topics in geometry

MATH 348 TOPICS IN GEOMETRYCourse Outline, Lecture Notes, Further Reading, Assignments, Tests & Examinations

A B

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A B

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Norman Do Mathematics and Statistics

May 3 to June 1 Faculty of Science

Summer 2010 McGill University

MATH 348 TOPICS IN GEOMETRY CONTENTS

Contents

0 COURSE OUTLINE 3

1 EUCLIDEAN GEOMETRY 6

1.1 Euclids Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.2 Triangles and Circles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

1.3 Triangle Centres . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

1.4 Geometric Gems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

2 SYMMETRY IN GEOMETRY 50

2.1 Isometries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

2.2 Symmetry and Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

2.3 Symmetry in the Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

2.4 Crystals, Friezes and Wallpapers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

3 POLYHEDRA, GRAPHS AND SURFACES 93

3.1 From Polyhedra to Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93

3.2 Platonic Solids and Beyond . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102

3.2 Surfaces and Topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

3.4 The Classification of Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

4 TILING AND DISSECTION 125

4.1 Tiling Rectangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125

4.2 Scissors Congruence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136

5 FURTHER READING 142

6 ASSIGNMENTS, TESTS AND EXAMINATIONS 146

Assignment 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146

Assignment 1 Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148

Assignment 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157


Assignment 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169


Sample Midterm Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185

Sample Midterm Test Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192

Midterm Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204

Midterm Test Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211

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MATH 348 TOPICS IN GEOMETRY CONTENTS

Sample Final Examination . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223

Final Examination . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234

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0. COURSE OUTLINE

General Information

COURSE MATH 348 Topics in GeometrySEMESTER Summer 2010 Section 001PREREQUISITE MATH 133 or equivalent or permission of instructorCREDITS 3.000

INSTRUCTOR Norman DoOFFICE Burnside Hall 1125TELEPHONE 514 398 2998EMAIL [email protected] HOURS Tue 1400-1500, Thu 14001500 or by appointment

CLASSROOM Burnside Hall 1B24CLASS HOURS Mon 11051325, Tue 11051325, Wed 11051325, Thu 11051325

Monday 3 May to Tuesday 1 June except for Monday 24 MayWEBPAGE http://www.math.mcgill.ca/ndo/MATH-348.html

Learning Outcomes

I hope that, by the end of the course, students should

understand all of the topics in geometry described below;

know a little about some of the major players in mathematical history;

appreciate the mysterious notions of theorem and proof; and

think that maths is awesome.

Instructional Method

The course includes 16 sessions, each one lasting for an epic 140 minutes. These will generally consist of alecture in which I talk about the course material, a tutorial in which we talk about solving maths problems,and some much needed break in between.

Course Materials

There is no required textbook for the course, so you can save your hard-earned money. Instead, I will publishconcise notes of varying quality after each lecture on the course website, outlining the content of the lecture.However, these in no way constitute a replacement for going to the lectures. At various stages throughoutthe course, I might suggest books or websites which reinforce and/or complement the material delivered inlectures. The course website, which will also include electronic copies of any handouts, can be found at thefollowing URL.

http://www.math.mcgill.ca/ndo/MATH-348.html

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0. COURSE OUTLINE

Assignments and Evaluation

Your evaluation for this course will be based on assignments, a test, and an examination.

AssignmentsThere will be three assignments, each graded out of 20. You are not discouraged from talking aboutassignment problems with other students, but every solution that you hand in must be your ownwork.1 Every page submitted should clearly indicate your name, your student number, the coursenumber, and the assignment number.2 Late assignments will not be accepted, unless under particularlyextreme circumstances.

Assignment 1 will be handed out at the end of class on Tuesday 4 May and is due at 11:00am onTuesday 11 May. It will test the material delivered in lectures 1.1 to 1.4.



TestThere will be a test graded out of 40. You will not be allowed to use calculators, computers, notes, orother aids. No provision will be made for a student who is absent on the day of the test.

The test will occur in class on Thursday 20 May.It will test the material delivered in lectures 1.1 to 3.3.

ExaminationThe examination will be graded out of 100. You will not be allowed to use calculators, computers, notes,or other aids. There will not be a supplemental examination in this course.

The examination will occur on Wednesday 2 June or Thursday 3 June.It will test the material delivered in lectures 1.1 to 4.4.

Your final mark will consist of

50% examination + 30% assignments + 20% test OR 100% examination,

whichever is greater. Your final grade will be a letter grade based on your final mark.

Course Content

We will be looking at the following four rather different topics in geometry.

Euclidean Geometry

Symmetry in Geometry

Polyhedra, Graphs and Surfaces

Tiling and Dissection

In an ideal world, we would be able to cover all of the course content outlined below. More realistically, wemight have to deviate from the plan a little, depending on time constraints and student response.

1McGill University values academic integrity. Therefore, all students must understand the meaning and consequences ofcheating, plagiarism and other academic offences under the Code of Student Conduct and Disciplinary Procedures. Please seewww.mcgill.ca/students/srr/honest/ for more information.

2In accord with McGill Universitys Charter of Students Rights, students in this course have the right to submit in English or inFrench any written work that is to be graded.

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0. COURSE OUTLINE

1. EUCLIDEAN GEOMETRY

Mon 3 May 1.1. Euclids Elements. . . in which we start at the very beginning, where Euclid started,with the axioms and build the world of geometry

Tue 4 May 1.2. Triangles and Circles. . . in which we study two of the most basic and fundamentalobjects of geometry namely, triangles and circles

Wed 5 May 1.3. Triangle Centres. . . in which we discover that the humble triangle has at least fourvery special points, each with very special properties

Thu 6 May 1.4. Geometric Gems. . . in which we use our geometric knowledge to discover wondrousthings before returning to the very beginning, where Euclid started, with the axioms

2. SYMMETRY IN GEOMETRY

Mon 10 May 2.1. Isometries. . . in which we learn the mathematics of flipping, sliding, turning andgliding, as well as what this all has to do with symmetry

Tue 11 May 2.2. Symmetry and Groups. . . in which we begin with the simple notion of symmetryand end with the abstract algebraic notion of a group

Wed 12 May 2.3. Symmetry in the Plane. . . in which we examine symmetry in the plane and prove atheorem of Leonardo da Vinci

Thu 13 May 2.4. Crystals, Friezes and Wallpapers. . . in which we consider the symmetry of crystals,friezes those long decorations found where the wall meets the ceiling and wallpapers

3. POLYHEDRA, GRAPHS AND SURFACES

Mon 17 May 3.1. From Polyhedra to Graphs. . . in which we glue together polygons to make polyhedraand introduce the notion of a graph

Tue 18 May 3.2. Platonic Solids and Beyond. . . in which we hunt down the Platonic solids and seethat theres more to life than polyhedra

Wed 19 May 3.3. Surfaces and Topology. . . in which we define the notion of a hole, discover themysterious one-sided surfaces and play with the rubbery world of topology

Thu 20 May 3.4. The Classification of Surfaces. . . in which we hunt down all of the surfaces and thenconsider what its like to live in three dimensions

4. TILING AND DISSECTION

Tue 25 May 4.1. Tiling Rectangles. . . in which we decide whether or not a rectangle can be tiled withsmaller shapes, using coloured pencils and other tricks

Wed 26 May 4.2. Scissors Congruence. . . in which we explore the seemingly simple notions of areaand volume with the help of scissors and glue

Thu 27 May 4.3. Tiling the Plane. . . in which we find shapes which tile the plane and encounter somecrazy aperiodic tilings

Mon 31 May 4.4. My Favourite Problems. . . in which we see a handful of my favourite problems ingeometry, both solved and unsolved

REVIEW

Tue 1 June Review

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1. EUCLIDEAN GEOMETRY 1.1. Euclids Elements

Reductionism for Dummies

Today, Wikipedia told me that reductionism is an approach to understand the nature of complex things byreducing them to the interactions of their parts, or to simpler or more fundamental things. Here are twoexamples of reductionism at work which you may or may not be familiar with.

Prime factorisationThe Fundamental Theorem of Arithmetic says that every positive integer can be constructed bymultiplying prime numbers together. In fact, it even tells us that, for every positive integer, there isreally only one way to do this. So we can reduce every positive integer to its prime factorisation, whichacts like a fingerprint for that number. You probably know that one way to find the prime factorisationof a number is to write down its factor tree, like Ive done below for 48 = 2 2 2 2 3.

48

8 6

4 2 3 2

22

So in this case, the fundamental things are the prime numbers and it turns out that there are infinitelymany of them. The oldest known proof of this fact is a beautiful piece of thinking by a guy namedEuclid who lived a really really long time ago.

PhysicsTo understand the universe, it makes sense to explore the stuff that we observe this is calledmatter. Quite a while back, we discovered that matter is made up of molecules and, a little bit later, thatmolecules are made up of atoms. These atoms are, in turn, made up of smaller particles like electrons,neutrons and protons. And more recently, physicists have found that these particles are made up oftiny tiny things called quarks. There are six different flavours of quark known as up, down, charm,strange, top, and bottom.

In this overly simplistic view of physics, the fundamental things are the quarks and it turns out thatthere are only finitely many of them. I should mention that, even more recently still, some physicistsbelieve that quarks are actually just incredibly tiny pieces of vibrating string.

Wheres the Geometry?

So what does all this have to do with geometry? Well, suppose that youre trying to convince me of thesimple geometric fact that all three angles in an equilateral triangle are 60. Your argument would probablylook something like this. . .

First, you might use the fact that the base angles in an isosceles triangle are equal. But theequilateral triangle is isosceles all three ways, so this means that all three angles are equal toeach other. And then you might invoke the well-known fact that the angles in a triangle alwaysadd up to 180. Since you now know that the three angles in an equilateral triangle are equal andadd up to 180, every single one of them must be 60, and youre done.

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Now imagine that Ive been listening really carefully to your argument, but that Im not a particularlyintelligent nor knowledgeable individual.3 In this purely hypothetical world, I probably dont know thatthe base angles in an isosceles triangle are equal. Nor would I believe the well-known fact that the anglesin a triangle always add up to 180. So you would have to prove these facts to me as well. And then in theprocess of proving these, you would probably use even more basic geometric facts and I would complainthat I have never heard of them before, and on and on it goes. But where does it all end?

Of course, the game ends once youve broken the proof down into geometric facts which are so simple that Imust know them to be true. What are these fundamental things in geometry? Do you need infinitely manyof them or only finitely many of them? And if the answer is finitely many, then just how many do you needexactly?

Euclids Elements

The fundamental things, the LEGO blocks, the basic truths from which we build up all of geometry orany mathematical theory, in fact are called axioms. The interesting thing about axioms is that you simplycannot prove them, so you just have to assume that theyre true.

The first person to successfully apply the reductionist approach to mathematics was a rather clever guy bythe name of Euclid, whom Ive already mentioned. He was a Greek mathematician who lived around 300 BCand is most famous for his series of thirteen books known as the Elements. Even though youve probablynever heard of them, theyre some of the most successful and influential books ever written, even more sothan the Harry Potter or Twilight series. For example, Euclids Elements is supposed to be second only tothe Bible in the number of editions published. In fact, it was used as the basic text on geometry throughoutthe Western world for over two millennia, up until around one hundred years ago. In those days, if youdidnt know Euclids Elements, then people didnt think that you were well-educated. So to stop people fromthinking that were ignorant, lets have a look at Book 1 of Euclids Elements, the most famous of them all.

Axioms and Common Notions

In Book 1, Euclid states ten axioms from which he deduces everything that he knows about geometry. Sothankfully, unlike the reductionist approach applied to prime factorisation, there are only finitely manyfundamental things. If there had been infinitely many, then Euclid would never have been able to writethem all down.

The five main axioms of Euclidean geometry which are often referred to as postulates are stated below.Ive taken some licence in paraphrasing Euclids old-fashioned Ancient Greek into more modern English.

A1. You can draw a unique line segment between any two given points.

A2. You can extend a line segment to produce a unique line.

A3. You can draw a unique circle with a given centre and a given radius.

A4. Any two right angles are equal to each other.

A5. Suppose that a line ` meets two other lines, making two interior angles on one side of ` which sum toless than 180. Then the two lines, when extended, will meet on that side of `.

3Of course, this is going to be a very difficult thing for you to imagine, but please try.

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`

a

b

these lines will meet on thisside of ` if a+ b < 180

Euclid supplemented these with another set of five axioms of a slightly different nature, which we refer to ascommon notions.

C1. If A = C and B = C, then A = B.

C2. If A = X and B = Y, then A + B = X +Y.

C3. If A = X and B = Y, then A B = XY.C4. If A and B coincide, then A = B.

C5. The whole of something is greater than a part of something.

Surely you agree that each of these ten axioms is so obvious that its self-evident. Essentially, Euclid wasdeclaring his ten axioms to be the rules of the game from which he would build the world of Euclideangeometry. To really appreciate Euclids work, you have to remember that before he entered the picture,geometry consisted of a bunch of useful rules, like the fact that a triangle with side lengths 3, 4, 5 has a rightangle. You knew a fact like this was true because if you made a triangle with these side lengths out of ropeand used it as a protractor, then the home that you were building for your family would appear to be vertical.Euclid came along and said that if you believe my ten axioms and Im pretty sure that you all do then Ican show by logic alone that you have to believe the more complicated things that Im going to talk about. Inthis way, Euclid shows us what it means exactly for a theorem in mathematics to be true. In the remainder ofBook 1, Euclid proceeds to deduce, one by one, forty-eight propositions, the proof of each one dependingonly on the axioms and on previously proven propositions.

The First Few Propositions

In the timeless words of Maria von Trapp from The Sound of Music, Lets start at the very beginning, a verygood place to start. To make digesting the proofs a little easier, Ill write which axiom or proposition Imusing to deduce each statement. Its generally a good habit when youre constructing geometry proofs orany proofs, as a matter of fact to provide reasoning for every single statement that you write down.

Proposition (Proposition I). Given a line segment, you can draw an equilateral triangle on it.

A B

C

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Proof. Let AB be the given line segment.

Draw the circle with centre A and radius AB. [A3]

Now draw the circle with centre B and radius BA. [A3]

If the circles meet at a point C, then draw the line segments CA and CB. [A1]

Since A is the centre of one circle, AC = AB. And since B is the centre of the other circle, BC = BA. Butthese two statements together imply that AC = BC. [C1]

So the line segments AB, BC, CA are all equal which simply means that triangle ABC is equilateral.

Hopefully, you can appreciate that this is a fully rigorous proof and that its pretty hard to find fault with anysingle part of it. Because for every statement written down, Euclid can assert its truth, merely by pointing toone of his axioms. In some sense, Euclids axioms are like the ten commandments, and thou shalt not arguewith them. One down and forty-seven to go. . .

Proposition (Proposition II). Given a line segment and a point, you can draw a line segment from the given point,equal in length to the given line segment.

AB

C

D

G

E

F

H

Proof. Let A be the given point and BC the given line segment.

Draw the line segment AB. [A1]

Now draw the equilateral triangle DAB. [P1]

Extend the line segments DA and DB to obtain lines DE and DF. [A2]

Draw the circle with centre B and radius BC and let it meet DF at G. [A3]

Now draw the circle with centre D and radius DG and let it meet DE at H. [A3]

Since D is the centre of a circle, we know that DH = DG. And because we constructed DA = DB, wemay now subtract to obtain DH DA = DG DB, or equivalently, AH = BG. [C3]Since B is the centre of a circle, we know that BC = BG. So we can deduce that AH = BC. [C1]

One thing youve probably noticed is that these propositions are ridiculously simple and seem pretty obvious.Youve hopefully also noticed that it takes some ingenuity to prove them using only the axioms.

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Proposition (Proposition III). Given two line segments of unequal lengths, you can divide the longer one into twoparts, one of which is equal in length to the shorter one.

AB

C

D

E

F

Proof. Let AB and CD be the two given line segments, where the longer one is AB.

Draw the line segment AE equal to the line segment CD. [P2]

Draw the circle with centre A and radius AE and let it meet AB at F. [A3]

This means that AE = AF, but we also know that AE = CD, so we can deduce that AF = CD. [C1]

Therefore, we have divided AB into two parts using the point F, one of which is equal to CD.

Hopefully, youre getting the hang of things now. . .

Proposition (Proposition IV). If two triangles have two pairs of equal sides and the angles formed by these sides areequal, then the two triangles are congruent in other words, they have the same side lengths and the same angles.

A

B C

D

E F

Proof. Let the two triangles be ABC and DEF, where AB = DE, AC = DF and BAC = EDF. Placetriangle ABC so that A coincides with D and the line AB coincides with the line DE. Since AB = DE, itmust be the case that B coincides with E. The equal angles BAC = EDF guarantee that the line AC willcoincide with the line DF while the equal lengths AC = DF guarantee that C will coincide with F.

The line segment BC must now coincide with the line segment EF. [A1]

Therefore, we know that BC = EF. [C4]

Furthermore, the angles of triangle ABC coincide with the angles of triangle DEF, so ABC = DEFand ACB = DFE. [C4]

We now know that triangles ABC and DEF are congruent.

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Pons Asinorum

We now come to Euclids fifth proposition from Book 1 of his Elements, a proposition which has historicallybeen given the nickname Pons Asinorum. If youre well-versed in Latin, youll know that this means theBridge of the Asses. Why is it called this? One reason that has been proposed is that the diagram used inthe proof looks like a steep bridge which can be crossed by an ass but not by the fuller-figured horse. Themore commonly accepted reason is that this proposition is the first real test of intelligence. At this point,the intelligent people are able to cross over to the harder propositions while the unintelligent asses get leftbehind. So that we dont feel like unintelligent asses, lets now try to understand Euclids fifth proposition.

Proposition (Proposition V Pons Asinorum). In an isosceles triangle, the base angles are equal and the anglesunder the base angles are equal.

A

B C

D E

F G

Proof. Let the isosceles triangle be ABC, where AB = AC.

Extend the sides AB and AC to produce the lines AD and AE, respectively. [A2]

Now choose a random point F on BD and let the point G divide the segment AE in such a way thatAF = AG. [P3]

Draw the line segments FC and GB. [A1]

Since AF = AG and AB = AC, the two line segments FA and AC are equal to the two line segmentsGA and AB, respectively. They also make a common angle FAC = GAB. So the two triangles AFCand AGB are congruent. [P4]

We know that AB = AC, but also that AF = AG, so we can subtract to obtain AF AB = AG AC.Of course, this is the same thing as BF = CG. [C3]

Remember that we already have FC = GB. So the two line segments BF and FC are equal to the twoline segments CG and GB, respectively. Weve also shown that CGB = BFC, so triangle BFC iscongruent to triangle CGB. [P4]

Hence, we have the equal angles ABG = ACF as well as the equal angles CBG = BCF. Aftersubtracting, we obtain ABGCBG = ACFBCF or, equivalently, ABC = ACB, which arethe base angles. [C3]

Furthermore, weve already shown that FBC = GCB, which are precisely the angles under the baseangles.

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Pythagoras Theorem

Now that Ive led you over the Pons Asinorum, it seems like a reasonably safe assumption that if you hadenough time and energy, you could understand the rest of Euclids Book 1. Rather than spend our timeworking through the remaining propositions, lets fast forward to the second last proposition in EuclidsBook 1, a theorem which you should already know and love.

Proposition (Proposition XLVII Pythagoras Theorem). In a right-angled triangle, the square of the hypotenuseis equal to the sum of the squares of the other two sides.

A

B

C

D

E

F

G

H K

L

M

Proof. Let ABC be the right-angled triangle with BAC = 90.Draw the square BDEC on BC, the square AGFB on AB, and the square CKHA on CA. [P46]

Draw the line segment AL parallel to BD, where L lies on DE. [P31]

Draw the line segments AD and FC. [A1]

Since BAC = BAG = 90, the line segment CG passes through A and, similarly, the line segmentBH passes through A. [P14]

Note that CBD = FBA, so adding ABC to both sides, we obtain ABD = FBC. [C2]We also have BD = BC and AB = FB, so triangles ABD and FBC are congruent. [P4]

The parallelogram BDLM is twice the area of triangle ABD because they have the same base BD andlie between the same parallel lines BD and AL. [P41]

The square AGFB is twice the area of triangle FBC because they have the same base FB and are betweenthe same parallel lines FB and GC. [P41]

Thus, the parallelogram BDLM is equal in area to the square AGFB. In a similar fashion, we can deduce thatthe parallelogram CELM is equal in area to the square CKHA. Thus, the area of the square BDEC is equal tothe sum of the areas of the two squares AGFB and CKHA exactly what we set out to prove.

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Of course, Euclids proof of Pythagoras Theorem relies on all sorts of propositions that we havent seenyet. So, for the sake of begin complete, here is a list of all forty-eight propositions from Book 1 of EuclidsElements.

Euclids Propositions

P1. Given a line segment, you can draw an equilateral triangle on it.

P2. Given a line segment and a point, you can draw a line segment from the given point, equal in length tothe given line segment.

P3. Given two line segments of unequal lengths, you can divide the longer one into two parts, one of whichis equal in length to the shorter one.

P4. If two triangles have two pairs of equal sides and the angles formed by these sides are equal, then thetwo triangles are congruent in other words, they have the same side lengths and the same angles.

P5. In an isosceles triangle, the base angles are equal and the angles under the base angles are equal.

P6. If a triangle has two equal angles, then the two sides opposite these angles are equal.

P7. Given a triangle ABC, there is no other point P on the same side of AB as C such that AC = AP andBC = BP.

P8. If two triangles have three pairs of equal sides, then they also have three pairs of equal angles.

P9. Given a right angle, you can draw a line which bisects it.

P10. Given a line segment, you can draw its midpoint.

P11. Given a line and a point on that line, you can draw another line perpendicular to the given line andpassing through the given point.

P12. Given a line and a point not on the line, you can draw another line perpendicular to the given line andpassing through the given point.

P13. If a line intersects another line, then it creates two angles which sum to 180.

P14. Given a point on a line, if two line segments drawn from the point lie on opposite sides of the line andform adjacent angles which sum to 180, then the line segments lie on a line.

P15. If two lines meet, then the two vertical angles are equal.

P16. In any triangle, if one of the sides is extended, then the exterior angle formed is greater than either ofthe two interior opposite angles.

P17. In any triangle, the sum of any two angles is less than 180.

P18. In any triangle, the angle opposite the longest side is the largest.

P19. In any triangle, the side opposite the largest angle is the longest.

P20. In any triangle, the sum of any two sides is longer than the remaining side.

P21. If P is a point inside triangle ABC, then AP + BP < AC + BC and APB > ACB.P22. Given three line segments, it is possible to draw a triangle whose sides are equal in length to the given

line segments whenever the sum of any two is longer than the remaining one.

P23. Given an angle and a line with a point on it, you can draw a line passing through the given point whichcreates an angle with the given line equal to the given angle.

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P24. If two triangles have two pairs of equal sides and the angle formed by these two sides is larger in onetriangle, then the third side is longer in that triangle.

P25. If two triangles have two pairs of equal sides and the third side is longer in one triangle, then the angleformed by these two sides is larger in that triangle.

P26. If two triangles have two pairs of equal angles and one pair of corresponding equal sides, then the twotriangles are congruent.

P27. If a line meets two lines and forms equal alternate angles, then the two lines are parallel.

P28. If a line meets two lines and forms an exterior angle equal to the interior opposite angle on the sameside, or the sum of the interior angles on the same side is equal to 180, then the lines are parallel.

P29. If a line meets two parallel lines, then alternate angles are equal, the exterior angle is equal to theinterior opposite angle, and the interior angles on the same side sum to 180.

P30. Lines parallel to the same line are also parallel to each other.

P31. Given a line and a point, you can draw a line through the given point parallel to the given line.

P32. In any triangle, if one of the sides is extended, then the exterior angle formed equals the sum of the twointerior opposite angles, and the sum of the three interior angles of the triangle equals 180.

P33. Line segments which join the ends of equal parallel line segments are equal and parallel.

P34. In a parallelogram, opposite sides and angles are equal to each other, and the diagonals bisect the area.

P35. Parallelograms with the same base and equal heights have equal areas.

P36. Parallelograms with equal bases and equal heights have equal areas.

P37. Triangles with the same base and equal heights have equal areas.

P38. Triangles with equal bases and equal heights have equal areas.

P39. Triangles with equal areas and the same base have equal heights.

P40. Triangles with equal areas and equal bases have equal heights.

P41. If a parallelogram has the same base and equal height with a triangle, then the parallelogram is twicethe area of the triangle.

P42. Given an angle and a triangle, you can draw a parallelogram with area equal to the area of the giventriangle and with one angle equal to the given angle.

P43. Consider a point on the diagonal of a parallelogram. If lines are drawn through this point parallelto the sides of the parallelogram, then four triangles and two parallelograms are formed. The twoparallelograms have equal area.

P44. Given a triangle, a line segment and an angle, you can draw a parallelogram on the line segment withone angle equal to the given angle and with area equal to the area of the given triangle.

P45. Given a polygon and an angle, you can draw a parallelogram with one angle equal to the given angleand with area equal to the area of the given polygon.

P46. You can draw a square on a given line segment.

P47. In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the othertwo sides.

P48. If the square of one side of a triangle is equal to the sum of the squares of the other two sides, then thetriangle is right-angled.

14


The Problem with Reductionism

At this point, were going to leave Euclids Elements behind and make our own way through the world ofgeometry. We do this for three main reasons.

So far, we have followed Euclids reductionist approach and carefully proved the first five of hispropositions. At this rate, it would take us more than half the course to finish Book 1 of the Elementsand wed still have twelve books left to go. As Willy Wonka from Willy Wonka and the Chocolate Factoryonce said, So much time and so little to do! Strike that, reverse it.

To be honest, I think that working through the remainder of Book 1 of Euclids Elements would notonly be incredibly laborious, but also excruciatingly boring.

Whenever you get too involved in the reductionist approach, you run the risk of missing the biggerpicture. For example, it is far from true that a physicist who unlocks the mysteries of the tiny tiny quarksand strings has a grasp of the universe around them. They would still have no further knowledgeabout topics such as beauty, music, linguistics, hockey and personal hygiene, to name a few. I think weshould take a broader overview of geometry, whereas if we continued with Euclid, we would not see theforest for the trees.

Still, there is great beauty in Euclids work, but it doesnt lie in the individual propositions themselves. It liesin the fact that the reductionist approach lets us truly understand what it means for a mathematical theoremto be true, the fact that the Elements was the paradigm of mathematical rigour for centuries upon centuries,and the fact that Euclid has done a lot of the dirty work for us and we can use his results as a stepping stoneto look at some much more interesting facets of geometry.

Problems

At the end of each lecture, Ill usually state a couple of example problems and solve them, to help you learnhow to approach problems in geometry. However, Im going to end todays lecture with the are-you-smarter-than-Euclid challenge.

Problem. As I mentioned earlier, Euclids fifth proposition otherwise known as Pons Asinorum is the first realtest of intelligence in Euclids Elements. This is because Euclids proof is relatively long and intricate. However, if youlook at the first four propositions very carefully, you might notice that these can be used to give a much slicker proof tothe Pons Asinorum. Try to find the proof that Euclid missed and write it out carefully.

15


Euclid

As I mentioned earlier, Euclid was a Greek mathemati-cian who lived a really really long time ago around300 BC, to be precise. But we cant be any more pre-cise than that, because theres very little known aboutEuclids life. In fact, its been stated that any of thefollowing three possibilities may be the actual truthabout Euclid.

Euclid was a person who wrote the Elementsas well as various other works which are at-tributed to him.

Euclid led a team of mathematicians in Alexan-dria, who all contributed to writing the com-plete works of Euclid, even continuing after Eu-clid died.

Euclid was entirely fictional and his completeworks were written by a team of mathemati-cians in Alexandria who borrowed the nameof Euclid from an actual person Euclid ofMegara who had lived about 100 years ear-lier.

Most people who arent conspiracy theorists tend tobelieve the first possibility, in which case Euclid wasa very clever guy indeed. He has quite rightly beenreferred to as the Father of Geometry. You might

be wondering, if we dont even know whether or notEuclid existed, then why is there a picture of himstudying geometry below? Well, this particular por-trait is just the product of an artists imagination.

16

1. EUCLIDEAN GEOMETRY 1.2. Triangles and Circles

Warming Up with Parallel Lines

Now that weve decided to leave Euclids Elements behind us, lets embark on a much less-detailed, thoughfar more exciting, geometric journey. Well warm up with a fact about parallel lines Euclid proved it, butwell assume it.

Proposition. If a line meets two parallel lines and you label one angle x as shown in the diagram below, then you canlabel all eight angles as shown.

xx

180 x180 x

xx

180 x180 x

Basic Facts about Triangles

Now its time for some basic facts about triangles, facts which you should know like the back of your handand be able to use when solving geometry problems.

The angles in a triangle add up to 180.

If a triangle ABC satisfies AB = AC, then ABC = ACB. On the other hand, if a triangle ABCsatisfies ABC = ACB, then AB = AC. If this is true, then we say that the triangle is isosceles, a wordwhich comes from the Greek words iso, meaning same, and skelos, meaning leg.

The area of a triangle is given by 12 b h, where b denotes the length of the base and h denotes theheight of the triangle.

You might be wondering why we care so much about triangles. One reason is because a triangle is thesimplest shape which encloses a region that you can draw using only straight line segments. Such shapes of which triangles, quadrilaterals and pentagons are examples are usually called polygons. In fact, wecan consider triangles to be the fundamental things which we can glue together to build any polygon. Forexample, if you wanted to know what the sum of the angles are in a quadrilateral, then you could simplydraw one of its diagonals to split it up into two smaller triangles. Each triangle on its own has angles whichadd to 180 so together, their angles add to 360. Since the angles of the individual triangles account for all ofthe angles in the quadrilateral, we have shown that every quadrilateral has angles which add to 360. Thistrick doesnt just work for quadrilaterals though, as we will now see.

Proposition. In a polygon with n sides, the angles add to (n 2) 180.

Proof. We already saw that a quadrilateral can be cut along a diagonal into two triangles. If you draw apentagon, you will notice that it can be cut along diagonals into three triangles. And if you draw a hexagon,you will notice that it can be cut along diagonals into four triangles. In general, a polygon with n sides can becut along diagonals into n 2 triangles. This is not a particularly easy fact to prove, so well just take it forgranted.

17


The figure above shows what you might get if you try this at home. Of course, each triangle on its own hasangles which add to 180 so together, their angles add to (n 2) 180. Since the angles of the individualtriangles account for all of the angles in the polygon, we now know that every polygon with n sides hasangles which add to (n 2) 180.

Congruence and Similarity

Congruence and similarity are two of the most important notions in geometry. We say that two shapes arecongruent if it is possible to pick one of them up and place it precisely on top of the other one. On the otherhand, we say that two shapes are similar if it is possible to pick one of them up, enlarge or shrink it by acertain factor, and then place it precisely on top of the other one. If the triangles ABC and XYZ are congruent,then we write this using the shorthand ABC = XYZ and if they are similar, then we write this using theshorthand ABC XYZ. When we use this notation, we always mean that the first vertex from one trianglecorresponds to the first vertex from the other, the second vertex from one triangle corresponds to the secondvertex from the other, and the third vertex from one triangle corresponds to the third vertex from the other.

There are four simple rules to determine whether or not two triangles are congruent. Each one comes with acatchy TLA4 which should hopefully be self-explanatory.

SSS (side-side-side)If ABC and XYZ are two triangles such that AB = XY, BC = YZ and CA = ZX, then the two trianglesare congruent.

SAS (side-angle-side)If ABC and XYZ are two triangles such that AB = XY, BC = YZ and ABC = XYZ, then the twotriangles are congruent.

ASA (angle-side-angle)If ABC and XYZ are two triangles such that ABC = XYZ, ACB = XZY and BC = YZ, then thetwo triangles are congruent.

RHS (right-hypotenuse-side)If ABC and XYZ are two triangles such that AB = XY, BC = YZ and BAC = YXZ = 90, then thetwo triangles are congruent.

You have to be really careful with SAS, because the equal angles must be enclosed by the two pairs of equalsides for the rule to work. In other words, the fictitious rule SSA (side-side-angle) cannot be used to showthat two triangles are congruent.

4Three Letter Acronym

18


To see this, we should be able to find two triangles ABC and XYZ such that AB = XY, BC = YZ andBCA = YZX such that the two triangles are not congruent. The following diagram should convince youthat such pairs of triangles certainly do exist.

A

B C

X

Y Z

There are also three simple rules to determine whether or not two triangles are similar. Each one comes witha catchy TLA which once again should hopefully be self-explanatory.

AAA (angle-angle-angle)If ABC and XYZ are two triangles such that ABC = XYZ, BCA = YZX and CAB = ZXY,then the two triangles are similar. In fact, since we know that the angles in a triangle add to 180, weonly need to know two of these equations and we get the third one for free.

PPP (proportion-proportion-proportion)If ABC and XYZ are two triangles such that the fractions ABXY =

BCYZ =

CAZX are equal, then the two

triangles are similar.

PAP (proportion-angle-proportion)If ABC and XYZ are two triangles such that the fractions ABXY =

BCYZ are equal and ABC = XYZ,

then the two triangles are similar.

Once again, you have to be really careful with PAP, because the angle must be enclosed by the two propor-tional sides for the rule to work. In other words, the fictitious rule PPA (proportion-proportion-angle) cannotbe used to show that two triangles are similar. Lets now apply our newfound knowledge about congruenceand similarity to prove the following simple, but extremely useful, theorem.

Theorem (Midpoint Theorem). Let ABC be a triangle where the midpoints of the sides BC, CA, AB are X, Y, Z,respectively. Then the four triangles AZY, ZBX, YXC and XYZ are all congruent to each other and similar to triangleABC.

A B

C

XY

Z

19


Proof. First, lets prove that triangle ABC is similar to triangle AZY. Its clear that CAB = YAZ becausethey actually coincide. We also have the equal fractions ABAZ =

ACAY = 2, so PAP tells us precisely that triangle

ABC is similar to triangle AZY and is twice the size.

An entirely similar argument no pun intended can be used to prove that triangle ABC is similar totriangles ZBX and YXC and is twice the size. So what we have deduced is that the triangles AZY, ZBX andYXC are all congruent to each other. In particular, we have the three equations XY = AZ, YZ = YZ andZX = AY which together imply that triangle XYZ is congruent to triangle AZY by SSS.

Pythagoras Theorem (Reprise)

Weve already seen Euclids proof of Pythagoras Theorem lets reword the theorem a little differently nowand consider a much slicker proof.

Theorem (Pythagoras Theorem). Consider a right-angled triangle with side lengths a, b, c where c is the length ofthe hypotenuse. Then a2 + b2 = c2.

Proof. The entire proof is captured by the two diagrams below. On the left, we see a square of side lengtha + b. Weve removed four right-angled triangles, each with side lengths a, b, c and the remaining shadedregions clearly have a combined area of a2 + b2.

a

a

a

a

b

b

b

b

a

a

a

a

b

b

b

b

On the right, we also see a square of side length a + b. Weve once again removed four right-angled triangles,each with side lengths a, b, c, but in a slightly different way. The remaining shaded region clearly has area c2

and it follows that a2 + b2 = c2.

Theorem (Converse of Pythagoras Theorem). Consider a triangle with side lengths a, b, c where a2 + b2 = c2.Then the triangle is right-angled and the hypotenuse has length c.

Proof. Construct a right-angled triangle whose legs have lengths a and b, and let its hypotenuse have length d.We can now invoke Pythagoras theorem since we proved it just a little bit earlier. It tells us that a2 + b2 = d2.But put this piece of information together with our assumption that a2 + b2 = c2 and you have the equationc2 = d2. This implies that c = d.

Therefore, the triangle with side lengths a, b, c that we were given possesses exactly the same side lengthsas the right-angled triangle that we have constructed. Because of SSS, this means that the two triangles are,in fact, congruent. So the given triangle with side lengths a, b, c is indeed right-angled, as we intended toprove.

20


Basic Facts about Circles

We now know lots about triangles, so lets move on to circles we start with some basic facts.

A circle is the set of points which are the samedistance r from some centre O. You should al-ready know by now that r is called the radius ofthe circle and O is called the centre of the circle.

A chord is a line segment which joins two pointson a circle.

An arc of a circle is the part of the circumferencecut off by a chord.

A diameter is a chord which passes through thecentre of the circle. Note that its length is twicethe radius.

diameter

chord

arc

This is all we need to know about circles in order to prove some pretty cool results, like the following.

Proposition. The diameter of a circle subtends an angle of 90. In other words, if AB is the diameter of a circle and Cis a point on the circle, then ACB = 90.

Proof. Let O be the centre of the circle. The beauty of considering the centre is that we have the three equalradii OA = OB = OC. Equal lengths, for obvious reasons, often lead to isosceles triangles, and our diagramhappens to have two of them. There is the isosceles triangle OAC which means that we can label the equalangles OAC = OCA = x. Theres also the isosceles triangle OBC, which means that we can label theequal angles OBC = OCB = y. Labelling equal angles like this is an extremely common and extremelyuseful trick.

A B

C

O

x y

x y

Now its time for some angle chasing.5 In particular, lets consider the sum of the angles in triangle ABC.

BAC +ACB +CBA = 1805Angle chasing is the art of chasing angles. As Wikipedia puts it, the term is used to describe a geometrical proof that involves

finding relationships between the various angles in a diagram.

21


We can replace all of these confusing angles with xs and ys in the following way.

x + (x + y) + y = 180

This equation is, of course, the same thing as 2(x + y) = 180 or, equivalently, x + y = 90. All we have to donow is recognise that ACB = x + y so we have proven that ACB = 90.

Proposition. The angle subtended by a chord at the centre is twice the angle subtended at the circumference, on thesame side. In other words, if AB is a chord of a circle with centre O and C is a point on the circle on the same side of ABas O, then AOB = 2ACB.

Proof. The beauty of considering the centre is that we have the three equal radii OA = OB = OC. Equallengths, for obvious reasons, often lead to isosceles triangles, and our diagram happens to have three of them.There is the isosceles triangle OAB which means that we can label the equal angles OAB = OBA = x.Theres also the isosceles triangle OBC, which means that we can label the equal angles OBC = OCB = y.And theres also the isosceles triangle OCA, which means that we can label the equal angles OCA =OAC = z. Labelling equal angles like this is an extremely common and extremely useful trick.

A B

C

O

x x

y

yz

z180 2x

Now its time for some angle chasing. In particular, lets consider the sum of the angles in triangle ABC.

BAC +ACB +CBA = 180

We can replace all of these confusing angles with xs, ys and zs in the following way.

(x + z) + (z + y) + (y + x) = 180

This equation is, of course, the same thing as 2(x + y + z) = 180. Lets keep this equation in the back ofour minds while we try and remember what it is exactly that were trying to do. We want to prove thatAOB = 2ACB. Using the angle sum in triangle OAB, we can write AOB = 180 2x. We can alsowrite ACB = y + z. So what were actually aiming for is the following equation.

180 2x = 2(y + z)

But after rearranging, this is just the same thing as 2(x + y + z) = 180, which we already proved.

22


Hopefully, youve managed to spot the similarity between these two proofs. Theyre both indicative of thegeneral strategies that well be using to solve tougher geometry problems. More specifically, I guess whatweve used here is the old find isosceles triangleslabel equal anglessum up the angles in a triangleworkout what youre trying to provethen prove it trick.

I should probably mention that some of the proofs Ive provided are a little incomplete. In particular, if youcheck the previous proof very carefully, youll notice that it only works when O lies inside triangle ABC. Itcould be possible that O lies on or even outside triangle ABC. However, the proofs in these other cases arequite similar, so Ill leave it as a fun exercise for you to find them.

The Hockey Theorem

Suppose that a hockey coach wanted to see which player on their team had the best shot. They could lineall the players parallel to the goal and ask them to shoot to see who scores and who misses. But this wouldcertainly be unfair to the players on the ends who have to shoot further and have a smaller angle to aim for.The coach could try and fix the problem by placing everyone on a circle whose centre coincides with themiddle of the goal. Of course, this means that everyone is the same distance from the goal now, but someplayers have a much greater angle to aim at than others. In fact, due to the frictionless nature of the sport,distance is no problem when shooting at a hockey goal. What we would rather test is the accuracy of eachplayer. So it makes sense to place everyone somewhere where they all have the same angle to aim at. In thatcase, where should we put all the players? The following proposition tells us the answer they should allstand on the arc of a circle whose end points are the goal posts.

goal goal goal

Proposition (Hockey Theorem). Angles subtended by a chord on the same side are equal. In other words, ifA, B, C, D are points on a circle with C and D lying on the same side of the chord AB, then ACB = ADB.

Proof. The proof to this is delightfully simple we start by letting O be the centre of the circle. Wehave already proved that AOB = 2ACB and also that AOB = 2ADB. So it must be the case thatACB = ADB.

A B

C

D

O

x

x

2x

23


Cyclic Quadrilaterals

If I give you one point and ask you to draw a circle through it, then thats a pretty easy thing to do, right?I could make life slightly more difficult for you by giving you two points and asking you to draw a circlethrough both of them. And even if I give you three points and ask you to draw a circle through all of them,then you could almost always do it, as long as they dont lie on a line. To see this, consider a massive circle, somassive that all three points that I give you lie inside it. Now start shrinking the circle. Sooner or later, yourcircle has to hit one of the points. Now keep that point on the circle, but keep shrinking the circle. Sooner orlater, your circle has to hit another of the points. We are now in the happy situation of having the circle passthrough two of the given points. Now we either shrink or expand our circle, while keeping it in contact withthese two points, until it finally passes through the third given point. Hopefully, this should convince youthat there is actually only one circle which passes through any three points, as long as they dont lie on a line.

So now Im going to make life particularly difficult for you by giving you four points and asking you to drawa circle through all of them. The shrink/expand trick we used above shows that there is a unique circle whichpasses through three of them. So to be able to accomplish the task, the fourth point must already lie on thiscircle. The probability that I was nice enough to actually give you four points where this is true is incrediblysmall. What Im trying to get at here is that if I give you a random quadrilateral, then it is extremely rare for acircle to pass through all four of its vertices. So if a circle does pass through all four of its vertices, then thequadrilateral must be very special indeed so special that we should give it a special name. In fact, we referto such a quadrilateral as a cyclic quadrilateral. Were now going to prove a very important fact about cyclicquadrilaterals.

Proposition. The opposite angles in a cyclic quadrilateral add up to 180. In other words, if ABCD is a cyclicquadrilateral, then ABC +CDA = 180 and BCD +DAB = 180.

Proof. If we draw in the diagonals AC and BD, the hockey theorem tells us that there are equal anglesgalore. For example, we can label ACB = ADB = w, BDC = BAC = x, CAD = CBD = y andDBA = DCA = z. You can go crazy labelling equal angles like this whenever theres a cyclic quadrilateralsomewhere in your diagram.

A B

C

D

x zy

wz

xw

y

Now were going to play a similar trick to one we played before were going to add up all of the angles inthe quadrilateral and the answer should be 360.

DAB +ABC +BCD +CDA = 360

24


We can replace all of these confusing angles with ws, xs, ys and zs in the following way.

(x + y) + (y + z) + (z + w) + (w + x) = 360

This equation is, of course, the same thing as 2(w + x + y + z) = 360 or, equivalently, w + x + y + z = 180.All we have to do now is recognise that ABC = y + z and CDA = w + x, so that

ABC +CDA = (y + z) + (w + x) = 180.

You could also have chosen to recognise that BCD = z + w and DAB = x + y, so that

BCD +DAB = (z + w) + (x + y) = 180.

How to Find a Cyclic Quadrilateral

Something were going to learn is that its incredibly useful to keep your eyes open for cyclic quadrilateralswhen solving problems in Euclidean geometry. If the problem happens to mention a circle which has fourpoints on it, then of course, those four points form a cyclic quadrilateral. But quite often, cyclic quadrilateralscan be hidden somewhere in your diagram, even when there are no circles involved. In those cases, youwould probably use one of the two facts below to prove that the quadrilateral is cyclic.

We proved earlier that the opposite angles in a cyclic quadrilateral add up to 180. One thing you might bewondering is whether the converse is true in other words, if someone gives you a quadrilateral where theopposite angles add up to 180, then is it necessarily true that the quadrilateral must have been cyclic? Thefollowing proposition tells you that the converse certainly is true.

Proposition. If the opposite angles in a quadrilateral add up to 180, then the quadrilateral is cyclic.

Another thing we proved earlier was the hockey theorem that if A, B, C, D are points on a circle withC and D lying on the same side of the chord AB, then ACB = ADB. And another thing you might bewondering is whether the converse is true in other words, if someone gives you a quadrilateral ABCDwhere ACB = ADB, then is it necessarily true that the quadrilateral must have been cyclic? The followingproposition tells you that the converse certainly is true, once again.

Proposition. If ABCD is a convex quadrilateral such that ACB = ADB, then the quadrilateral is cyclic.

A particularly useful case occurs when you are given four points A, B, C, D such that ABC = CDA = 90.If ABCD is a convex quadrilateral, then we can use the first proposition above to deduce that it must actuallybe a cyclic quadrilateral. On the other hand, if ABCD is a convex quadrilateral, then we can use the secondproposition above to deduce that it must actually be a cyclic quadrilateral. So either way, the four pointsA, B, C, D lie on a circle. This means that when right angles appear, you can often expect cyclic quadrilateralsto appear as well.

You can solve many many geometry problems by searching for cyclic quadrilaterals and using what youknow about them. The beauty of cyclic quadrilaterals is that if you find one pair of equal angles, then you getthree more for free. This is because if you find that ACB = ADB, then the quadrilateral ABCD is cyclic,in which case we can apply the hockey theorem to also obtain the equal angles

BDC = BAC, CAD = CBD, and ABD = ACD.

25


Tangents

A tangent is a line which touches a circle at preciselyone point. Given a point outside a circle, you candraw two tangents, thereby creating a diagram whichlooks very much like an ice cream cone.

Theorem (Ice Cream Cone Theorem). The pictureof the ice cream cone on the right is symmetric so thatAP = AQ and the line OA bisects the angle PAQ. A

P Q

O

The proof of the ice cream cone theorem is a very simple application of congruent triangles combined withthe fact that the radius drawn from the centre of a circle to a point on its circumference is perpendicular tothe tangent at that point. The following is another extremely useful fact about tangents.

Theorem (Alternate Segment Theorem). Suppose that AT is a chord of a circle and that PQ is a line tangent to thecircle at T. If B lies on the circle, on the opposite side of the chord AT from P, then ABT = ATP.

T

A

B

P Q

Proof. The hockey theorem guarantees that, as long as C lies on the circle, on the same side of AT as B,then ACT = ABT. So the trick here is to choose a particular point C on the circle and prove thatACT = ATP. One nice way to choose the location of the point C is so that TC is a diameter of the circle.

T

O

A

B

C

P Q

x

26


Choosing C in this way is great because weve introduced two right angles into the diagram. We haveCAT = 90, because its an angle subtended by the diameter TC and we also have PTO = 90, becauseits an angle created by a radius and a tangent at the point T.

So if we label ATP = x, then we have OTA = 90 x. Now use the fact that the angles in triangle CATadd to 180 and you find that ACT = x. However, we already mentioned that ABT = ACT = x by thehockey theorem, so we can now conclude that ABT = ATP.

It turns out that the converse of the alternate segment theorem is also true but Ill let you try to write downexactly what it states.

Problems

Many geometry problems can be solved by angle chasing. This means labelling some but not too many well-chosen angles in your diagram and using what you know to determine other ones. Here are some verycommon ways to relate different angles in your diagram.

If two angles are next door to each other, then they add up to 180.

Parallel lines give you equal angles and angles which add up to 180.

The angles in a triangle add up to 180.

Congruent or similar triangles give you equal corresponding angles.

In any cyclic quadrilateral, the opposite angles add up to 180.

In any cyclic quadrilateral, you can apply the hockey theorem to give you four pairs of equal angles.

Apart from these tips, possibly one of the most useful things I can tell you is to always draw a very accurate,very large diagram, preferably in multiple colours. Below are two example problems and their solutions. Youshould try to read the solutions carefully to make sure that you understand them and then reread them untilyou think you can reconstruct the proofs on your own.

Problem. Two circles intersect at P and Q. A line through P meets the circles C1 and C2 at A and B, respectively. LetY be the midpoint of AB and suppose that the line QY meets the circles C1 and C2 at X and Z, respectively. Prove thattriangle XYA is congruent to triangle ZYB.

A

BP

Q

X

Y

Z

x

C1 C2

27


Proof. The first thing to notice is that the two triangles have the angles XYA = ZYB as well as the sidesAY = BY in common. So to prove that theyre congruent, we could try to use either ASA or SAS. It turns outthat the first choice is better for this problem, since the circles in the diagram will give us equal angles.

So lets label YAX = x and note that the problem is completely solved if we can only show that YBZ = xas well. Since the angle YAX = PAX is subtended by the chord PX, we can use the hockey theorem todeduce that PQX = PAX = x. And since the angle PQZ = PQX is subtended by the chord PZ, wecan use the hockey theorem again to deduce that PBZ = PQZ = x. However, PBZ = YBZ so weveproved that YBZ = x and the problem is solved.

Problem. Consider a semicircle with diameter AB. Suppose that D is a point such that AB = AD and AD intersectsthe semicircle at the point E. Let F be the point on the chord AE such that DE = EF and extend BF to meet thesemicircle at the point C. Prove that BAE = 2EAC.

A B

D

E

FC

Proof. A good start would be to label EAC = x and aim to prove that BAE = 2x. Given that there is acircle or at least half of one there is probably a cyclic quadrilateral lurking about. Hopefully, you can seethat ABEC is a cyclic quadrilateral, so we can apply the hockey theorem to deduce that EBC = EAC = xas well.

Note that AB is a diameter and there is a certain fact that we know about diameters namely, they subtendangles of 90. This means that AEB = ACB = 90. Since EBC = x, considering the angle sum intriangle EFB yields the fact that EFB = 90 x.Now if youve drawn a nice accurate picture, then youll see that the two triangles FEB and DEB looksuspiciously congruent. These suspicions can be confirmed by observing that that FE = ED, EB = EB,FEB = DEB = 90 and using SAS. One consequence of this is the fact that EDB = EFB = 90 x.One piece of information we havent used yet is the fact that AB = AD or, in other words, that triangleABD is isosceles. The equal side lengths imply that there are equal angles in the diagram namely,ABD = ADB = EDB = 90 x.Now we note that we have labelled two of the angles in triangle ABD, so we can definitely determine whatthe third angle is.

BAD = 180 ABDADB = 180 (90 x) (90 x) = 2x

However, this tells us that BAE = 2x as well, exactly what we set out to prove.

28


Pythagoras

Pythagoras was a Greek mathematician who livedeven before Euclid was born from around 570 BCto around 475 BC. Even though we know and lovehim for his theorem about right-angled triangles, itwas essentially known before Pythagoras arrived onthe scene and he never even managed to prove it. Ac-tually, Pythagoras was not a great mathematician atall, but was the first person to call himself a philoso-pher a word which in ancient Greek literally meansa lover of wisdom.

Pythagoras founded a religious cult whose membersbelieved that everything was related to mathematicsand that numbers were the ultimate reality. Pythago-ras himself was pretty serious about this belief, asyou can tell from the following story. When his fellowcult member Hippasus of Metapontum managed toprove that the number

2 was irrational a rather

ingenious and important mathematical feat whichdidnt sit well with the Pythagorean philosophy Pythagoras supposedly could not accept the resultand sentenced Hippasus to death by drowning.

The Pythagorean cult also adhered to various rules,some mildly practical but most simply bizarre thefollowing are examples.

Never eat beans.

Dont pick up something that has fallen.

Dont walk on highways.

When you get out of bed, take the sheets androll them all together.

Apart from these facts, there isnt a great deal which isknown about Pythagoras because no written work ofhis has survived to this day. Its commonly believedthat many of the accomplishments attributed to himwere simply accomplishments of people who were inhis cult.

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1. EUCLIDEAN GEOMETRY 1.3. Triangle Centres

Triangle Centres

The humble triangle, with its three vertices and three sides, is actually a very remarkable object. For evidenceof this fact, just look at the Encyclopedia of Triangle Centers6 on the internet, which catalogues literally thousandsof special points that every triangle has. Well only be looking at the big four namely, the circumcentre,the incentre, the orthocentre, and the centroid. While exploring these constructions, well need all of ournewfound geometric knowledge from the previous lecture, so lets have a quick recap.

Triangles

Congruence : There are four simple rules to determine whether or not two triangles are congruent.They are SSS, SAS, ASA and RHS.

Similarity : There are also three simple rules to determine whether or not two triangles are similar.They are AAA, PPP and PAP.

Midpoint Theorem : Let ABC be a triangle where the midpoints of the sides BC, CA, AB are X, Y, Z,respectively. Then the four triangles AZY, ZBX, YXC and XYZ are all congruent to each otherand similar to triangle ABC.

Circles

The diameter of a circle subtends an angle of 90. In other words, if AB is the diameter of a circleand C is a point on the circle, then ACB = 90.

The angle subtended by a chord at the centre is twice the angle subtended at the circumference, onthe same side. In other words, if AB is a chord of a circle with centre O and C is a point on thecircle on the same side of AB as O, then AOB = 2ACB.

Hockey Theorem : Angles subtended by a chord on the same side are equal. In other words, ifA, B, C, D are points on a circle with C and D lying on the same side of the chord AB, thenACB = ADB.

Cyclic Quadrilaterals

The opposite angles in a cyclic quadrilateral add up to 180. In other words, if ABCD is a cyclicquadrilateral, then ABC +CDA = 180 and BCD +DAB = 180.

If the opposite angles in a quadrilateral add up to 180, then the quadrilateral is cyclic. Hockey Theorem : If ABCD is a convex quadrilateral such that ACB = ADB, then the quadrilat-

eral is cyclic.

Tangents

Ice Cream Cone Theorem : A picture of an ice cream cone is symmetric so that the two tangents havethe same length and the line joining the centre of the circle and the tip of the cone bisects the coneangle.

Alternate Segment Theorem : Suppose that AT is a chord of a circle and that PQ is a line tangent to thecircle at T. If B lies on the circle, on the opposite side of the chord AT from P, thenABT = ATP.

These facts should all be burned into your memory, so that you can recall and use them, whenever youencounter a problem in Euclidean geometry.

6http://faculty.evansville.edu/ck6/encyclopedia/ETC.html

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The Circumcentre

A perpendicular bisector of a triangle is a line which passes through the midpoint of one side and is perpendic-ular to that side. Note that three randomly chosen lines will almost never ever meet at a point and yet, forany particular triangle we choose, well see that its three perpendicular bisectors always do.

Proposition. The three perpendicular bisectors of a triangle meet at a point.

Proof. Our proof relies crucially on the following lemma, which can be proven using congruent triangles.

Lemma. A point P lies on the perpendicular bisector of AB if and only if AP = BP.

So take a triangle ABC and the perpendicular bisectors of the sides AB and BC. If we suppose that these twolines meet at a point O, then it must be the case that AO = BO and also that BO = CO. These two equationstogether imply that AO = CO, in which case O lies on the perpendicular bisector of the side CA as well.

In the previous proof, we noted that if the perpendicular bisectors of triangle ABC meet at O, then thedistances from O to the vertices are all equal. Another way to say this is that theres a circle with centre Owhich passes through the vertices A, B, C. This circle is called the circumcircle of the triangle and the pointO is called the circumcentre.7 Furthermore, the radius of the circumcircle is known as the circumradius forobvious reasons. We now know that every triangle has exactly one circumcircle and that its centre lies on theperpendicular bisectors of the triangle.

Something interesting to note is that when triangle ABC is acute, O lies inside the triangle; when triangleABC is right-angled, O lies on the hypotenuse of the triangle, at its midpoint; and when triangle ABC isobtuse, O lies outside the triangle.

Lets draw an acute triangle ABC and draw in the three perpendicular bisectors XO, YO, ZO, just like Ivedone below. There are three cyclic quadrilaterals lurking in the diagram surely you can spot them.

A B

C

O

XY

Z

The cyclic quadrilaterals are

AZOY the opposite angles AZO and OYA add to 180;BXOZ the opposite angles BXO and OZB add to 180; andCYOX the opposite angles CYO and OXC add to 180.

7In Latin, the word circum means around and this makes sense because the circumcentre goes around the triangle.

31


Lets now draw in the three line segments AO, BO, CO as well as the triangle XYZ. An extremely usefulexercise is to label all of the thirty-six angles in the diagram in terms of a = CAB, b = ABC and c = BCA.

A B

C

OXY

Z

The midpoint theorem states that the triangles AZY and ABC are similar, so we have YZA = b. Youcan use this strategy to label six of the angles in the diagram.

Since the quadrilateral AZOY is cyclic, the hockey theorem tells us that YOA = YZA = b. You canuse this strategy to label six more of the angles in the diagram.

Since the sum of the angles in triangle YOA is 180, we must have YAO = 90 YOA = 90 b.You can use this strategy to label six more of the angles in the diagram.

Since the quadrilateral AZOY is cyclic, the hockey theorem tells us that YZO = YAO = 90 b.You can use this strategy to label six more of the angles in the diagram.

The remaining twelve angles can be labelled by using the fact that the angles in a triangle add to 180.

The Incentre

An angle bisector of a triangle is a line which passes through a vertex and bisects the angle at that vertex. Notethat three randomly chosen lines will almost never ever meet at a point and yet, for any particular trianglewe choose, well see that its three angle bisectors always do.

Proposition. The three angle bisectors of a triangle meet at a point.

Proof. Our proof relies crucially on the following lemma, which can be proven using congruent triangles.

Lemma. A point P lies on the angle bisector of ABC if and only if the distance from P to AB isequal to the distance from P to CB.

So take a triangle ABC and the angle bisectors at the vertices A and B. If we suppose that these two linesmeet at a point I, then it must be the case that the distance from I to CA equals the distance from I to AB andthe distance from I to AB equals the distance from I to BC. These two statements together imply that thedistance from I to CA equals the distance from I to BC, in which case I lies on the angle bisector at the vertexC as well.

In the previous prof, we noted that if the angle bisectors of triangle ABC meet at I, then the distances from Ito the three sides are all equal. Another way to say this is that theres a circle with centre I which touchesthe sides AB, BC, CA. This circle is called the incircle of the triangle and the point I is called the incentre.8

8In Latin, the word in means inside and this makes sense because the incentre goes inside the triangle.

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Furthermore, the radius of the incircle is known as the inradius for obvious reasons. We now know that everytriangle has exactly one incircle and that its centre lies on the angle bisectors of the triangle.

Lets draw a triangle ABC and draw in the three radii of the incircle PI, QI, RI, just like Ive done below.There are three cyclic quadrilaterals lurking in the diagram surely you can spot them.

A B

C

I

PQ

R


ARIQ the opposite angles ARI and IQA add to 180;BPIR the opposite angles BPI and IRB add to 180; andCQIP the opposite angles CQI and IPC add to 180.

Lets now draw in the three line segments AI, BI, CI as well as the triangle PQR. An extremely useful exerciseis to label all of the thirty-six angles in the diagram in terms of a = CAB, b = ABC and c = BCA.

A B

C

I

PQ

R

The ice cream cone theorem states that the angles QAI and RAI are equal, so we have QAI =RAI = a2 . You can use this strategy to label six of the angles in the diagram.Since the quadrilateral ARIQ is cyclic, the hockey theorem tells us that QRI = QAI = a2 . You canuse this strategy to label six more of the angles in the diagram.

Since the sum of the angles in triangle QAI is 180, we must have QIA = 90 QAI = 90 a2 .You can use this strategy to label six more of the angles in the diagram.

Since the quadrilateral ARIQ is cyclic, the hockey theorem tells us that QRA = QIA = 90 a2 .You can use this strategy to label six more of the angles in the diagram.

The remaining twelve angles can be labelled by using the fact that the angles in a triangle add to 180

and, in fact, they are all equal to 90.

33


The Orthocentre

An altitude of a triangle is a line which passes through a vertex and is perpendicular to the opposite side. Ifthe triangle happens to have an angle greater than 90, then you will need to extend the sides in order todraw all three altitudes. Note that three randomly chosen lines will never ever meet at a point yet, for anyparticular triangle we choose, well see that its three altitudes always do.

Proposition. The three altitudes of a triangle meet at a point.

Proof. Draw the triangle ABC such that AB is parallel to AB and C lies on AB, BC is parallel to BC andA lies on BC, and CA is parallel to CA and B lies on CA.

A B

C

D

E

F

AB

C

This creates the parallelograms ABCB and ABAC, so that we have the equal lengths BC = AB = CA.Similarly, we have the equations CA = BC = AB and AB = CA = BC. So the points A, B, C are simplythe midpoints of the sides BC, CA, AB, respectively. This is precisely the setup for the midpoint theorem.

Since CF is perpendicular to AB, its also parallel to BA in other words, CF is the perpendicular bisectorof AB. Similarly, we know that AD is the perpendicular bisector of BC and BE is the perpendicular bisectorof CA. But we proved earlier that the three perpendicular bisectors of a triangle meet at a point. Therefore,the three altitudes AD, BE, CF of triangle ABC meet at a point.

The three altitudes AD, BE, CF of triangle ABC meet at a single point H called the orthocentre.9 Somethinginteresting to note is that when triangle ABC is acute, H lies inside the triangle; when triangle ABC isright-angled, H lies at a vertex of the triangle; and when triangle ABC is obtuse, H lies outside the triangle.

Lets draw an acute triangle ABC and draw in the three altitudes AD, BE, CF, just like Ive done below.Amazingly, there are six cyclic quadrilaterals lurking in the diagram can you find them all?

9In ancient Greek, the word ortho means vertical and this makes sense because the altitude of a triangle is vertical with respect tothe base.

34


A B

C

D

E

F

H


AFHE the opposite angles AFH and HEA add to 180;BDHF the opposite angles BDH and HFB add to 180;CEHD the opposite angles CEH and HDC add to 180;ABDE the angles ADB and AEB are equal;BCEF the angles BEC and BFC are equal; andCAFD the angles CFA and CDA are equal.

Lets now draw in the triangle DEF. An extremely useful exercise is to label all of the thirty-six angles in thediagram in terms of a = CAB, b = ABC and c = BCA.

A B

C

D

E

F

H

Since the sum of the angles in triangle ABD is 180, we must have BAD = 90 DBA = 90 b.You can use this strategy to label six of the angles in the diagram.

Since the quadrilateral AFHE is cyclic, the hockey theorem tells us that FEH = FAH = 90 b.You can use this strategy to label six more of the angles in the diagram.

Since HEA is a right angle, we must have FEA = 90 FEH = b. You can use this strategy tolabel six more of the angles in the diagram.

Since the quadrilateral AFHE is cyclic, the hockey theorem tells us that FHA = FEA = b. You canuse this strategy to label six more of the angles in the diagram.

The remaining twelve angles can be labelled by using the fact that the angles in a triangle add to 180.

35


The Centroid

A median of a triangle is a line which passes through a vertex and the midpoint of the opposite side.

Proposition. The three medians of a triangle meet at a point.

Proof. Suppose that the medians AX and BY meet at G1. By the midpoint theorem, we know that XY isparallel to AB which implies that the triangles ABG1 and XYG1 are similar by AAA. We also know by themidpoint theorem that ABXY = 2, so the constant of proportionality is 2. This means that

AG1G1X

= 2. So themedian BY cuts the median AX at a point G1 such that AG1 is twice as long as G1X.

A B

C

G1

XY

A B

C

G2

X

Z

Now suppose that the medians AX and CZ meet at G2. By the midpoint theorem, we know that XZ isparallel to AC which implies that the triangles ACG2 and XZG2 are similar by AAA. We also know by themidpoint theorem that ACXZ = 2, so the constant of proportionality is 2. This means that

AG2G2X

= 2. So themedian CZ cuts the median AX at a point G2 such that AG2 is twice as long as G2X.

Putting these two pieces of information together, we deduce that the points G1 and G2 must actually be thesame point. In other words, the medians BY and CZ meet the median AX at the same point.

The three medians AX, BY, CZ of triangle ABC meet at a single point G called the centroid. It is the centre ofgravity of ABC in the sense that if you cut the triangle out of cardboard, then it should theoretically balanceon the tip of a pencil placed at the point G. One consequence of our proof above is the fact that we have theequal fractions

AGGX

=BGGY

=CGGZ

= 2.

More Fun with Triangle Centres

There is so much more to triangle centres than we have mentioned. Lets write down some interesting factshere, which you can try to prove on your own.

Proposition. If the orthocentre of triangle ABC is H, then the orthocentre of triangle HBC is A, the orthocentre oftriangle HCA is B and the orthocentre of triangle HAB is C.

Proposition. If you take a triangle ABC and draw in the three medians AX, BY, CZ, then the six resulting trianglesall have equal area.

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The following proposition shows that the four triangle centres we have looked at are related in variouspeculiar ways you should try to prove all of these statements.

Proposition. Given a triangle ABC, let the midpoints of the sides be X, Y, Z, let the incircle touch the sides at P, Q, R,and let the feet of the altitudes be D, E, F.

The circumcentre O of triangle ABC is the orthocentre of triangle XYZ.

The incentre I of triangle ABC is the circumcentre of triangle PQR.

The orthocentre H of triangle ABC is the incentre of triangle DEF.

The centroid G of triangle ABC is the centroid of triangle XYZ.

Problems

When solving difficult geometry problems, here are a few things to always keep a look out for. If you becomewell practised at spotting these objects in your geometry diagrams, then you are well on the way to becominga geometry guru.

cyclic quadrilaterals

isosceles triangles

equal angles and lengths

similar or congruent triangles

right angles

ice cream cones

Problem. Let ABC be a triangle with incentre I and extend AI until it meets the circumcircle of triangle ABC at X.Prove that X is the circumcentre of triangle BIC.

A

B C

X

I

a a

bb

cc

Proof. When the incentre of triangle ABC is involved, I like to let the angles at A, B, C be 2a, 2b, 2c, respectively.This is because I lies on the angle bisectors so I can label

BAI = CAI = a, CBI = ABI = b, ACI = BCI = c.

We want to prove that X is the circumcentre of triangle BIC or equivalently, that the lengths BX, IX, CX areall equal. Hopefully you can see that the problem is solved if we can prove that triangle BXI is isosceles withBX = IX. This is because the same reasoning will tell us that triangle CXI is isosceles with CX = IX.

37


So lets focus on proving that BX = IX. Since there are many relationships between angles in circles, it makessense to try to instead prove that IBX = BIX. Using our notation, we obtain that

IBX = CBX +IBC = CAX +IBC = a + b.

Here, weve used the hockey theorem on chord CX to deduce that CBX = CAX = a.Now observe that BIX +BIA = 180, since they form a straight line. And if we sum up the angles intriangle ABI, we obtain the equation BAI +ABI +BIA = 180 or equivalently, a + b +BIA = 180.These two facts imply that BIX = a + b, so we have deduced that IBX = BIX. As we mentionedearlier, it follows that BX = IX and similar reasoning will lead to CX = IX as well. Hence, the three lengthsBX, IX, CX are all equal and X is the circumcentre of triangle BIC.

Problem. If ABC is a triangle with a right angle at C, prove that the angle bisector from C bisects the angle formed bythe altitude from C and the median from C.

Proof. Lets call the altitude, the angle bisector and the median CF, CT and CZ, respectively. Then what wedlike to prove can be rephrased as FCT = TCZ. However, we already know that CT bisects the right angle,so that TCA = TCB. What this means is that we can rephrase the problem once again as FCA = ZCB.With this in mind, lets write a = CAB and try to determine the angles FCA and ZCB in terms of a.

A B

C

F ZTa

The first of these angles is easy since triangle CFA is right-angled, we can write FCA = 90 a. Nowif we consider the sum of the angles in triangle ABC, we obtain ABC = 90 a which is equivalent toZBC = 90 a. Remember that our goal is to prove that ZCB = FCA = 90 a. So what we shouldtry to prove now is that triangle ZBC is isosceles, with ZB = ZC. But remember that the diameter of a circlesubtends an angle of 90, so a circle with diameter AB passes through C. As Z is the midpoint of AB, thismeans that Z is the circumcentre of triangle ABC and ZB = ZC, as desired.

38


Fermat

Pierre de Fermat was actually a lawyer by day atthe Parlement of Toulouse and an amateur mathe-matician by night. He lived in the early seventeenthcentury from 1601 to 1665 and is often credited withthe development of a very early form of what we nowcall calculus. However, Fermat is probably most fa-mous for his work in number theory. One theorem ofhis Fermats Little Theorem says that if you pickyour favourite positive integer a and your favouriteprime number p, then the number ap a will be divis-ible by p. The most famous story about Fermat tellsof how he pencilled in the margin of a mathematicsbook the following problem.

It is impossible to find two perfect cubeswhich sum to a perfect cube, two per-fect fourth powers which sum to a perfectfourth power, or in general, two perfectnth powers which sum to a perfect nthpower, if n is an integer greater than 2.

Rather tantalisingly, Fermat also wrote that he hada marvellous proof which was too small to fit intothe margin. Of course, many people tried to recreatethe supposedly unwritten proof, but to no avail. Infact, Fermats Last Theorem as the result is com-monly called was not proved until 1995, the proofbeing over one hundred pages long and using ex-tremely technical mathematical tools which haventbeen around for very long. Almost every mathemati-cian believes that Fermat must have been either mis-taken or lying.

In geometry, there is a triangle centre known as theFermat point. Given a triangle ABC, it is the point Pwhich makes the sum of the distances PA + PB + PCas small as possible. If no angles of the triangleare greater than or equal to 120, the point P willbe the unique point inside the triangle such thatAPB = BPC = CPA = 120.

39

1. EUCLIDEAN GEOMETRY 1.4. Geometric Gems

The Nine Point Circle

Now we come to one of the real gems of geometry, a theorem which was discovered more than two thousandyears after Euclids day. Remember that for every three points which do not lie on a line, there is a uniquecircle which passes through them. For four points, there is very rarely going to be a circle which passesthrough them all, and if there is, then those four points form a very special type of quadrilateral called acyclic quadrilateral. Given that its already quite difficult for four points to lie on a circle, then it must benear impossible to find nine points which lie on a circle. And yet, this is precisely what the nine point circletheorem tells us we can find nine points which lie on a circle, associated to any particular triangle wechoose to think about.

Theorem (Nine Point Circle Theorem). Let ABC be a triangle with altitudes AD, BE, CF, medians AX, BY, CZ,and orthocentre H. If A, B, C are the midpoints of AH, BH, CH, respectively, then the nine points A, B, C, D, E, F,X, Y, Z all lie on a circle which is for obvious reasons called the nine point circle of triangle ABC.

A B

C

XY

Z

D

E

F

A B

C

H

Proof. There are

topics in geometry

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