.


The following theorem states in particular that hyperbolic trianglesare congruent if their corresponding angles are equal; in particular inhyperbolic geometry similar triangles have to be congruent.

12.6. AAA congruence condition. Two nondegenerate triangleshABC and hA

′B′C′ in the h-plane are congruent if ∡hABC == ±∡hA

′B′C′, ∡hBCA = ±∡hB′C′A′ and ∡hCAB = ±∡hC

′A′B′.

Proof. Note hat if ABh = A′B′h then the theorem follows from ASA.

A′

B′

C′

B′′

C′′

Assume contrary. Without loss of general-ity we may assume that ABh < A′B′

h. There-fore we can choose the point B′′ ∈ [A′B′]h suchthat A′B′′

h = ABh.Choose an h-half-line [B′′X) so that

∡hA′B′′X = ∡hA

′B′C′.

According to Exercise 10.5, (B′′X)h ‖ (B′C′)h.By Pasch’s theorem (3.10), (B′′X)h inter-

sects [A′C′]h. Denote by C′′ the point of intersection.According to ASA, hABC ∼= hA

′B′′C′′; in particular

➋ defect(hABC) = defect(hA′B′′C′′).

Applying Exercise 10.10 twice, we get

➌defect(hA

′B′C′) = defect(hA′B′′C′′)+

+ defect(hB′′C′′C′) + defect(hB

′′C′B′).

By Theorem 12.5, the defects has to be positive. Therefore

defect(hA′B′C′) > defect(hABC).

On the other hand,

defect(hA′B′C′) = |∡hA

′B′C′|+ |∡hB′C′A′|+ |∡hC

′A′B′| == |∡hABC|+ |∡hBCA|+ |∡hCAB| == defect(hABC),

a contradiction.

Recall that a bijection from plane to itself is called angle preservingif

∡ABC = ∡A′B′C′

for any triangle ABC and its image A′B′C′.

12.7. Exercise. Show that any angle-preserving transformation ofh-plane is a motion.

103

Conformal interpretation

Let us give an other interpretation of the h-distance.

12.8. Lemma. Consider h-plane with absolute formed by the unitcircle centered at O. Fix a point P and let Q be an other point in theh-plane. Set x = PQ and y = PQh then

limx→0

yx=

2

1−OP 2.

The above formula tells that the h-distance from P to a near bypoint Q is nearly proportional to the Euclidean distance with the co-efficient 2

1−OP 2 . The value λ(P ) = 21−OP 2 is called conformal factor

of h-metric.The value 1

λ(P ) =12 ·(1−OP 2) can be interpenetrated as the speed

limit at the given point P . In this case the h-distance is the minimaltime needed to travel from one point of h-plane to the other point.

Γ

O P

QQ′

P ′

Proof. If P = O, then according toLemma 11.10

➍y

x=

ln 1+x1−x

x→ 2

as x → 0.If P 6= O, denote by P ′ the inver-

sion of P in the absolute. Denote by Γthe circle with center P ′ perpendicularto the absolute.

According to Main Observation11.7 and Lemma 11.6 the inversion in Γ is a motion of h-plane whichsends P to O. In particular, if we denote by Q′ the inversion of Q inΓ then OQ′

h = PQh.Set x′ = OQ′ According to Lemma 9.2,

x′

x=

OP ′

P ′Q.

Since P ′ is the inversion of P in the absolute, we have PO·OP ′ = 1.Therefore

x′

x→ OP ′

P ′P=

1

1−OP 2

as x → 0.Together with ➍, it implies

y

x=

y

x′ ·x′

x→ 2

1−OP 2


as x → 0.

Here is an application of the lemma above.

12.9. Proposition. The circumference of an h-circle of h-radius ris

2·π· sh r,where sh r denotes hyperbolic sine of r; that is,

sh rdef==

er − e−r

2.

Before we proceed with the proof let us discuss the same problemin the Euclidean plane.

The circumference of the circle in the Euclidean plane can be de-fined as limit of perimeters of regular n-gons inscribed in the circle asn → ∞.

Namely, let us fix r > 0. Given a positive integer n considerAOB such that ∡AOB = 2·π

nand OA = OB = r. Set xn = AB.

Note that xn is the side of regular n-gon inscribed in the circle ofradius r. Therefore the perimeter of the n-gon is equal to n·xn.

A

B

O

2·πn

r

r

The circumference of the circle withradius r might be defined as the limit

➎ limn→∞

n·xn = 2·π·r.

(This limit can be taken as the defini-tion of π.)

In the following proof we repeat thesame construction in the h-plane.

Proof. Without loss of generality wecan assume that the center O of the cir-

cle is the center of absolute.By Lemma 11.10, the h-circle with h-radius r is formed by the

Euclidean circle with center O and radius

a =er − 1

er + 1.

Denote by xn and yn the Euclidean and hyperbolic side lengths ofthe regular n-gon inscribed in the circle.

Note that xn → 0 as n → ∞. By Lemma 12.8,

limn→∞

ynxn

=2

1− a2.

105

Applying ➎, we get that the circumference of the h-circle can befound the following way

limn→∞

n·yn =2

1− a2· limn→∞

n·xn =

=4·π·a1− a2

=

=4·π·

(

er−1er+1

)

1−(

er−1er+1

)2 =

= 2·π· er − e−r

2=

= 2·π· sh r.

12.10. Exercise. Denote by circumh(r) the circumference of the h-circle of radius r. Show that

circumh(r + 1) > 2· circumh(r)

for all r > 0.

Chapter 13

Affine geometry

Affine transformations

A bijection of Euclidean plane to itself is called affine transformationif it maps any line to a line.

We say that three points are collinear if they lie on one line. Notethat affine transformation sends collinear points to collinear; the fol-lowing exercise gives a converse.

13.1. Exercise. Assume f is a bijection from Euclidean plane toitself which sends collinear points to collinear points. Show that f isan affine transformation.

In other words, you need to show that f maps noncollinear pointsto noncollinear.

Affine geometry studies so called incidence structure of Euclideanplane. The incidence structure is the data about which points lie onwhich lines and nothing else; we can not talk about distance angles andso on. One may also say that affine geometry studies the propertiesof Euclidean plane which preserved under affine transformations.

13.2. Exercise. Show that affine transformation sends parallel linesto the parallel lines.

Constructions with parallel tool and ruler

Let us consider geometric constructions with ruler and parallel tool ;the later makes possible to draw a line through given point parallel togiven line. From the exercisers above it follows that any constructionwith these two tools are invariant with respect to affine transformation.

106

107

For example to solve the following exercise, it is sufficient to prove thatmidpoint of given segment can be constructed with ruler and paralleltool.

13.3. Exercise. Let M be the midpoint of segment [AB] in the Eu-clidean plane. Assume that an affine transformation sends points A,B and M to A′, B′ and M ′ correspondingly. Show that M ′ is themidpoint of [A′B′].

The following exercise will be used in the proof of Theorem 13.6.

13.4. Exercise. Assume that in Euclidean plane we have 4 pointswith coordinates (0, 0), (1, 0), (a, 0) and (b, 0). Use ruler and paralleltool to construct the points with coordinates (a·b, 0) and (a+ b, 0).

13.5. Exercise. Use ruler and parallel tool to construct the center ofgiven circle.

Matrix form

Since the lines are defined in terms of metric; any motion of Euclideanplane is also an affine transformation.

On the other hand, there are affine transformations of Euclideanplane which are not motions.

13.6. Theorem. Consider Euclidean plane with coordinate system;let us use the column notation for the coordinates; that is, we will write( xy ) instead of (x, y).

A map β from the plane to itself is affine transformation if andonly if

➊ β : ( xy ) 7→(

a bc d

)

· ( xy ) + ( vw )

for some fixed invertible matrix(

a bc d

)

and vector(

vw

)

.In particular, any affine transformation of Euclidean plane is con-

tinuous.

In the proof of “only if” part, we will use the following algebraiclemma.

13.7. Algebraic lemma. Assume f : R → R is a function such that

f(1) = 1,

f(x+ y) = f(x) + f(y),

f(x·y) = f(x)·f(y)

108 CHAPTER 13. AFFINE GEOMETRY

for any x, y ∈ R. Then f(x) = x for any x ∈ R.

Note that we do not assume that f is continuous.The function f satisfying three conditions in the lemma is called

field automorphism. Therefore the lemma states that the identity func-tion is the only automorphism of the field of real numbers.

On the other hand, the conjugation z 7→ z (see page 137) gives anexample of nontrivial automorphism of complex numbers.

Proof. Sincef(0) + f(1) = f(0 + 1),

we getf(0) + 1 = 1;

that is,

➋ f(0) = 0.

Further0 = f(0) = f(x) + f(−x).

Therefore

➌ f(−x) = −f(x) for any x ∈ R.

Further

f(2) = f(1) + f(1) = 1 + 1 = 2

f(3) = f(2) + f(1) = 2 + 1 = 3

. . .

Together with ➌ it implies that

f(n) = n for any integer n.

Sincef(m) = f(m

n)·f(n)

we getf(m

n) = m

n

for any rational number mn.

Assume a > 0. Then the equation x·x = a has a real solution x =√a. Therefore [f(

√a)]2 = f(

√a)·f(√a) = f(a). Whence f(a) > 0.

That is

➍ a > 0 =⇒ f(a) > 0.

109

Applying ➌, we also get

➎ a 6 0 =⇒ f(a) 6 0.

Finally, assume f(a) 6= a for some a ∈ R. Then there is a rationalnumber m

nwhich lies between a and f(a); i.e, the numbers x = a− m

n

and y = f(a) − mn

have opposite signs. Since f(mn) = m

n, we get

f(x) = y. The later contradicts ➍ or ➎.

13.8. Lemma. Assume γ is an affine transformation which fix threepoints ( 0

0 ), (10 ) and ( 01 ) on the coordinate plane. Then γ is the identity

map; that is, γ ( xy ) = ( xy ) for any point ( xy ).

Proof. Since affine transformation sends lines to lines, we get that eachaxes is mapped to itself.

According to Exercise 13.2, parallel lines are mapped to parallellines. Therefore we get that horizontal lines mapped to horizontallines and vertical lines mapped to vertical. In other words,

γ ( xy ) =(

f(x)h(y)

)

.

for some functions f, h : R → R.Note that f(1) = h(1) = 1 and according to Exercise 13.4, both

f and h satisfies the other two conditions of Algebraic lemma 13.7.Applying the lemma, we get that f and h are identity functions andso is γ.

Proof of Theorem 13.6. Recall that matrix(

a bc d

)

is invertible if

det(

a bc d

)

= a·d− b·c 6= 0;

in this case the matrix

1a·d−b·c ·

(

d −b−c a

)

is the inverse of(

a bc d

)

.Assume that the map β is described by ➊. Note that

➏ ( xy ) 7→ 1a·d−b·c ·

(

d −b−c a

)

·(

x−vy−w

)

.

is inverse of β. In particular β is a bijection.A line in the plane form the solutions of equations

➐ p·x+ q ·y + r = 0,


where p 6= 0 or q 6= 0. Find ( xy ) from its β-image by formula ➏ and

substitute the result in ➐. You will get the equation of the image of theline. The equation has the same type as ➐, with different constants, inparticular it describes a line. Therefore β is an affine transformation.

To prove “only if” part, fix an affine transformation α. Set

( vw ) = α ( 00 ) ,

( ac ) = α ( 10 )− α ( 00 ) ,(

bd

)

= α ( 10 )− α ( 00 ) .

Note that the points α ( 00 ), α ( 01 ), α ( 10 ) do not lie on one line.

Therefore the matrix(

a bc d

)

is invertible.For the affine transformation β defined by ➊ we have

β ( 00 ) = α ( 00 ) ,

β ( 10 ) = α ( 10 ) ,

β ( 01 ) = α ( 01 ) .

It remains to show that α = β or equivalently the compositionγ = α β−1 is the identity map.

Note that γ is an affine transformation which fix points ( 00 ), (10 )

and ( 01 ). It remains to apply Lemma 13.8.

Consider so called shear mapping ( xy ) 7→(

x+k·yy

)

. According toTheorem 13.6, shear mapping is an affine transformation. The shearmapping can change the angle between vertical and horizontal linesalmost arbitrary. The later can be used to prove impossibility of someconstructions with ruler and parallel tool; here is one example.

13.9. Exercise. Show that with ruler and parallel tool one can notconstruct a line perpendicular to the given line.

On inversive transformations

Recall that inversive plane is Euclidean plane with added a point atinfinity, denoted as ∞. We assume that any line pass through ∞. Theterm circline stays for circle or line;

The inversive transformation is bijection from inversive plane toitself which sends circlines to circlines. Inversive geometry can be de-fined as geometry which circline incidence structure of inversive plane;that is we can say which points lie on which circlines.

13.10. Theorem. A map from inversive plane to itself is an inversivetransformation if and only if it can be presented as a composition ofinversions and reflections.

111

Proof. According to Theorem 9.7 any inversion is a inversive transfor-mation. Therefore the same holds for composition of inversions.

To prove converse, fix an inversive transformation α.Assume α(∞) = ∞. Recall that any circline passing through ∞

is a line. If follows that α maps lines to lines; that is, it is an affinetransformation.

Further, α is not an arbitrary affine transformation, it maps circlesto circles.

Composing α with a reflection, say ρ1, we can assume that α′ =ρ1 α maps the unit circle with center at the origin to a concentriccircle.

Composing the obtained map α′ with a homothety χ with centerat the origin, we can assume that α′′ = χ α′ sends the unit circle toitself.

Composing the obtained map α′′ a reflection ρ2 in a line throughthe origin, we can assume that in addition α′′′ = ρ2 α′′ the point(1, 0) maps to itself.

By Exercise 13.5, α′′′ fixes the center of the circle; that is, it fixesthe origin.

The obtained map α′′′ is an affine transformation. Applying The-orem 13.6, together with the properties of α′′ described above we get

α′′′ : ( xy ) 7→(

1 b0 d

)

· ( xy )

for an invertible matrix(

1 b0 d

)

. Since the point (0, 1) maps to the unitcircle we get

b2 + d2 = 1.

Since the point ( 1√2, 1√

2) maps to the unit circle we get

(b + d)2 = 1.

It followsα′′′ : ( xy ) 7→

(

1 00 ±1

)

· ( xy ) ;

that is, either α′′′ is the identity map or reflection the x-axis.Note that the homothety χ is a composition of two inversions in

concentric circles. Hence the result follows.Now assume P = α(∞) 6= ∞. Consider an inversion β in a circle

with center at P . Note that β(P ) = ∞; therefore β α(∞) = ∞.Since β is inversive, so is β α. From above we get that β α is acomposition of reflections and inversions therefore so is α.

13.11. Exercise. Show that any reflection can be presented as a com-position of three inverses.


Note that exercise above together with Theorem 13.10, impliesthat any inversive map is a composition of inversions, no reflectionsare needed.

Chapter 14

Projective geometry

Real projective plane

In the Euclidean plane two distinct lines might have one or zero pointsof intersection (in the later case the lines are called parallel). Our aimis to extend Euclidean plane by ideal points so that any two distinctlines will have exactly one point of intersection.

A collection of lines in the Euclidean planeis called concurrent if they all intersect at singlepoint or all of them pairwise parallel. A set ofconcurrent lines passing through each point inthe plane is called pencil. There are two types ofpencils first is the set of all lines passing throughgiven point called the center of the pencil andsecond is the set of pairwise parallel lines.

Note that each point in Euclidean planeuniquely defines a pencil with center in it, butthe parallel pencils have no center. Also anytwo lines completely determine the pencil con-taining both.

Let us add one ideal point for each parallelpencil, and assume that all these ideal pointslie on one ideal line.

We obtain so called real projective plane.Each point in the real projective plane definedas a pencil of lines in Euclidean plane. We say that three points lie onone line the corresponding pencils contain a common line. (We assumethat the ideal line belongs to each parallel pencil).

113

114 CHAPTER 14. PROJECTIVE GEOMETRY

Euclidean space

Let us repeat the construction of metric d2 (page 11) in the space.We will denote by R3 the set of all triples (x, y, z) of real numbers.

Assume A = (xA, yA, zA) and B = (xB , yB, zB) are arbitrary points.Define the metric on R3 the following way

ABdef==√

(xA − xB)2 + (yA − yB)2 + (zA − zB)2.

The obtained metric space is called Euclidean space.Assume at least one of the real numbers a, b or c is distinct from

zero. Then the subset of points (x, y, z) ∈ R3 described by equation

a·x+ b·y + c·z + d = 0

is called plane; here d is a real number.It is straightforward to show that any plane in Euclidean space is

isometric to Euclidean plane. Further, any three points on the spacelie on one plane. And any nonempty intersection of two distinct planesforms a line in each of these planes.

The later statements makes possible to generalize many notionsand results from Euclidean plane geometry to Euclidean space by ap-plying plane geometry in the planes of the space.

Perspective projection

Consider two planes Π and Π′ in the Euclidean space. Let O be apoint which does not belong neither to Π nor Π′.

Consider the perspective projection from Π to Π′ with center at O.The projection of P ∈ Π is defined as the point P ′ ∈ Π′ which lies onthe line (OP ).

115

Note that the perspective projection sends collinear points to col-linear. Indeed, assume three points P , Q, R lie on one line ℓ in Π andP ′, Q′, R′ are their images in Π′. Then all the points P , Q, R, P ′, Q′,R′ lie in the plane, say Θ, containing O and ℓ. Therefore the pointsP ′, Q′, R′ lie in the line formed by intersection ℓ′ = Θ ∩ Π′.

The perspective projection is not a bijection between the planes;.Indeed, if the line (OP ) is parallel to Π′ (that is, if (OP ) ∩ Π′ = ∅)then the perspective projection is not defined. Also if (OP ′) ‖ Π fora point P ′ ∈ Π′ then point P ′ is not an image of the perspectiveprojection.

A similar story happened with inversion. If inversion is consideredfor the Euclidean plane then is not defied at the center of inversion;it is also not the image of any point. To deal with this problem wepassed to inversive plane which is Euclidean plane extended by oneideal point.

A similar strategy works for perspective projection Π → Π′, butthis time real projective plane is the right choice of extension. Denoteby Π and Π′ the corresponding real projective planes.

Note that there is a natural bijection between points in the realprojective plane Π and all the lines passing through O. If P ∈ Π thentake the line (OP ); if P is an ideal point of Π, so defined by a parallelpencil of lines then take the line through O which is parallel to eachlines in this pencil.

The same construction gives a bijection between points in the realprojective plane Π′ and all the lines passing through O. Composingthese bijections we get a bijection Π → Π′ which coincides with theperspective projection P 7→ P ′ where it is defined.

Note that the ideal line of Π maps to the line formed by intersectionof Π′ and the plane through O parallel to Π. Similarly the ideal line ofΠ′ is the image of the line formed by intersection of Π and the planethrough O parallel to Π′.

Strictly speaking this gives a transformation from one real projec-tive plane to another, but if we identify the two planes, say by fixinga coordinate system in each, we get a projective transformation fromthe plane to itself.

14.1. Exercise. Let O be the origin of (x, y, z)-coordinate space andthe planes Π and Π′ formed by the solutions of equations x = 1 andy = 1 correspondingly. The perspective projection from Π to Π′ withcenter at O sends P to P ′. Assume P has coordinates (1, y, z), findthe coordinates of P ′.

For which points P ∈ Π the perspective projection is undefined?Which points P ′ ∈ Π′ are not images of points under perspective pro-jection?


Projective transformations

A bijection from real projective plane to itself which sends lines tolines is called projective transformation.

Projective and affine geometries study incidence structure of Eu-clidean and real projective plane correspondingly. One may also saythat projective geometry studies the properties of real projective planewhich preserved under projective transformations.

Note that any affine transformation defines a projective transfor-mation on the corresponding real projective plane. We will call suchprojective transformations affine; these are projective transformationswhich send the ideal line to itself.

The perspective projection discussed in the previous section givesan example of projective transformation which is not affine.

14.2. Theorem. Any projective transformation can be obtained as acomposition of an affine transformation and a perspective projection.

Proof. Assume α is a projective transformation. If is α sends idealline to itself then it has to be affine. Hence the theorem follows.

Assume α sends the ideal line to line ℓ, choose a perspective pro-jection β which sends ℓ back to the ideal line. To do this we have toidentify our plane with a plane Π in the space, then fix a point O /∈ Πand then choose a plane Π′ which is parallel to the plane containing ℓand O.

The composition β α sends ideal line to itself, therefore it has tobe affine. Hence the result follows.

Desargues’ theorem

Loosely speaking, any statement in projective geometry can be formu-lated using only terms collinear points, concurrent lines.

Here is a classical example of a theorem in projective geometry.

14.3. Desargues’ theorem. Consider three concurrent lines (AA′),(BB′) and (CC′) in the real projective plane. Set

X = (BC) ∩ (B′C′),

Y = (CA) ∩ (C′A′),

Z = (AB) ∩ (A′B′)

Then the points X, Y and Z are collinear.

117

A

A′

B

B′

C C′

Z

X

Y

Proof. Without loss of generality, wemay assume that the line (XY ) is ideal.If not, apply a perspective projectionwhich sends the line (XY ) to the idealline.

In other words, we can assume that(BC) ‖ (B′C′) and (CA) ‖ (C′A′) andwe need to show that (AB) ‖ (A′B′).

Assume that the lines (AA′), (BB′)and (CC′) intersect at point O. Since(BC) ‖ (B′C′), Transversal prop-erty 6.18, implies that that ∡OBC ≡≡ ∡OB′C′ and ∡OCB ≡ ∡OC′B′.By AA-similarity condition, OBC ∼∼ OB′C′. In particular,

OB

OB′ =OC

OC′ .

A

A′

BB′

C C′

The same way we get OAC ∼∼ OA′C′ and

OA

OA′ =OC

OC′ .

Therefore

OA

OA′ =OB

OB′ .

By SAS-similarity condition, we getOAB ∼ OA′B′, in particular∡OAB ≡ ±∡OA′B′.

Note that ∡AOB ≡ ∡A′OB′.Therefore

∡OAB ≡ ∡OA′B′.

By Transversal property 6.18, (AB) ‖ (A′B′).The case when the lines (AA′) ‖ (BB′) ‖ (CC′) can be done sim-

ilarly. In this case the quadrilaterals B′BCC′ and A′ACC′ areparallelograms. Therefore

BB′ = CC′ = AA′.

Whence B′BAA′ is a parallelogram, in particular (AB) ‖ (A′B′).


Here is an other classical theorem of of projective geometry.

14.4. Pappus’s theorem. Assume that two triples of points A, B,C, and A′, B′, C′ are collinear. Set

X = (BC′) ∩ (B′C), Y = (CA′) ∩ (C′A), Z = (AB′) ∩ (A′B).

Then the points X, Y , Z are collinear.

A

A′

B

B′

C

C′

A

A′

B

B′

C

C′

XYZ

Pappus’s theorem can be proved the same way as Desargues’ the-orem.

Idea of the proof. Applying a perspective projection, we can assumethat X and Y lie on the ideal line. It remains to show that Z lies onthe ideal line.

In other words, assuming that (AB′) ‖ (A′B) and (AC′) ‖ (A′C),we need to show that (BC′) ‖ (B′C).

14.5. Exercise. Finish the proof of Pappus’s theorem using the ideadescribed above.

Duality

Let us fix a bijection between the set of lines and the set of points ofthe plane. Given a point P let us denote by lower case letter p thecorresponding line. Also the other way around, given a line s we willdenote upper case letter S the corresponding point.

The bijection between points and lines is called duality1 if

P ∈ s ⇐⇒ p ∋ S.

for any point P and line s.Existence of duality in a plane says that the lines and the points

in this plane have the same rights in terms of incidence.

1Usual definition of duality is more general; we consider a special case which isalso called polarity.

119

p

q r

s

A

D

C

BE

F

P Q

R

S

a

b

c

d

e

f

Dual configurations.

14.6. Exercise. Show that Euclidean plane does not admit a duality.

14.7. Theorem. The real projective plane admits a duality.

Proof. Consider a plane Π and a point O /∈ Π in the Euclidean space;denote by Π the corresponding real projective plane.

Recall that there is a natural bijection between points in the realprojective plane Π and all the lines passing through O. If P ∈ Π thenthe corresponding line is (OP ); if P is an ideal point of Π, so P isdefined as a parallel pencil of lines, take the line through O which isparallel to each lines in this pencil. Given P ∈ Π, denote the obtainedline P .

Similarly there is a natural bijection between lines in Π and all theplanes passing through O. If is s is a line in Π then take the planecontaining O and s; if s is the ideal line of Π, take the plane throughO parallel to Π. Given a line s in Π, denote the obtained plane as s.

It is straightforward to check that P ⊂ s if and only if P ∈ s;that is, the bijections P 7→ P and s 7→ s remember all the incidencestructure of the real projective plane Π.

It remains to construct a bijection s 7→ S between the set of planesand the set of lines passing through O such that

➊ r ⊂ S ⇐⇒ R ⊃ s

for any two lines line r and s passing through O.Set S to be the plane through O which is perpendicular to s. Note

that both conditions ➊ are equivalent to r ⊥ s; hence the result follows.

14.8. Exercise. Consider Euclidean plane with (x, y)-coordinates,denote by O the origin. Given a point P 6= O with coordinates (a, b)consider the line p formed by solution of the equation a·x+ b·y = 1.

Show that the correspondence P to p extends to duality of corre-sponding real projective plane.

Which line corresponds to O?


Which point of real projective plane corresponds to the line formedby solutions a·x+ b·y = 0?

The existence of duality in the real projective planes makes pos-sible to formulate an equivalent dual statement to any statement inprojective geometry. Say, the dual statement for “the points X , Y andZ lie on one line ℓ” would be the “lines x, y and z intersect at onepoint L”.

For example the dual statement for Desargues’ theorem 14.3 is itsconverse.

14.9. Dual Desargues’ theorem. Consider the collinear pointsX, Y and Z. Assume that X = (BC) ∩ (B′C′), Y = (CA) ∩ (C′A′)and Z = (AB) ∩ (A′B′). Then the lines (AA′), (BB′) and (CC′) areconcurrent.

In this theorem the points X , Y and Z are dual to the lines(AA′), (BB′) and (CC′) in the original formulation, and the otherway around.

14.10. Exercise. Formulate Dual Pappus’s theorem, see 14.4.

Proving Desargues’ theorem and applying duality (Theorem 14.7)we get Dual Desargues’ theorem. The same can be done the other wayaround.

The Desargues’ theorem and its dual can be packed in the followingstrong version of Desargues’ theorem.

14.11. Strong Desargues’ theorem. Assume that X = (BC) ∩∩ (B′C′), Y = (CA)∩ (C′A′) and Z = (AB)∩ (A′B′). Then the lines(AA′), (BB′) and (CC′) are concurrent if and only if the points X,Y and Z are collinear.

Axioms

Note that the real projective plane described above satisfies the fol-lowing set of axioms.

I. Any two distinct points lie on a unique line.II. Any two distinct lines pass through a unique point.III. There exist at least four points of which no three are collinear.Let us take these three axioms as a definition of projective plane; so

the real projective plane discussed above becomes a particular exampleof projective plane.

14.12. Exercise. Show that any line in projective plane contains atleast three points.

121

It turns out that there is a projective plane whichcontains exactly 3 points on each line. This is so calledFano plane which you can see on the diagram; it con-tains 7 points and 7 lines. This is an example of finiteprojective plane; that is, projective plane with finitelymany points.

Consider the following analog of Axiom III.

III′. There exist at least four lines of which no three are concurrent.

14.13. Exercise. Show that Axiom III ′ is equivalent to Axiom III.

Exercise above show that in the axiomatic system of projectiveplane, lines and points have the same rights. In fact one can switcheverywhere words “point” with “line”, “pass trough” with “lies on”,“collinear” with “concurrent” and we get an equivalent set of axioms.

14.14. Exercise. Assume that one of the lines in a finite projectiveplane contains exactly n+ 1 points.(a) Show that each line contains exactly n+ 1 points.(b) Show that the number of the points in the plane has to be

n2 + n+ 1.

(c) Show that in any finite projective plane the number of pointscoincides with the number of lines.

The number n in the above exercise is called order of finite pro-jective plane. For example Fano plane has order 2. Here is one of themost famous conjecture in finite geometry.

14.15. Conjecture. The order of a finite projective plane is thepower of a prime.

Chapter 15

Spherical geometry

Spherical geometry is the geometry of the surface of the unit sphere.This type of geometry has practical applications in cartography, nav-igation and astronomy.

The spherical geometry is a close relative of Euclidean and hy-perbolic geometries. Most of theorems of hyperbolic geometry havespherical analogs, but spherical geometry is easier to visualize.

We discuss few theorems in spherical geometry; the proofs are notcompletely rigorous.

Spheres in the space

Recall that Euclidean space is the set R3 of all triples (x, y, z) ofreal numbers such that the distance between a pair of points A =(xA, yA, zA) and B = (xB , yB, zB) is defined by the following formula

ABdef==√

(xA − xB)2 + (yA − yB)2 + (zA − zB)2.

The planes in the space defined as the set of solutions of equation

a·x+ b·y + c·z + d = 0

for real numbers a, b, c and d such that at least one of the numbersa, b or c is not zero. Equivalently plane can be defined as a subset ofEuclidean space which is isometric to to Euclidean plane.

Sphere in the space is the direct analog of circle in the plane. For-mally, sphere with center O and radius r is the set of points in thespace which lie on the distance r from O.

Let A and B be two points on the unit sphere centered at O. Thespherical distance from A to B (briefly ABs) is defined as |∡AOB|.

122

123

In the spherical geometry, the role of lines play the great circles ;that is, the intersection of the sphere with a plane passing through O.

Note that the great circles do not form lines in the sense of Defini-tion 1.8. Also any two distinct great circles intersect at two antipodalpoints. In particular, the sphere does not satisfy the axioms of absoluteplane.

Pythagorean theorem

Here is an analog of Pythagorean Theorems (6.10 and 16.8) in sphericalgeometry.

15.1. Theorem. Let sABC be a spherical triangle with right angleat C. Set a = BCs, b = CAs and c = ABs. Then

cos c = cos a· cos b.

In the proof we will use the notion of scalar product which we areabout to discuss.

Let A and B be two points in Euclidean space. Denote by vA == (xA, yA, zA) and vB = (xB , yB, zB) the position vectors of A and Bcorrespondingly. The scalar product of two vectors vA and vB in R3

is defined as

➊ 〈vA, vB〉 def== xA ·xB + yA ·yB + zA ·zB.

Assume both vectors vA and vB are nonzero and ϕ is the anglemeasure between these two vectors. In this case the scalar productcan be expressed the following way:

〈vA, vB〉 = |vA|·|vB |· cosϕ,

where

|vA| =√

x2A + y2A + z2A, |vB| =

√

x2B + y2B + z2B.

O

C

B A

x

y

zNow, assume the points A and B lie on

the unit sphere in R3 centered at the origin.In this case |vA| = |vB| = 1. By ➊ we get

➋ cosABs = 〈vA, vB〉.

Proof. Since the angle at C is right, we canchoose coordinates in R3 so that vC = (0, 0, 1),

124 CHAPTER 15. SPHERICAL GEOMETRY

vA lies in xz-plane, so vA = (xA, 0, zA) and vB lies in yz-plane, sovB = (0, yB, zB).

Applying, ➋, we get

zA = 〈vC , vA〉 = cos b,

zB = 〈vC , vB〉 = cos a.

Applying, ➋ again, we get

cos c = 〈vA, vB〉 == xA ·0 + 0·yB + zA ·zB =

= cos b· cos a.

15.2. Exercise. Show that if sABC be a spherical triangle withright angle at C and ACs = BCs =

π4 then ABs =

π3 .

Try to find two solutions, with and without using the spherical Py-thagorean theorem.

Inversion of the space

The inversion in the sphere defined the same way as we define inversionin the circle.

Formally, let Σ be the sphere with center O and radius r. Theinversion in Σ of a point P is the point P ′ ∈ [OP ) such that

OP ·OP ′ = r2.

In this case, the sphere Σ will be called the sphere of inversion andits center is called center of inversion.

We also add ∞ to the space and assume that the center of inversionis mapped to ∞ and the other way around. The space R3 with thepoint ∞ will be called inversive space.

The inversion of the space has many properties of the inversion ofthe plane. Most important for us is the analogs of theorems 9.6, 9.7,9.24 which can be summarized as follows.

15.3. Theorem. The inversion in the sphere has the following prop-erties:(a) Inversion maps sphere or plane into sphere or plane.(b) Inversion maps circle or line into circle or line.(c) Inversion preserves cross-ratio; that is, if A′, B′, C′ and D′ be

the inversions of the points A, B, C and D correspondingly then

AB ·CD

BC ·DA=

A′B′ ·C′D′

B′C′ ·D′A′ .

125

(d) Inversion maps arcs into arcs.

(e) Inversion preserves the absolute value of the angle measure be-tween tangent half-lines to the arcs.

We do not present the proofs here, but they nearly repeat thecorresponding proofs in plane geometry. To prove (a), you will needin addition the following lemma; its proof is left to the reader.

15.4. Lemma. Let Σ be a subset of Euclidean space which containsat least two points. Fix a point O in the space.

Then Σ is a sphere if and only if for any plane Π passing throughO, the intersection Π∩Σ is either empty set, one point set or a circle.

The following observation helps to reduce part (b) to part (a).

15.5. Observation. Any circle in the space can be presented as anintersection of two spheres.

Stereographic projection

Consider the unit sphere Σ centered at the origin (0, 0, 0). This spherecan be described by equation x2 + y2 + z2 = 1.

O

P

S

N

P ′

The plane throughP , O and S.

Denote by Π the xy-plane; itis defined by the equation z = 0.Clearly Π runs through the centerof Σ.

Denote by N = (0, 0, 1) the“North Pole” and by S = (0, 0,−1)be the “South Pole” of Σ; these arethe points on the sphere which haveextremal distances to Π. Denote byΩ the “equator” of Σ; it is the inter-section Σ ∩Π.

For any point P 6= S on Σ, con-sider the line (SP ) in the space.This line intersects Π in exactly onepoint, say P ′. We set in additionthat S′ = ∞.

The map P 7→ P ′ is called stereographic projection from Σ to Πfrom the South Pole. The inverse of this map P ′ 7→ P is called stereo-graphic projection from Π to Σ from the South Pole.

The same way one can define stereographic projections from theNorth Pole N .

Note that P = P ′ if and only if P ∈ Ω.

126 CHAPTER 15. SPHERICAL GEOMETRY

Note that if Σ and Π as above. Then the stereographic projectionsΣ → Π and Π → Σ from S are the restrictions of the inversion in thesphere with center S and radius

√2 to Σ and Π correspondingly.

From above and Theorem 15.3, it follows that the stereographicprojection preserves the angles between arcs; more precisely the abso-lute value of the angle measure between arcs on the sphere.

This makes it particularly useful in cartography. A map of a bigregion of earth can not be done in the constant scale, but using stere-ographic projection, one can keep the angles between roads the sameas on earth.

In the following exercises, we assume that Σ, Π, Ω, O, S and Nare as above.

15.6. Exercise. Show that the composition of stereographic projec-tions from Π to Σ from S and from Σ to Π from N is the inversionof the plane Π in Ω.

15.7. Exercise. Show that stereographic projection Σ → Π sends thegreat circles to circlines on the plane which intersects Ω at two oppositepoints.

15.8. Exercise. Fix a point P ∈ Π and let Q be yet an other pointin Π. Denote by P ′ and Q′ their stereographic projections to Σ. Setx = PQ and y = P ′Q′

s. Show that

limx→0

y

x=

2

1 +OP 2.

Compare with Lemma 12.8.

Central projection

Let Σ be the unit sphere centered at the origin which will be denotedas O. Denote by Π+ the plane described by equation z = 1. This planeis parallel to xy-plane and it pass through the North Pole N = (0, 0, 1)of Σ.

Recall that north hemisphere of Σ, is the subset of points (x, y, z) ∈∈ Σ such that z > 0. The north hemisphere will be denoted as Σ+.

Given a point P ∈ Σ+, consider half-line [OP ) and denote by P ′

the intersection of [OP ) and Π+. Note that if P = (x, y, z) thenP ′ = (x

z, yz, 1). It follows that P 7→ P ′ is a bijection between Σ+ and

Π+.The described map Σ+ → Π+ is called central projection of hemi-

sphere Σ+.

127

In spherical geometry, central projection is analogous to the pro-jective model of hyperbolic plane which is discussed in Chapter 16.

Note that the central projection sends intersections of great circleswith Σ+ to the lines in Π+. The later follows since great circles areformed by intersection of Σ with planes passing through the origin andthe lines in Π+ are formed by intersection of Π+ with these planes.

15.9. Exercise. Assume that N is the North Pole and sNBC hasright angle at C and lies completely in the north hemisphere. LetNB′C′ be the image of sNBC under central projection.

Observe that NB′C′ has right angle at C′.Set

a = BCs, b = CNs, c = NBs,

s = B′C′, t = C′N, u = NB′.

Show that

s = tg acos b , t = tg b, u = tg c.

Use these identities together with the Euclidean Pythagorean theo-rem

u2 = s2 + t2

for NB′C′ to prove spherical Pythagorean theorem

cos c = cos a· cos b

for sNBC.

The following exercise is analogous to Exercise 16.4 in hyperbolicgeometry.

15.10. Exercise. Let sABC be a nondegenerate spherical triangle.Assume that the plane Π+ is parallel to the plane passing through A,B and C. Denote by A′, B′ and C′ the central projections of A, Band C.(a) Show that the midpoints of [A′B′], [B′C′] and [C′A′] are central

projections of the midpoints of [AB]s, [BC]s and [CA]s corre-spondingly.

(b) Use part (a) to show that medians of spherical triangle intersectat one point.

Chapter 16

Projective model

The projective model is an other model of hyperbolic plane discov-ered by Beltrami; it is often called Klein model. The projective andconformal models are saying exactly the same thing but in two differ-ent languages. Some problems in hyperbolic geometry admit simplerproof using the projective model and others have simpler proof in theconformal model. Therefore it worth to know both.

Special bijection of h-plane to itself

Consider the conformal disc model with absolute at the unit circle Ωcentered at O. Choose a coordinate system (x, y) on the plane withorigin at O, so the circle Ω is described by the equation x2 + y2 = 1.

O P

N

S

P ′

P

The plane through P , O and S.

Let us think of our plane Π as itlies in the Euclidean space as the xy-plane. Denote by Σ the unit spherecentered at O; it is described by theequation

x2 + y2 + z2 = 1.

Set S = (0, 0,−1) and N = (0, 0, 1);these are the South and North Polesof Σ.

Consider stereographic projec-tion Π → Σ from S; given pointP ∈ Π denote its image in Σ as P ′.Note that the h-plane is mapped to

the North Hemisphere; that is, to the set of points (x, y, z) in Σ de-scribed by inequality z > 0.

128

129

For a point P ′ ∈ Σ consider its foot point P on Π; this is the closestpoint on Π from P ′.

The composition P 7→ P of these two maps is a bijection of h-planeto itself.

Note that P = P if and only if P ∈ Ω or P = O.

16.1. Exercise. Show that the map P 7→ P described above can bedescribed the following way: set O = O and for any other point pointP take P ∈ [OP ) such that

OP =2·x

1 + x2,

where x = OP .

16.2. Lemma. Let (PQ)h be an h-line with the ideal points A andB. Then P , Q ∈ [AB].

Moreover

➊AQ·BP

QB ·PA=

(

AQ·BP

QB ·PA

)2

.

In particular, if A,P,Q,B appear on the line in the same order then

PQh = 12 · ln

AQ·BP

QB ·PA.

Proof. Consider the stereographic projection Π → Σ from the SouthPole. Denote by P ′ and Q′ the images of P and Q.

A BP

P ′

The plane Λ.

According to Theorem 15.3(c),

➋AQ·BP

QB ·PA=

AQ′ ·BP ′

Q′B ·P ′A.

By Theorem 15.3(e), each circline in Πwhich is perpendicular to Ω is mapped to acircle in Σ which is still perpendicular to Ω.It follows that the stereographic projectionsends (PQ)h to the intersection of the northhemisphere of Σ with a plane, say Λ, perpen-dicular to Π.

Consider the plane Λ. It contains points A, B, P ′, P and the circleΓ = Σ∩Λ. (It also contains Q′ and Q but we will not use these pointsfor a while.)

Note that

130 CHAPTER 16. PROJECTIVE MODEL

⋄ A,B, P ′ ∈ Γ,

⋄ [AB] is a diameter of Γ,⋄ (AB) = Π ∩ Λ,

⋄ P ∈ [AB]⋄ (P ′P ) ⊥ (AB).

Since [AB] is the diameter of Γ, by Corollary 8.6, the angle ∠AP ′Bis right. Hence APP ′ ∼ AP ′B ∼ P ′PB. In particular

AP ′

BP ′ =AP

P ′P=

P ′P

BP.

Therefore

➌AP

BP=

(

AP ′

BP ′

)2

.

The same way we get

➍AQ

BQ=

(

AQ′

BQ′

)2

.

Finally note that ➋+➌+➍ imply ➊.The last statement follows from ➊ and the definition of h-distance.

Indeed,

PQhdef== ln

AQ·BP

QB ·PA=

= ln

(

AQ·BP

QB ·PA

)1

2

=

= 12 · ln

AQ·BP

QB ·PA.

A1

B1

A2

B2

A3

B3

Γ1

Γ2

Γ3

Ω

16.3. Exercise. Let Γ1, Γ2 andΓ3 be three circles perpendicularto the circle Ω. Let us denoteby [A1B1], [A2B2] and [A3B3] thecommon chords of Ω and Γ1, Γ2,Γ3 correspondingly. Show that thechords [A1B1], [A2B2] and [A3B3]intersect at one point inside Ω ifand only if Γ1, Γ2 and Γ3 inter-sect at two points.

131

Projective model

The following picture illustrates the map P 7→ P described in theprevious section. If you take the picture on the left and apply the mapP 7→ P , you get the picture on the right. The picture on the right givesa new way to look at the hyperbolic plane, which is called projectivemodel. One may think of the map P 7→ P as about translation fromone language to the other.

Conformal model Projective model

In the projective model things look different; some become simpler,other things become more complicated.

Lines. The h-lines in the projective model are formed by chords.More precisely, they are formed by the intersections of chords of theabsolute with the h-plane.

Circles and equidistants. The h-circles and equidistants in theprojective model are formed by the certain type of ellipses and theirarcs. It follows since the stereographic projection sends circles one theplane to the circles on the unit sphere and the foot point projectionof circle back to plane is formed by ellipse. (One may define ellipseas the foot point projection of a circle which lies in the space to theplane.)

Distance. To find the h-distance between the points P and Q in theprojective model, you have to find the points of intersection, say A andB, of the Euclidean line (PQ) with the absolute; then, by Lemma 16.2,

PQh = 12 · ln

AQ·BP

QB ·PA,

assuming the points A,P,Q,R appear on the line in the same order.

Angles. The angle measures in the projective model are very differentfrom the Euclidean angles and it is hard to figure out by looking on


the picture. For example all the intersecting h-lines on the picture areperpendicular. There are two useful exceptions

⋄ If O is the center of absolute then

∡hAOB = ∡AOB.

⋄ If O is the center of absolute and ∡OAB = ±π2 then

∡hOAB = ∡OAB = ±π2 .

A B

P Q

To find the angle measure in the pro-jective model, you may apply a motion ofh-plane which moves the vertex of the an-gle to the center of absolute; once it is donethe hyperbolic and Euclidean angles havethe same measure.

Motions. The motions of h-plane in theconformal and projective models are rele-vant to inversive transformations and pro-jective transformation in the same way.Namely:

⋄ Any inversive transformations which preserve h-plane describe amotion of h-plane in the conformal model.

⋄ Any projective transformation which preserve h-plane describesa motion in the projective model.

The following exercise is hyperbolic analog of Exercise 15.10. Thisis the first example of a statement which admits an easier proof usingthe projective model.

16.4. Exercise. Let P and Q be the point in h-plane which lie onthe same distance from the center of absolute. Observe that in theprojective model, h-midpoint of [PQ]h coincides with the Euclideanmidpoint of [PQ]h.

Conclude that if an h-triangle is inscribed in an h-circle then itsmedians intersect at one point.

Think how to prove the same for a general h-triangle.

m

ℓs

t

16.5. Exercise. Let ℓ and m are h-lines inthe projective model. Denote by s and t the Eu-clidean lines tangent to the absolute at the idealpoints of ℓ and denote by m the Euclidean linewhich contains m. Show that if the lines s, tand m meet at one point then ℓ and m are per-

pendicular h-lines in the projective model.

133

16.6. Exercise. Give a proof of Proposition 12.1 using projectivemodel.

16.7. Exercise. Use projective model to find the inradius of the idealtriangle.

Hyperbolic Pythagorean theorem

16.8. Theorem. Assume that hACB is a triangle in h-plane withright angle at C. Set a = BCh, b = CBh and c = ABh. Then

➎ ch c = ch a· ch b.

where ch denotes hyperbolic cosine; that is, the function defined thefollowing way

chxdef== ex+e−x

2 .

A

B

C

s

t

u

X

Y

Proof. We will use projective model of h-plane witha unit circle as the absolute.

We can assume that A is the center of absolute.Therefore both ∠hACB and ∠ACB are right.

Set s = BC, t = CA, u = AB. According toEuclidean Pythagorean theorem (6.10),

u2 = s2 + t2.

Note that

b = 12 · ln

1 + t

1− t;

therefore

➏ch b =

(

1+t1−t

)1

2

+(

1−t1+t

)1

2

2=

=1√

1− t2.

The same way we get

c = 12 · ln

1 + u

1− u


and

➐ch c =

(

1+u1−u

)1

2

+(

1−u1+u

)1

2

2=

=1√

1− u2.

Let X and Y are the ideal points of (BC)h. Applying the Pythag-orean theorem (6.10) again, we get

CX = CY =√

1− t2.

Therefore

a = 12 · ln

√1− t2 + s√1− t2 − s

and

➑

ch a =

(√1−t2+s√1−t2−s

)1

2

+(√

1−t2−s√1−t2+s

)1

2

2=

=

√1− t2√

1− t2 − s2

=

√1− t2√1− u2

Finally note that ➏+➐+➑ implies ➎.

Bolyai’s construction

Assume we need to construct a line asymptotically parallel to the givenline through the given point. The initial configuration is given by threepoints, say P , A and B and we need to construct a line through Pwhich is asymptotically parallel to ℓ = (AB).

Note that ideal points do not lie in the h-plane, so there is no wayto use them in the construction.

The following construction was given by Bolyai. We assume thatyou know a compass-and-ruler construction of the perpendicular line tothe given line through the given point; see the solution of Exercise 5.23.

16.9. Bolyai’s construction.

1. Construct the line m through P which perpendicular to ℓ. Denoteby Q the foot point of P on ℓ.

2. Construct the line n through P which perpendicular to m.

135

3. Draw the circle Γ1 with center Q through P and mark by R apoint of intersection of Γ1 with ℓ.

4. Construct the line k through R which perpendicular to n.5. Draw the circle Γ2 with center P through Q and mark by T a

point of intersection of Γ2 with k.6. The line (PT )h is asymptotically parallel to ℓ.

16.10. Exercise. Explain what happens if one performs the Bolyaiconstruction in the Euclidean plane.

To prove that Bolyai’s construction gives the asymptotically par-allel line in h-plane, it is sufficient to show the following.

16.11. Proposition. Assume P , Q, R, S, T be points in h-planesuch that

⋄ S ∈ (RT )h,⋄ (PQ)h ⊥ (QR)h,⋄ (PS)h ⊥ (PQ)h,⋄ (RT )h ⊥ (PS)h and⋄ (PT )h and (QR)h are asymptotically parallel.

Then QRh = PTh.

Proof. We will use the projective model. Without loss of generality,we may assume that P is the center of absolute. As it was notedon page 132, in this case the corresponding Euclidean lines are alsoperpendicular; that is, (PQ) ⊥ (QR), (PS) ⊥ (PQ) and (RT ) ⊥(PS).

Denote by A be the ideal point of (QR)h and (PT )h. Denote by Band C the remaining ideal points of (QR)h and (PT )h correspondingly.

P

QR

S

T

AB

C

Note that the Euclidean lines (PQ),(TR) and (CB) are parallel.

Therefore

AQP ∼ ART ∼ ABC.

In particular,

AC

AB=

AT

AR=

AP

AQ.

It follows that

AT

AR=

AP

AQ=

BR

CT=

BQ

CP.

In particular,AT ·CP

TC ·PA=

AR·BQ

RB ·QA;

hence QRh = PTh.

Chapter 17

Complex coordinates

In this chapter we give an interpretation of inversive geometry usingcomplex coordinates. The results of this chapter will not be used inthis book, but they lead to deeper understanding of both concepts.

Complex numbers

Informally, a complex number is a number that can be put in the form

➊ z = x+ i·y,where x and y are real numbers and i2 = −1.

The set of complex numbers will be further denoted by C. If x, yand z as in ➊, then x is called the real part and y the imaginary partof the complex number z. Briefly it is written as

x = Re z and y = Im z.

On the more formal level, a complex number is a pair of real num-bers (x, y) with addition and multiplication described below. Theformula x+ i·y is only a convenient way to write the pair (x, y).

(x1 + i·y1) + (x2 + i·y2) def== (x1 + x2) + i·(y1 + y2);

(x1 + i·y1)·(x2 + i·y2) def== (x1 ·x2 − y1 ·y2) + i·(x1 ·y2 + y1 ·x2).

Complex coordinates

Recall that one can think of Euclidean plane as the set of all pairs ofreal numbers (x, y) equipped with the metric

AB =√

(xA − xB)2 + (yA − yB)2

136

137

where A = (xA, yA) and B = (xB , yB).One can pack coordinates (x, y) of a point in the Euclidean plane,

in one complex number z = x + i·y. This way we get one-to-onecorrespondence between points of Euclidean plane and C. Given apoint Z = (x, y), the complex number z = x + i·y is called complexcoordinate of Z.

Note that if O, E and I are the points in the plane with complexcoordinates 0, 1 and i then ∡EOI = ±π

2 . Further we assume that∡EOI = π

2 ; if not, one has to change the direction of the y-coordinate.

Conjugation and absolute value

Let z = x + i·y and both x and y are real. Denote by Z the point inthe plane with complex coordinate z.

If y = 0, we say that z is a real and if x = 0 we say that z is animaginary complex number. The set of points with real and imaginarycomplex coordinates form lines in the plane, which are called real andimaginary lines which will be denoted as R and i·R.

The complex number z = x− iy is called complex conjugate of z.Note that the point Z with complex coordinate z is the reflection

of Z in the real line.It is straightforward to check that

➋ x = Re z =z + z

2, y = Im z =

z − z

i·2 , x2 + y2 = z ·z.

The last formula in ➋ makes possible to express the quotient wzof

two complex numbers w and z = x+ i·y:w

z= 1

z ·z ·w· z = 1x2+y2 ·w·z.

Note that

z + w = z + w, z − w = z − w, z ·w = z ·w, z/w = z/w;

that is, complex conjugation respects all the arithmetic operations.The value

|z| =√

x2 + y2 =

=√z ·z

is called absolute value of z. If |z| = 1 then z is called unit complexnumber.

Note that if Z and W are points in the Euclidean plane and z andw their complex coordinates then

ZW = |z − w|.

138 CHAPTER 17. COMPLEX COORDINATES

Euler’s formula

Let α be a real number. The following identity is called Euler’s for-mula.

➌ ei·α = cosα+ i· sinα.

In particular, ei·π = −1 and ei·π

2 = i.

0 1

i

ei·α

α

Geometrically Euler’s formula meansthe following. Assume that O and Eare the point with complex coordinates 0and 1 correspondingly. Assume OZ = 1and ∡EOZ ≡ α then ei·α is the com-plex coordinate of Z. In particular, thecomplex coordinate of any point on theunit circle centered at O can be uniquelyexpressed as ei·α for some α ∈ (−π, π].

Why should you think that ➌ is

true? The proof of Euler’s identity de-pends on the way you define exponent. If you never had to takeexponent of imaginary number, you may take the right hand side in➌ as the definition of the ei·α.

In this case formally nothing has to be proved, but it is better tocheck that ei·α the satisfies familiar identities. For example

ei·α ·ei·β = ei·(α+β).

The later can be proved using the following trigonometric formulas,which we assume to be known:

cos(α + β) = cosα· cosβ − sinα· sinβsin(α + β) = sinα· cosβ + cosα· sinβ

If you know power series for sine, cosine and exponent, the followingmight also convince that ➌ is the right definition.

ei·α = 1 + i·α+(i·α)22!

+(i·α)33!

+(i·α)44!

+(i·α)55!

+ · · · =

= 1 + i·α− α2

2!− i·α

3

3!+

α4

4!+ i·α

5

5!− · · · =

=

(

1− α2

2!+

α4

4!− · · ·

)

+ i·(

α− α3

3!+

α5

5!− · · ·

)

=

= cosα+ i· sinα.

139

Argument and polar coordinates

As above, assume that O and E denote the points with complex co-ordinates 0 and 1 correspondingly.

Let Z be the point distinct form O. Set ρ = OZ and ϑ = ∡EOZ.The pair (ρ, ϑ) is called polar coordinates of Z.

If z is the complex coordinate of Z then then ρ = |z|. The value ϑis called argument of z (briefly, ϑ = arg z). In this case

z = ρ·ei·ϑ = ρ·(cosϑ+ i· sinϑ).

Note that

arg(z ·w) ≡ arg z + argw and arg zw≡ arg z − argw

if z, w 6= 0. In particular, if Z, V , W be points with complex coordi-nates z, v and w correspondingly then

➍∡V ZW = arg

(

w − z

v − z

)

≡

≡ arg(w − z)− arg(v − z)

once the left hand side is defined.

17.1. Exercise. Use the formula ➍ to show that

∡ZVW + ∡VWZ + ∡WZV ≡ π

for any ZVW in the Euclidean plane.

17.2. Exercise. Assume that points V , W and Z have complex coor-dinates v, w and v·w correspondingly and the point O and E as above.Show that

OEV ∼ OWZ.

The following Theorem is a reformulation of Theorem 8.11 whichuse complex coordinates.

17.3. Theorem. Let UVWZ be a quadrilateral and u, v, w and zbe the complex coordinates of its vertices. Then UVWZ is inscribedif and only if the number

(v − u)·(z − w)

(v − w)·(z − u)

is real.


The value (v−u)·(w−z)(v−w)·(z−u) is called complex cross-ratio, it will be dis-

cussed in more details below.

17.4. Exercise. Observe that the complex number z 6= 0 is real ifand only if arg z = 0 or π; in other words, 2· arg z ≡ 0.

Use this observation to show that Theorem 17.3 is indeed a refor-mulation of Theorem 8.11.

Mobius transformations

17.5. Exercise. Watch video “Mobius Transformations Revealed”by Douglas Arnold and Jonathan Rogness. (It is 3 minutes long andavailable on YouTube.)

The complex plane C extended by one ideal number ∞ is calledextended complex plane. It is denoted by C, so C = C ∪ ∞

Mobius transformation of C is a function of one complex variablez which can be written as

f(z) =a·z + b

c·z + d,

where the coefficients a, b, c, d are complex numbers satisfying a·d−− b·c 6= 0. (If a·d − b·c = 0 the function defined above is a constantand is not considered to be a Mobius transformation.)

In case c 6= 0, we assume that

f(−d/c) = ∞ and f(∞) = a/c;

and if c = 0 we assumef(∞) = ∞.

Elementary transformations

The following three types of Mobius transformations are called ele-mentary.

1. z 7→ z + w,2. z 7→ w·z for w 6= 0,3. z 7→ 1

z.

The geometric interpretations. As before we will denote by O thepoint with complex coordinate 0.

The first map z 7→ z +w, corresponds to so called parallel transla-tion of Euclidean plane, its geometric meaning should be evident.

http://youtu.be/0z1fIsUNhO4

141

The second map is called rotational homothety with center at O.That is, the point O maps to itself and any other point Z maps to apoint Z ′ such that OZ ′ = |w|·OZ and ∡ZOZ ′ = argw.

The third map can be described as a composition of inversion inthe unit circle centered at O and the reflection in R (the compositioncan be taken in any order). Indeed, arg z ≡ − arg 1

ztherefore

arg z = arg(1/z);

that is, if the points Z and Z ′ have complex coordinates z and 1/z thenZ ′ ∈ [OZ). Clearly OZ = |z| and OZ ′ = |1/z| = 1

|z| . Therefore Z ′ is

inversion of Z in the unit circle centered at O. Finally the reflectionof Z ′ in R, has complex coordinate 1

z= (1/z).

17.6. Proposition. A map f : C → C is a Mobius transformation ifand only if it can be expressed as a composition of elementary Mobiustransformation.

Proof; (⇒). Consider, the Mobius transformation

f(z) =a·z + b

c·z + d.

It is straightforward to check that

➎ f(z) = f4 f3 f2 f1(z),

where⋄ f1(z) = z + d

c,

⋄ f2(z) =1z,

⋄ f3(z) = −a·d−b·cc2

·z,⋄ f4(z) = z + a

c

if c 6= 0 and⋄ f1(z) =

ad·z,

⋄ f2(z) = z + bd,

⋄ f3(z) = f4(z) = zif c = 0.

(⇐). We need to show that composing elementary transformations,we can only get Mobius transformations. Note that it is sufficient tocheck that composition of a Mobius transformations

f(z) =a·z + b

c·z + d.

with any elementary transformation is a Mobius transformations.


The later is done by means of direct calculations.

a·(z + w) + b

c·(z + w) + d=

a·z + (b + a·w)c·z + (d+ c·w) ,

a·(w·z) + b

c·(w·z) + d=

(a·w)·z + b

(c·w)·z + d,

a· 1z+ b

c· 1z+ d

=b·z + a

d·z + c.

17.7. Corollary. The image of circline under Mobius transformationis a circline.

Proof. By Proposition 17.6, it is sufficient to check that each elemen-tary transformation sends circline to circline.

For the first and second elementary transformation the later isevident.

As it was noted above, the map z 7→ 1zis a composition of inversion

and reflection. By Theorem 9.11, inversion sends circline to circline.Hence the result follows.

17.8. Exercise. Show that inverse of Mobius transformation is aMobius transformation.

17.9. Exercise. Given distinct values z0, z1, z∞ ∈ C, construct aMobius transformation f such that f(z0) = 0, f(z1) = 1 and f(z∞) == ∞. Show that such transformation is unique.

17.10. Exercise. Show that any inversion is complex conjugate to aMobius transformation.

Complex cross-ratio

Given four distinct complex numbers u, v, w, z, the complex number

(u− w)·(v − z)

(v − w)·(u− z)

is called complex cross-ratio; it will be denoted as (u, v;w, z).If one of the numbers u, v, w, z, is ∞, then the complex cross-ratio

has to be defined by taking the appropriate limit; in other words, weassume that ∞

∞ = 1. For example,

(u, v;w,∞) =(u− w)

(v − w).

143

Assume that U , V , W and Z be the points with complex coordi-nates u, v, w and z correspondingly. Note that

UW ·V Z

VW ·UZ= |(u, v;w, z)|,

∡WUZ + ∡ZVW = argu− w

u− z+ arg

v − z

v − w≡

≡ arg(u, v;w, z).

It makes possible to reformulate Theorem 9.6 using the complexcoordinates the following way.

17.11. Theorem. Let UWV Z and U ′W ′V ′Z ′ be two quadrilat-erals such that the points U ′, W ′, V ′ and Z ′ are inversions of U , W ,V , and Z correspondingly. Assume u, w, v, z, u′, w′, v′ and z′ be thecomplex coordinates of U , W , V , Z, U ′, W ′, V ′ and Z ′ correspond-ingly.

Then(u′, v′;w′, z′) = (u, v;w, z).

The following Exercise is a generalization of the Theorem above.It admits a short and simple solution which use Proposition 17.6.

17.12. Exercise. Show that complex cross-ratios are invariant un-der Mobius transformations. That is, if a Mobius transformation mapsfour distinct complex numbers u, v, w, z to complex numbers u′, v′, w′, z′

respectively, then

(u′, v′;w′, z′) = (u, v;w, z).

Schwarz–Pick theorem

The following theorem shows that the metric in the conformal discmodel naturally appears in other branches of mathematics. We donot give a proof, but it can be found in any textbook on geometriccomplex analysis.

Let us denote by D the unit disc in the complex plane centered at0; that is, a complex number z belongs to D if and only if |z| < 1.

Let us use the disc D as h-plane in the conformal disc model; theh-distance between z, w ∈ D will be denoted as dh(z, w).

A function f : D → C is called holomorphic if for every z ∈ D thereis a complex number s such that

f(z + w) = f(z) + s·w + o(|w|).


In other words, f is complex-differentiable at any z ∈ D. The numbers above is called derivative of f at z and denoted as f ′(z).

17.13. Schwarz–Pick theorem. Assume f : D → D is a holomor-phic function. Then

dh(f(z), f(w)) 6 dh(z, w)

for any z, w ∈ D.Moreover if equality holds for one pair of distinct numbers z, w ∈ D

then it holds for any pair and f : D → D is a motion of the h-plane.

17.14. Exercise. Show that Schwarz lemma stated below follows fromSchwarz–Pick theorem.

17.15. Schwarz lemma. Let f : D → D be a holomorphic functionand f(0) = 0. Then |f(z)| 6 |z| for any z ∈ D.

Moreover if equality holds for some z 6= 0 then there is a unitcomplex number u such that f(z) = u·z for any z ∈ D.

Chapter 18

Geometric constructions

Geometric constructions has great pedagogical value as an introduc-tion to mathematical proofs. We were using construction problemseverywhere starting from Chapter 5.

In this chapter we briefly discuss the classical results in geometricconstructions.

Classical problems

Here we list couple of classical construction problems; each known formore than thousand years. The solutions of the following two problemsare quite nontrivial.

18.1. Brahmagupta’s problem. Construct an inscribed quadri-lateral with given sides.

18.2. Apollonius’s problem. Construct a circle which is tangentto three given circles in a plane.

The following exercise is a simplified versionof Apollonius’s problem, which is still nontrivial.

18.3. Exercise. Construct a circle which passthrough given point and tangent to two intersect-ing lines.

The following three problems can not be solved in principle; thatis, the needed compass-and-ruler construction does not exist.

Doubling the cube. Construct the side of a new cube, which hasvolume twice bigger than the volume of the given cube.

145

146 CHAPTER 18. GEOMETRIC CONSTRUCTIONS

In other words, given a segment of length a one needs to constructa segment of length 3

√2·a.

Squaring the circle. Construct a square with the same area as agiven circle.

If r is the radius of the given circle, we need to construct a segmentof length

√π ·r.

Angle trisection. Divide the given angle into three equal angles.

In fact there is no compass-and-ruler construction which trisectsangle with measure π

3 . Existence of such construction would implyconstructability of regular 9-gon which is prohibited by the followingfamous result.

18.4. Gauss–Wantzel theorem. A regular n-gon is constructablewith ruler and compass if and only if n is the product of a power of 2and any number of distinct Fermat primes.

The Fermat prime is a prime numbers of the form 2k +1 for someinteger k. Only five Fermat primes are known today:

3, 5, 17, 257, 65537.

For example,⋄ one can construct a regular 34-gon since 34 = 2·17 and 17 is aFermat prime;

⋄ one can not construct a regular 7-gon since 7 is not a Fermatprime;

⋄ one can not construct a regular 9-gon; although 9 = 3·3 is aproduct of two Fermat primes, these primes are not distinct.

The impossibility of these constructions was proved only in 19thcentury. The method used in the proofs is indicated in the next section.

Constructable numbers

In the classical compass-and-ruler constructions initial configurationcan be completely described by a finite number of points; each line isdefined by two points on it and each circle is described by its centerand a point on it (equivalently, you may describe a circle by tree pointson it).

The same way the result of construction can be described by afinite collection of points.

Choose a coordinate system, such that one of the initial points asthe origin (0, 0) and yet an other initial point has coordinates (1, 0). In

147

this coordinate system, the initial configuration of n points is describedby 2·n − 4 numbers, their coordinates, say (x3, y3, x4, y4, . . . , xn, yn).It is straightforward to show that coordinates of any point constructedcan be written through the numbers x3, y3, x4, y4, . . . , xn, yn using thefour arithmetic operations “+”, “−”, “·”, “/” and the square root“√

”.For example, assume we want to find the points X1 = (x1, y1) and

X2 = (x2, y2) of intersection of a line passing through A = (xA, yA)and B = (xB , yB) and the circle with center O = (xO , yO) which passthrough a point W = (xW , yW ). Let us write the equations of thesecircle and line in the coordinates (x, y):

[

(x − xO)2 + (y − yO)

2 = (xW − xO)2 + (yW − yO)

2

(x− xA)·(yB − yA) = (y − yA)·(xB − xA)

Expressing y from the second equation and substituting the result inthe first one, gives us a quadratic equation in x, which can be solvedusing “+”, “−”, “·”, “/” and “

√” only.

The same can be performed for the intersection of two circles. Theintersection of two lines is even simpler; it described as a solution oftwo linear equations and can be expressed using only four arithmeticoperations; the square root “

√” is not needed.

Summarizing, the coordinates of any point which can be obtainedby a compass-and-ruler construction can be expressed in terms ofx3, y3, x4, y4, . . . , xn, yn and integers using only “+”, “−”, “·”, “/”and “

√”.

On the other hand, it is easy to produce compass-and-ruler con-structions which produce a segment of length a+ b and a− b from twogiven segments of lengths a > b.

A BD

CTo perform “·”, “/” and “

√”

consider the following diagram. Let[AB] be diameter of a circle; Con-sider point C on the circle and let Dbe the foot point of C on [AB]. Notethat

ABC ∼ ACD ∼ BDC.

It follows that AD·DC = BD2.Using this diagram, one should

guess the compass-and-ruler constructions which produce a segments

of lengths√a·b and a2

b. Say to construct

√a·b one can do the fol-

lowing: (1) construct points A, B and D ∈ [AB] such that AD = aand BD = b (2) construct a circle Γ with diameter [AB] (3) draw the


perpendicular ℓ through D to (AB) (4) let C be an intersection of Γand ℓ. Then DC =

√a·b.

Taking 1 for a or b above we can produce√a, a2, 1

b. Further com-

bining these constructions we can produce a·b = (√a·b)2, a

b= a· 1

b. In

other words we produced a compass-and-ruler calculator, which cando “+”, “−”, “·”, “/” and “

√”.

The discussion above gives a rough sketch of the proof of the fol-lowing theorem:

18.5. Theorem. Assume that initial configuration of geometric con-struction is given by points A1 = (0, 0), A2 = (1, 0), A3 = (x3, y3), . . .. . . , An = (xn, yn). Then a point X = (x, y) can be constructed usinga compass-and-ruler construction if and only if both coordinates x andy can be expressed from x3, y3, x4, y4, . . . , xn, yn using the arithmeticoperations “+”, “−”, “·”, “/” and the square root “

√”.

This theorem translates any compass-and-ruler construction prob-lem into purely algebraic language — we need to decide if the givennumber is constructable. For example:

⋄ Impossibility of solution for doubling cube problem states that3√2 is not a constructable number.

⋄ Impossibility of solution for squaring the circle states that√π,

or equivalently π, is not a constructable number.

⋄ Gauss–Wantzel theorem tells for which integers n the numbercos 2·π

nis constructable.

Some of these statements might look evident, but a rigorous proofsrequire some knowledge of field theory which is out of scope of thisbook.

In the next section we discuss a similar but simpler examples ofimpossible constructions with unusual tools.

Constructions with set square

Set square is a construction tool which can produce aline through the given point which makes angle π

2 or±π

4 with the given line.

18.6. Exercise. Trisect the given segment with rulerand set square.

Consider ruler-and-set-square construction. Using the same ideaas in the previous section, we can define ruler-and-set-square con-structable numbers and prove the following analog of Theorem 18.5.

149

18.7. Theorem. Assume that initial configuration of geometric con-struction is given by points A1 = (0, 0), A2 = (1, 0), A3 = (x3, y3), . . .. . . , An = (xn, yn). Then a point X = (x, y) can be constructed using aruler-and-set-square construction if and only if both coordinates x andy can be expressed from x3, y3, x4, y4, . . . , xn, yn using the arithmeticoperations “+”, “−”, “·”, “/”.

Note that if all the coordinates x2, y2, . . . , xn, yn are rational num-bers then the theorem above implies that with ruler and set squareone can only construct the points with rational coordinates. A pointwith both rational coordinates is called rational, and if at least one ofthe coordinates is irrational, then the point is called irrational.

18.8. Exercise. Show that an equilateral triangle in Euclidean planehas at least one irrational point.

Conclude that with ruler and set square one can not construct anequilateral triangle.

18.9. Exercise. Make a ruler-and-set-square construction which ver-ifies if the given triangle is equilateral. (We assume that we can “ver-ify” if two constructed points coincide.)

More impossible constructions

In this section we discuss yet an other source of impossible construc-tions.

Recall that a circumtool produces a circle passing through giventhree points or a line if all three points lie on one line. Let us restateExercise 9.9.

Exercise. Show that with circumtool only, it is impossible to constructthe center of given circle Γ.

Remark. In geometric constructions we allow to choose some freepoints, say any point on the plane, or a point on a constructed line,or a point which does not lie on a constructed line and so on.

In principle, when you make such a free choice it is possible to markthe center of Γ by an accident. Nether the less, we do not accept suchcoincidence as true construction; we say that a construction producethe center if it produce it for any free choices.

Solution. Arguing by contradiction, assume we have a construction ofthe center.

Apply to whole construction an inversion in a circle perpendicularto Γ. According to Corollary 9.15, Γ maps to itself. Since inversion


sends circline to circline, we get that all construction is mapped to anequivalent construction; that is, a constriction with different choice offree points.

According to Exercise 9.8, the inversion sends the center of Γ toan other point. That is, following the same construction, we can endup at a different point, a contradiction.

18.10. Exercise. Show that there is no circumtool-only constructionwhich verifies if the given point is the center of given circle. (Weassume that we can only “verify” if two constructed points coincide.)

A similar example of impossible constructions for ruler and paralleltool is given in Exercise 13.9.

Let us discuss yet an other example for ruler-only construction.Note that ruler-only constructions are invariant with respect to theprojective transformations. In particular, to solve the following exer-cise, it is sufficient to construct a projective transformation which fixtwo points A and B and moves its midpoint.

18.11. Exercise. Show that midpoint of given segment can not beconstructed with ruler only.

The following theorem is a stronger version of the exercise above.

18.12. Theorem. The center of given circle can not be constructedwith ruler only.

Proof. To prove this theorem it is sufficient to construct a projectivetransformation which sends the given circle Γ to a circle, say Γ′ butthe center of Γ′ is not the image of center of Γ.

Let Γ be a circle which lies in the plane Π in the Euclidean space.By Theorem 15.3, inversion of circle in a sphere is a circle or a line.

Fix a sphere Σ with center O so that the inversion Γ′ of Γ is a circleand the plane Π′ containing Γ′ is not parallel to Π; any sphere Σ ingeneral position will do.

Denote by Z and Z ′ the centers of Γ and Γ′. Note that Z ′ /∈ (OZ).It follows that the perspective projection Π → Π′ with center at Osends Γ to Γ′, but Z ′ is not the image of Z.

Construction of polar

Assume Γ is a circle in the plane. Given point P consider two chords[XX ′] and [Y Y ′] of Γ such that (XX ′)∩ (Y Y ′) = P . Let Z = (XY )∩(X ′Y ′) and Z ′ = (XY ′) ∩ (X ′Y ) and p = (ZZ ′). If P ∈ Γ we assumethat p is the tangent to Γ at P .

151

18.13. Claim. The line p = (ZZ ′) does not depend on the choice ofchords. Moreover P 7→ p is a duality (see page 118).

The line p is called polar of P with respect to Γ. The same waythe point P is called polar of the line p with respect to Γ.

P

p

X X ′

YY ′

Z

Z ′

We will not give a proof of this claim, butwill try to use it in the constructions.

18.14. Exercise. Let p be the polar line ofpoint P with respect to circle Γ. Assume thatp intersects Γ at points V and W . Show thatthe lines (PV ) and (PW ) are tangent to Γ.

Come up with a ruler-only construction ofthe tangent lines to the given circle through thegiven point.

18.15. Exercise. Assume two concentric circles Γ and Γ′ are given.Construct the common center of Γ and Γ′ with ruler only.

Chapter 19

Area

Any rigorous introduction to area is tedious and long. On the otherhand, there is a more general notion called Lebesgue measure. Theconstruction of Lebesgue measure is typically using the method ofcoordinates and it is included in any textbook in Real Analysis.

We use this construction without proof; namely we omit the proofTheorem 19.6 and it easily follows from the properties of Lebesguemeasure.

Solid triangles

We say that a point X lies inside a nondegenerate triangle ABC ifthe following three condition hold:

⋄ A and X lie on the same side from the line (BC);

⋄ B and X lie on the same side from the line (CA);

⋄ C and X lie on the same side from the line (AB).

AB

C

X

The set of all points inside ABC and on its sides[AB], [BC], [CA] will be called solid triangle and de-noted by ABC.

19.1. Exercise. Show that solid triangle is convex;that is if X,Y ∈ ABC then [XY ] ⊂ ABC.

The notations ABC and ABC look similar, they also haveclose but different meanings, which better not to be confused. Recallthat ABC is an ordered triple of distinct points (see page 17), whileABC is an infinite set of points.

In particular ABC = BAC; indeed any point which belongto the set ABC also belongs to the set BAC and the other way

152

153

around. On the other hand, ABC 6= BAC simply because thesequence of points (A,B,C) is distinct from the sequence (B,A,C).

Also, in general ABC ≇ BAC, but it is always true thatABC ∼= BAC, where congruence of the sets ABC and BACunderstood the following way.

19.2. Definition. Two sets S and T in the plane are called congruent(briefly S ∼= T ) if T = f(S) for some motion f of the plane.

If ABC is not degenerate and

ABC ∼= A′B′C′,

then after appropriate relabeling of the vertices of ABC we will have

ABC ∼= A′B′C′.

The proof follow from the following exercise.

19.3. Exercise. Let ABC be nondegenerate and X ∈ ABC.Show that X is a vertex of ABC if and only if there is a line ℓwhich intersects ABC at the single point X.

Polygonal sets

Elementary set on the plane is a set of one of the following three types:

⋄ one-point set;

⋄ segment;

⋄ solid triangle.

A set in the plane is called polygonal set if itcan be presented as a union of finite collection ofelementary sets.

A polygonal set is called degenerate if it can bepresented as union of finite number of one-point setsand segments.

According to this definition, empty set∅ is a polygonal set. Indeed,∅ is a union of empty collection of elementary sets.

If X and Y lie on the opposite sides of the line (AB) then theunion of solid triangles AXB ∪ BY A is a polygonal set whichis called solid quadrilateral and denoted as AXBY . In particular,we can talk about solid parallelograms, rectangles and squares in theEuclidean plane.

154 CHAPTER 19. AREA

One polyhedral set typically admitsmany presentation as union of finite col-lection of elementary sets. For example, ifAXBY is a parallelogram, then

AXBY = AXB ∪ BY A = XAY ∪ Y BX.

19.4. Exercise. Show that a circle is not a polygonal set.

Definition of area

19.5. Claim. For any two polygonal sets P and Q, the union P ∪Qas well as the intersection P ∩Q are also polygonal sets.

Semi-proof. Indeed, let us present the P and Q as a union of finite col-lection of elementary sets P1, . . . ,Pk and Q1, . . . ,Qn correspondingly.

Note that

P ∪Q = P1 ∪ · · · ∪ Pk ∪ Q1 ∪ · · · ∪ Qn.

Therefore P ∪ Q is polygonal.Note that the union of all sets Pi∩Qj forms P∩Q.Therefore in order to show that P∩Q is polygonal,

it is sufficient to show that each Pi∩Qj is polygonal.The diagram should suggest an idea of proof.

The following theorem defines the area as a function which returnsa real number for any polygonal set and satisfying certain conditions.

19.6. Theorem. For each polyhedral set P in the Euclidean planethere is a real number s called area of P (briefly s = areaP) such that

area∅ = 0 and areaK = 1

where K the solid square with unit side and the conditions

P ∼= Q ⇒ areaP = areaQ;

P ⊂ Q ⇒ areaP 6 areaQ;

areaP + areaQ = area(P ∪ Q) + area(P ∩ Q)

hold for any two polyhedral sets P and Q.Moreover the above conditions uniquely define the function

P 7→ areaP .

155

We omit the proof of this theorem. It follows from the constructionof Lebesgue measure which can be found in any text book on realanalysis.

Further you will see that based on this theorem, the concept ofarea can be painlessly developed to a dissent level without cheating.

19.7. Proposition. For any polyhedral set P in the Euclidean plane,we have

areaP > 0.

Proof. Since ∅ ⊂ P , we get

area∅ 6 areaP .

Since area∅ = 0 the result follows.

Vanishing area and subdivisions

19.8. Proposition. Any one-point set as well as any segment in theEuclidean plane have vanishing area.

Proof. Fix a line segment [AB]. Consider a sold square ABCD.Note that given a positive integer n, there are n disjoint segments

[A1B1], . . . , [AnBn] in ABCD, such that each [AiBi] is congruent to[AB] in the sense of the Definition 19.2.

A1

B1

. . .

. . .

An

Bn

Applying the last identity in Theorem 19.6 fewtimes, we get

n· area[AB] = area ([A1B1] ∪ · · · ∪ [AnBn]) 6

6 area(ABCD)

That is

area[AB] 6 1n· area(ABCD)

for any positive integer n. Therefore area[AB] 6 0.On the other hand, by Proposition 19.7,

area[AB] > 0.

Hence the result follows.For any one-point set A we have

∅ ⊂ A ⊂ [AB].


Therefore

0 6 areaA 6 area[AB] = 0.

Whence areaA = 0.

19.9. Corollary. Any degenerate polygonal set in the Euclidean planehas vanishing area.

Proof. Let P be a degenerate set, say

P = [A1B1] ∪ · · · ∪ [AnBn] ∪ C1, . . . , Ck.

Applying Theorem 19.6 together with Proposition 19.7, we get

areaP 6 area[A1B1] + · · ·+ area[AnBn]+

+ areaC1+ · · ·+ areaCk.

By Proposition 19.8, the right hand side vanish. Hence the statementfollows.

We say that polygonal set P is subdivided into two polygonal setsQ1 and Q2 if P = Q1∪Q2 and the intersection Q1∩Q2 is degenerate.(Recall that the intersection Q1 ∩Q2 has to be a polygonal set.)

19.10. Proposition. Assume polygonal set P is subdivided into twopolygonal set Q1 and Q2. Then

areaP = areaQ1 + areaQ2.

Proof. By Theorem 19.6,

areaP = areaQ1 + areaQ2 − area(Q1 ∩Q2).

Since Q1 ∩ Q2 is degenerate, by Corollary 19.9,

area(Q1 ∩ Q2) = 0.


Area of solid rectangles

19.11. Theorem. The solid rectangle in the Euclidean plane withsides a and b has area a·b.

157

19.12. Algebraic lemma. Assume that a function s returns a non-negative real number s(a, b) for any pair of positive real numbers (a, b)and satisfies the following identities

s(1, 1) = 1;

s(a, b+ c) = s(a, b) + s(a, c)

s(a+ b, c) = s(a, c) + s(b, c)

for any a, b, c > 0. Then

s(a, b) = a·b

for any a, b > 0.

The proof is similar to the proof of Lemma 13.7; we omit it.

Proof of Theorem 19.11. Denote by Ra,b the solid rectangle with sidesa and b. Set

s(a, b) = areaRa,b.

By theorem 19.6, s(1, 1) = 1. That is, the first identity in theAlgebraic Lemma holds.

Ra,c Rb,c

Ra+b,c

Note that the rectangle Ra+b,c can be subdi-vided into two rectangle congruent to Ra,b andRa,c. Therefore by Proposition 19.10,

areaRa+b,c = areaRa,c + areaRb,c

That is, the second identity in the Algebraic Lemma holds. The proofof the third identity is analogues.

It remains to apply Algebraic lemma.

Area of solid parallelograms

19.13. Proposition. Let ABCD be a parallelogram in the Eu-clidean plane, a = AB and h be the distance between the lines (AB)and (CD). Then

area(ABCD) = a·h.

Proof. Let us denote by A′ and B′ the foot points of A and B on theline (CD).

Note that ABB′A′ is a rectangle with sides a and h. Therefore, byProposition 19.11,

➊ area(ABB′A′) = h·a.


A B

CDA′ B′ Without loss of generality we may as-sume that ABCA′ contains ABCDand ABB′A′.

In this case ABB′D admits twosubdivisions. First into ABCD and

BB′C. Second into ABB′D and AA′D.By Proposition 19.10,

➋area(ABCD) + area(BB′C) =

= area(ABB′D) + area(AA′D).

Note that

➌. AA′D ∼= BB′C

Indeed, sinceABB′A′ andABCD are parallelogram, by Lemma 6.22,we have AA′ = BB′ AD = BC. Further by Transversal property 6.18,

∡DAA′ = ∡BAA′ − ∡BAD =

= ∡CBA− ∡B′BA =

= ∡CBB′.

Applying SAS congruence condition, we get ➌.In particular,

➍ area(BB′C) = area(AA′D).

A

B

C

D

B′

C′D′

Subtracting ➍ from ➋, we get

area(ABCD) = area(ABB′D).

From ➊, the statement follows.

19.14. Exercise. Assume ABCD and AB′C′D′

are two parallelograms such that B′ ∈ [BC] and D ∈∈ [C′D′]. Show that

area(ABCD) = area(AB′C′D′).

Area of solid triangles

19.15. Theorem. Given a triangle ABC in the Euclidean plane,set a = BC and hA to be the altitude from A. Then

area(ABC) = 12 ·a·hA.

159

Remark. SinceABC completely determines the solid triangle ABC,it is acceptable to write area(ABC) for area(ABC).

Proof. Draw the line m through A which is parallel to (BC) and linen through C parallel to (AB). Note that m ∦ n. Therefore we canchoose D = m ∩ n; by construction ABCD is a parallelogram.

A B

CD

m

nNote that parallelogram ABCD admits a sub-

division into ABC and CDA. Therefore

area(ABCD) = area(ABC) + area(CDA)

SinceABCD is a parallelogram, by Lemma 6.22we have

AB = CD and BC = DA.

Therefore by SSS congruence condition, we have ABC ∼= CDA.In particular

area(ABC) = area(CDA).

Therefore, from above and Proposition 19.13,

area(ABC) = 12 · area(ABCD) =

= 12 ·hA ·a

19.16. Exercise. Denote by hA, hB and hC the altitudes of ABCfrom vertices A, B and C correspondingly. Note that from Theo-rem 19.15, it follows that

hA ·BC = hB ·CA = hC ·AB.

Give a proof of this statement without using area.

19.17. Exercise. Assume M lies inside the parallelogram ABCD;that is M belongs to the solid parallelogram ABCD, but does not lieon its sides. Show that

area(ABM) + area(CDM) = 12 · area(ABCD).

19.18. Exercise. Assume that diagonals of a nondegenerate quadri-lateral ABCD intersect at point M . Show that

area(ABM)· area(CDM) = area(BCM)· area(DAM).

19.19. Exercise. Let r be the inradius of ABC and p be its semiperime-ter; that is p = 1

2 ·(AB +BC + CA). Show that

area(ABC) = p·r.


Area method

Slim proofs using area are often used to impress students. We willshow couple of such examples in this section, but one should be aware— these proofs are not truly elementary; to do these proofs, one payshigh price of introducing area..

Proof of the Pythagorean theorem via the area method. We need toshow that if a and b are legs and c is the hypotenuse of right trianglethen

a2 + b2 = c2.

Denote by the solid right triangle with legs a and b and by x

be the solid square with side x.Let us construct two subdivisions of a+b.

1. Subdivide a+b into two solidsquares congruent to a and b and 4solid triangles congruent to , see theleft diagram.

2. Subdivide a+b into one solidsquare congruent to c and 4 solidright triangles congruent to , see the

right diagram.Applying Proposition 19.10 few times, we get.

areaa+b = areaa + areab + 4· area =

= areac + 4· area.

Thereforeareaa + areab = areac.

Sinceareax = x2,

the statement follows.

19.20. Exercise. Build an other proof of Pythagoreantheorem based on the diagram.

(In the notations above it shows a subdivision of c

into a−b and 4 solid triangles congruent to .)

19.21. Exercise. Show that the sum of distances from a point tothe sides of an equilateral triangle is the same for all points inside thetriangle.

The following claim is simple but very useful.

161

19.22. Claim. Assume that two triangles ABC and A′B′C′ inthe Euclidean plane have equal altitudes dropped from A and A′ cor-respondingly. Then

A′B′C′

ABC=

B′C′

BC.

In particular, the same identity holds if A = A′ and the bases [BC]and [B′C′] lie on one line.

Proof. Let h be the altitude. By Theorem 19.15,

A′B′C′

ABC=

12 ·h·B′C′

12 ·h·BC

=B′C′

BC.

A

BC D

Lemma 7.7 via Area method. We have to showthat if ABC is nondegenerate and the bisectorof ∠BAC intersects [BC] at the point D. Then

AB

AC=

DB

DC.

Applying Claim 19.22, we get

area(ABD)

area(ACD)=

BD

CD.

By Lemma 5.13 the triangles ABD and ACD have equal alti-tudes from D. Applying Claim 19.22 again, we get

area(ABD)

area(ACD)=

AB

AC.


19.23. Exercise. Let X lies inside a nondegenerate triangle ABC.Show that X lies on the median from A if and only if

area(ABX) = area(ACX).

19.24. Exercise. Build a proof of Theorem 7.5 via area method basedon the exercise above.

Namely, show that medians of nondegenerate triangle intersect atone point and the point of their intersection divides each median inthe ration 1:2.


Remark on area in

absolute planes and spheres

The theorem Theorem 19.6 will hold in the absolute planes and spheresif the solid unit squareK exchanged to a fixed nondegenerate polygonalset. One has to make such change for good reason — hyperbolic planeand sphere have no unit squares.

The set K in this case plays role of is unit measure for the area andchanging K will require conversion of area units.

A

B C

D

Kn

1n

1n

According to the standard convention, the set K istaken so that on small scales area behaves like area inthe Euclidean plane. Say if Kn denotes the solid quadri-lateral ABCD in the with right angles at A, B and Cand AB = BC = 1

nthen we may assume that

➎ n2 · areaKn → 1 as n → ∞.

This convention works equally well for spheres and absolute planes,including Euclidean plane. In spherical geometry equivalently we mayassume that if r is the radius of the sphere then the area of wholesphere is 4·π·r2.

Index

∠, 14∞, 70∡, 14, 153, 152, 52, 17∼=, 17, 153≡, 15‖, 45⊥, 36∼, 46(PQ), [PQ), [PQ], 13(PQ)h, [PQ)h,[PQ]h, 87(u, v;w, z), 142

AA similarity condition, 47absolute, 87absolute plane, 77absolute value, 137acute

angle, 36triangle, 48

affine transformation, 106, 116altitude, 56angle, 14

angle of parallelism, 97negative angle, 25obtuse angle, 36positive angle, 25right angle, 36straight angle, 22vertical angles, 23

angle measure, 20hyperbolic angle measure, 88

arc, 64

area, 154ASA congruence condition, 32asymptotically parallel lines, 97

base, 33between, 22bijection, 12bisector

angle bisector, 40external bisector, 40perpendicular bisector, 36

center, 42center of the pencil, 113

central projection, 126centroid, 57ch, 133chord, 42circle, 42

circle of Apollonius, 68circle arc, 64circline, 70circumtool, 149collinear, 106collinear points, 116complex conjugate, 137concurrent, 113conformal disk model, 86conformal factor, 103congruent

sets, 153congruent triangles, 17consistent, 83convex set, 152cross-ratio, 68

complex cross-ratio, 140, 142

163

164 INDEX

curvature, 84

d1, d2, d∞, 11defect of triangle, 81degenerate

triangle, 23polygonal set, 153quadrilateral, 52

diagonals of quadrilateral, 52diameter, 42direct motion, 39discrete metric, 11distance, 11distance-preserving map, 12duality, 118

elementary set, 153elementary transformation, 140endpoint of arc, 64equidistant, 100equilateral triangle, 33equivalence relation, 47Euclidean metric, 11Euclidean plane, 20Euclidean space, 114Euler’s formula, 138excenter, 174

Fano plane, 121Fermat prime, 146field automorphism, 108finite projective plane, 121foot point, 37

great circle, 123

h-angle measure, 88h-circle, 91h-half-line, 87h-line, 87h-plane, 87h-point, 87h-radius, 91h-segment, 87half-plane, 27holomorphic function, 143horocycle, 100hyperbolic angle measure, 88

hyperbolic cosine, 133hyperbolic geometry, 83hyperbolic plane, 87hyperbolic sine, 104hypotenuse, 48

ideal point, 87identity map, 187imaginary complex number, 137imaginary line, 137incenter, 58incidence structure, 106incircle, 58indirect motion, 39inradius, 58inscribed triangle, 61inside

inside a circle, 42inside a triangle, 152

intersecting lines, 45inverse, 12inversion, 67

center of inversion, 67, 124circle of inversion, 67inversion in circline, 73inversion in a sphere, 124inversion in the line, 73sphere of inversion, 124

inversive plane, 70inversive space, 124inversive transformation, 110irrational point, 149isometry, 12isosceles triangle, 33

Klein mode, 128

leg, 48line, 13Lobachevskian geometry, 83

Mobius transformation, 140Manhattan metric, 11maximum metric, 11metric, 10metric space, 11motion, 12

INDEX 165

neutral plane, 77

obtuse angle, 36order of finite projective plane, 121orthic triangle, 59orthocenter, 56outside a circle, 42

parallel lines, 45ultra parallel lines, 97

parallel tool, 106parallel translation, 140parallelogram, 52

solid parallelogram, 153pencil, 113perpendicular, 36perpendicular bisector, 36perpendicular circlines, 72perspective projection, 114pint at infinity, 69plane

absolute plane, 77Euclidean plane, 20h-plane, 87hyperbolic plane, 87inversive plane, 70neutral plane, 77plane in the space, 114

Poincare disc mode, 86point, 11

ideal point, 87polar coordinates, 139polarity, 118polygonal set, 153

degenerate polygonal set, 153projective model, 131projective plane, 120projective transformation, 116

quadrilateral, 52degenerate quadrilateral, 52inscribed quadrilateral, 63solid quadrilateral, 153

radius, 42rational point, 149real complex number, 137real line, 11, 137

real projective plane, 113rectangle, 53

solid rectangle, 153reflection, 38rhombus, 53rotational homothety, 141ruler-and-compass construction, 43

SAA congruence condition, 79SAS congruence condition, 32SAS similarity condition, 47secant line, 42sh, 104side

side of quadrilateral, 52side of the triangle, 28

similar triangles, 46solid

quadrilateral, parallelogram, rect-angle, square, 153

triangle, 152sphere, 122spherical distance, 122square, 53

solid square, 153SSS congruence condition, 34SSS similarity condition, 47stereographic projection, 125subdivision of polygonal set, 156

tangentcircles, 42half-line, 64line, 42

transversal, 51triangle, 17

congruent triangles, 17degenerate triangle, 23ideal triangle, 99orthic triangle, 59right triangle, 48similar triangles, 46solid triangle, 152

unit complex number, 137

vertex of the angle, 14vertical angles, 23

Used resources

[1] Euclid’s Elements.[2] Eugenio Beltrami, Teoria fondamentale degli spazii di curvatura

costante, Annali. di Mat., ser II, 2 (1868), 232–255.[3] Birkhoff, George David, A Set of postulates for plane geometry,

based on scale and protractors, Annals of Mathematics 33 (1932),329–345.

[4] Marvin J. Greenberg, Euclidean and Non-Euclidean Geometries:Development and History, 4th ed., New York: W. H. Freeman,2007.

[5] Kiselev, A. P., Kiselev’s Geometry. Book I. Planimetry, Adaptedfrom Russian by Alexander Givental.

[6] Lambert, Johann Heinrich, Theorie der Parallellinien, F. Engel,P. Stackel (Eds.) (1786) Leipzig.

[7] Legendre, Adrien-Marie, Elements de geometrie, 1794.[8] Лобачевский, Никлай Иванович, О началах геометрии, Ка-

занский вестник, вып. 25–28 (1829–1830 гг.).[9] Lobachevsky, N. I., Geometrische Untersuchungen zur Theorie

der Parallellinien. Berlin: F. Fincke, 1840.[10] Moise, Edwin, Elementary Geometry From An Advanced Stand-

point, 3rd ed. Boston: Addison-Wesley. 1990.[11] Prasolov, Viktor, Problems in Plane and Solid Geometry, trans-

lated and edited by Dimitry Leites.[12] Saccheri, Giovanni Girolamo, Euclides ab omni nævo vindicatus,

1733.[13] Шарыгин, Игорь Федорович, Геометрия 7–9, М.: Дрофа,

1997.

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Documents

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triangle geometry

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hyperbolic plane