2 Euclidean Geometry

2.1 Euclid’s Postulates and Book I of the Elements

Euclid’s Elements (c. 300 BC) was a core part of European and Arabic curricula until the mid 20th

century. Pictures of the oldest fragment and full copy are below, along with various textbook editions.

Earliest Fragment c. AD 100 Full copy, Vatican, 9thC Pop-up edition, 1500’s

Latin translation, 1572 Color edition, 1847 Textbook, 1903

Euclid’s proofs can be found online, and you can read Byrne’s 1847 edition here: the cover is Euclid’sproof of Pythagoras’ Theorem. We now turn to Euclid’s system itself and an overview of Book I.

Undefined Terms E.g., point, line, etc.1

Axioms/Postulates A1 If a = b and b = c then a = c (= is transitive)

A2 If a = b and c = d then a + c = b + d

A3 If a = b and c = d then a− c = b− d

A4 Things that coincide are equal (in magnitude)

A5 The whole is greater than the part

P1 A pair of points may be joined to create a line

P2 A line may be extended

P3 Given a center and a radius, a circle may be drawn

P4 All right angles are equal

P5 If a straight line crosses two others and the angles on one side sum to less than two rightangles then the two lines (when extended) meet on that side.

1In fact Euclid attempted to define these: ‘A point is that which has no part,’ and ‘A line has length but no breadth.’

1

Euclid’s system doesn’t quite fit the modern standard of an axiomatic system. The axioms are a littlevague (what are a, b?!), and there are several more serious shortcomings. We’ll consider some of theselater, but for now we clarify two issues and introduce some notation:

Line segments In Euclid a line was finite in extent. We will refer to these solely as line segments anddenote the segment joining points A, B by AB. In modern geometry, a line extends infinitely.

Congruence Euclid uses equal where modern mathematicians say congruent. We’ll write, say,∠ABC ∼=∠DEF rather than ∠ABC = ∠DEF.

The first three postulates describe the intuitive ruler and compass contructions, which can be physicallyproduced via simple tools. P4 allows Euclid to compare angles located at different locations. P5 isusually known as the parallel postulate.

Basic Theorems a la Euclid

Many theorems were phrased as a problem for which Euclid first provides a construction (postulatesP1–P3) before proving that his construction really solves the problem.

Theorem 2.1 (I. 1). Problem: to construct an equilateral triangle on a given segment.

Proof. Given a line segment AB:By P3, construct circles centered at A and B with radius AB.Call one of the intersection points C; by P1, construct AC and BC.Then4ABC is equilateral.

To prove this, observe that AB and AC are radii of the circle cen-tered at A, while AB and BC are radii of the circle centered at B.By Axiom A1, the three sides of4ABC are congruent.

A B

C

Euclid rapidly proceeds to develop some well-known constructions and properties of triangles.

• (I. 4) Side-angle-side (SAS) congruence: if two triangles have two pairs of congruent sides withthe angles between these being congruent also, then the remaining sides and angles are con-gruent in pairs. Formulaically, this might be written:

AB ∼= DE∠ABC ∼= ∠DEFBC ∼= EF

=⇒

AC ∼= DF∠BCA ∼= ∠EFD∠CAB ∼= ∠FDE

• (I. 5) An isosceles triangle has equal base angles.

• (I. 9) To bisect an angle.

• (I. 10) To find the midpoint of a line.

• (I. 15) If two lines cut one another, opposite angles are equal

Have a look at some of the proofs of these by following the above links.

2

Parallel lines, their construction and uniqueness

Particularly important for our purposes is Euclid’s discussion of parallels.

Definition 2.2. Lines are parallel if they do not intersect. Segments are parallel if no extensions ofthem intersect.a

aThis is not necessarily the familiar definition of parallel: in this context, a line is not parallel to itself.

The next result is one of the most important in Euclidean geometry: it describes how to create aparallel line through a point which is not on any extension of that line.

Theorem 2.3 (I. 16 Exterior Angle Theorem). If one side ofa triangle is extended, then the exterior angle is larger thaneither of the opposite interior angles.In the language of the picture (although Euclid did not quan-tify angles), we have δ > α and δ > β.

α

β δ

A

B C

M

EProof. Draw the bisector BM of AC (Thm I. 10).Extend BM the same distance beyond M to E (Thm I. 2).Connect CE (P1).The opposite angles at M are congruent (Thm I. 15).SAS (Thm I. 4) applied to 4AMB and 4CME says that∠BAM ∼= ∠EMC, which is clearly less than the exteriorangle.

Bisecting BC and repeating the argument shows that β < δ.

Theorem I. 16 essentially constructs a parallel line to AB through a given point C not on the line. Itremains to prove that this line really is parallel.

Theorem 2.4 (I. 27). If a line falls on two other lines in sucha way that the alternate angles (labelled α, β) are congruent,then the original lines are parallel.

α

β

Proof. If the lines were not parallel, then they would meeton one side. WLOG suppose they meet on the right side at apoint C.The angle β at B, being external to 4ABC must be greaterthan the angle α at A (Theorem I. 16): contradiction.

α

β

A

B C

It follows that the line CE constructed in the proof of the Exterior Angle Theorem really is parallel toAB. We have an immediate corollary.

3

Theorem 2.5 (I. 28). If a line falling on two other linesmakes congruent angles, then the original lines are parallel.

Thusfar, Euclid uses only the first four of his postulates. In any model for which the first four pos-tulates are true, it is therefore possible to create parallel lines using the construction of the ExteriorAngle Theorem. Euclid now proves the converse to Theorem I. 27, invoking the parallel postulate forthe first time to show that the alternate angle construction is the only way to create parallel lines.

Theorem 2.6 (I. 29). If a line falls on two parallel lines, then the alternate angles are congruent.

Proof. Given the picture, we must prove that α ∼= β.Suppose not and WLOG that α > β.But then β + γ < α + γ, which is a straight edge.By the parallel postulate, the lines `, m meet on the left sideof the picture, whence ` and m are not parallel.

m

`

αγ

β

n

Angles in a triangle Euclid is finally in a position to prove the most well-known result about trian-gles: that the interior angles sum to a straight edge (180°). Euclid words this slightly differently.

Theorem 2.7 (I. 32). If one side of a triangle is protruded, the exterior angle is equal to the sum ofthe interior opposite angles.

In the language of the picture, the result claims that∠ACD ∼= ∠BAC +∠ABC. For clarity we’ll label the an-gles with Greek letters.

Proof. Construct CE parallel to BA as in Theorem I. 16, sothat ∠ACE ∼= ∠CAB.BD falls on two parallel lines, whence ∠ABC ∼= ∠ECD(Corollary of I. 29).But then ∠ACD ∼= ∠ACE +∠ECD ∼= ∠CAB +∠ABC.

A

B C D

E

β βα

α

Non-Euclidean Geometry Euclid’s proof that the angle sum in a triangle is 180° relies on the par-allel postulate. That he waited so long suggests that he was trying to establish as much as he couldabout triangles and basic geometry in its absence. For centuries, many mathematicians assumed thatsuch a fundamental fact about triangles must be true independently of the parallel postulate.

With the development of hyperbolic geometry in the 17–1800’s, a model was found in which Euclid’sfirst four postulates hold but for which the parallel postulate is false.2 We shall eventually see thatevery triangle in hyperbolic geometry has angle sum less than 180°!

2This shows that the parallel postulate is independent: in fact all the postulates are independent. They are also consis-tent (the ‘usual’ points and lines in the plane are a model), but incomplete. For an sample undecidable, see the exercises.

4

To get this far in hyperbolic geometry will require a lot of work. For amore easily visualized non-Euclidean geometry3 consider elliptic geom-etry, whose simplest example is the round sphere.

Imagine stretching a rubber band between three points on a sphere: theresult is a spherical triangle. In the picture, one corner of an equilateraltriangle is placed at the top of the sphere and the other two corners lieon the equator: the sum of the angles is clearly 270°!

Elliptic geometry is not enough to show that the parallel postulate isindependent in Euclidean geometry since it does not satisfy all of thefirst four of Euclid’s postulates: for instance, one cannot extend a line indefinitely on the sphere.

A similar game can be played on a saddle-shaped surface: as in hyperbolic geometry, triangles willhave angle sum less than 180°.

Playfair’s Postulate

Euclid’s parallel postulate is stated in the negative (angles don’t sum to a straight edge, therefore linesare not parallel). Euclid possibly chose this formulation to facilitate proofs by contradiction, but theeffect is to make the parallel postulate hard to understand. Here is a more modern interpretation.

Axiom 2.8 (Playfair’s Postulate). Given a line ` and apoint P not on `, there exists exactly one parallel line mthrough P.

m

`

P

It should immediately be clear that the parallel postulateimplies Playfair’s postulate.

• Let A, B ∈ ` and construct the triangle4ABP.

• The Exterior Angle Theorem (I. 16) constructs Q andthus a parallel m to ` by Theorem I. 27.

• Theorem I. 29 invokes the parallel postulate to provethat this is the only such parallel.

`

m

B A

P Q

In fact the postulates are equivalent:

Theorem 2.9. In the presence of Euclid’s first four postulates, Playfair’s postulate and the parallelpostulate are equivalent.

Proof. We proved that P5 =⇒ Playfair above. The converse (Playfair =⇒ P5) requires more work.

We prove the contrapositive. Start by assuming Euclid’s postulates P1–P4 are true and that P5 isfalse. We therefore need to consider the negation of the parallel postulate. Using quantifiers, and withreference to the picture in Theorem I. 29, the parallel postulate may be stated:

∀ pairs of lines `, m and ∀ crossing lines n, β + γ < straight edge =⇒ `, m not parallel.

3Loosely, any geometry in which the parallel postulate is false.

5

Its negation is therefore:

∃ parallel lines `, m and a crossing line n, suchthat β + γ < straight edge

The assumption is without loss of generality: if β + γ weregreater than a straight edge, consider the angles on theother side of n.

Use the construction of the Exterior Angle Theorem (I. 16)to build a parallel line ˆ to ` through the intersection P ofm and n: in the picture, β ∼= β. This only requires resultsprior to I. 29 and so is perfectly legitimate in this context.

`

m

γ

β

n

ˆ

`

mPβ

γ

β

Observe that ˆ and m are distinct since β + γ is less than a straight edge. It follows that there exists aline ` and a point P not on that line, though which pass (at least) two parallels to `: we conclude thatPlayfair’s postulate is false.

Pythagoras’ Theorem

Following his discussion of parallels, Euclid shows that parallelograms with the same base andheight are equal (in area) (Thms I. 33–41), before providing explicit constructions of parallelogramsand squares (Thms I. 42–46). Some of this is covered in the exercises. Immediately afterwards, Euclidpresents the crowning achievement of Book I.

Theorem 2.10 (I. 47). The square on the hypotenuse of a right triangle equals (has the same area as)the sum of the squares on the other sides.

Proof. The given triangle4ABC is assumed to have a right angle at A.

1. Construct squares on each side of4ABC (Thm. I. 46)and a parallel AL to BD (e.g. Thm. I. 16).

2. We have AB ∼= FB and BD ∼= BC (sides of squaresare congruent). We also have ∠ABD ∼= ∠FBC sinceboth contain ∠ABC and a right angle.

3. By Side-Angle-Side (Thm. I. 4), 4ABD and 4FBCare congruent (identical up to rotation by 90°).

4. By Thm I. 41,

Area(�ABFG) = 2 Area(4FBC)= 2 Area(4ABD)

= Area(

BOLD)

5. Similarly Area(�ACKH) = Area(

OCEL)

.

A

B CO

LD E

F

G

H

K

6. Sum the rectangles to obtain the square �BCED and complete the proof.

6

Euclid finishes Book I with the converse.

Theorem 2.11 (I. 48). If the squares on two sides of a triangle equal the square on the third side, thetriangle has a right angle opposite the third side.

The remaining books of the Elements contain further geometric constructions, including in three di-mensions, as well as discussions of basic number theory such as what is now known as the Euclideanalgorithm. This is enough pure-Euclid for us however; we now turn to a more modern descriptionof Euclidean geometry, courtesy of David Hilbert.

Exercises. 2.1.1. (a) Prove the ‘vertical angle theorem’ (I. 15): if two lines cut one another, oppositeangles are congruent.(Hint: This is one place where you will need to use postulate 4 regarding right angles)

(b) Use part (a) to complete the proof of the Exterior Angle Theorem: i.e. explain why β < δ.

2.1.2. To prove Pythagoras’ Theorem, Euclid needs to compare areas of triangles and parallelogramsand to construct squares. Prove the following as best as you can: the pictures will help!

(a) (Thm I. 11) At a given point on a line, to construct a perpendicular.(b) (Thm I. 46) To construct a square on a given segment.(c) (Thm I. 35) Parallelograms on the same base and with the same height have equal area.(d) (Thm I. 41) A parallelogram has twice the area of a triangle on the same base and with the

same height.

A B

D

C

A B

CD

A B

CD EF

Thm I. 11 Thm I. 46 Thm I. 35

2.1.3. (a) State the negation of Playfair’s postulate.(b) Argue that in spherical geometry no parallels exist (a line is a great circle. . . )(c) (Hard) Consider the construction from the Exterior Angle Theorem for creating a paral-

lel. Where does the proof fail in spherical geometry?

2.1.4. Prove that Playfair’s Postulate is equivalent to the following statement: Whenever a line isperpendicular to one of two parallel lines, it must be perpendicular to the other.

2.1.5. The line-circle continuity property states:

If point P lies inside and Q lies outside a circle α, then the segment PQ intersects α.

By considering the set of rational points in the plane Q2 = {(x, y) : x, y ∈ Q}, and making asensible definition of line and circle, show that the line-circle continuity property is undecid-able within Euclid’s system.

7

2.2 Hilbert’s Axioms, part I: Incidence and Order

While Euclid’s Elements provided the first serious attempt at an axiomatization of basic geometry, hisapproach contains several errors and omissions. Over the centuries, mathematicians identified theseand worked towards a correct axiomatic system for Euclidean Geometry. The culmination of thiswork came with the publication of David Hilbert’s Grundlagen der Geometrie (Foundations of Geometry)in 1899.Hilbert’s axioms for planar geometry4 are listed on the following page. Our purpose is not to followHilbert strictly, but rather to explore how his system can address some of the difficulties/omissionsin Euclid.

The undefined terms consist of two types of objects (points and lines), and three relations (between, onand congruence.) At various places, definitions and notations are needed: we summarise here.

Definition 2.12. Throughout, A, B, C denote points and `, m lines.

1.←→AB denotes the line through distinct A and B. This exists and is unique by axioms I-1 and I-2.

2. AB denotes the segment consisting of distinct A and B together with all C lying between A andB. This exists and is unique by axioms O-1, O-2 and O-3.

3. The ray−→AB is the segment AB together with all C such that B lies between A and C: that is

A ∗ B ∗ C.

4. An angle ∠BAC with vertex A consists of A together with its sides: two rays−→AB and

−→AC.

5. A triangle 4ABC consists of segments AB, BC and CA where A, B, C are non-collinear. Twotriangles are congruent if their sides and angles are congruent in pairs.

6. Lines ` and m intersect if there exists a point lying on both. Lines are parallel if they do notintersect. Segments and/or rays are parallel if and only if the corresponding lines are parallel.

7. Distinct A and B (not on `) lie on the same side of ` if the segment AB does not intersect `.Otherwise A and B lie on opposite sides of `.

In the usual model of plane geometry, a line is assumed to extend infinitely in both directions, asegment is bounded, and a ray extends infinitely in one direction.

A

B

Line←→AB

A

B

Segment AB

A

B

Ray−→AB

A B

C

Angle ∠BAC

4Hilbert in fact axiomatized 3D geometry: we only give the axioms relevant to planar geometry. These are completein the sense that there is essentially only one model, the ‘usual’ Cartesian Geometry: this is not quite the same as beingcomplete, though it is about as good as one can hope for! The axioms have been shown to be consistent in the absenceof the continuity axiom, though consistency cannot be proved once this is included. As stated, the axioms are not quiteindependent. In particular, axiom O-3 does not require existence (follows from Pasch’s axiom), C-1 does not requireuniqueness (follows from the uniqueness in C-4) and C-6 can be weakened. Hilbert’s axioms were revised after theirinitial publication and we’ve only stated one version.

8

Hilbert’s Axioms for Plane Geometry

Undefined terms

1. Points (use capital letters, e.g., A, B, C)

2. Lines (use lower case letters, e.g., `)

3. On (a point A lies on a line `)

4. Between (a point B lies between points A, C,written A ∗ B ∗ C)

5. Congruence ∼= (of segments/angles)

Axioms of Incidence

I-1 For any distinct A, B there exists a line ` onwhich lie A, B.

I-2 There is at most one line through distinctA, B (A and B both on the line).

Notation: line←→AB through A and B

I-3 On every line there exist at least two dis-tinct points. There exist at least threepoints not all on the same line.

Axioms of Order/Betweenness

O-1 If A ∗ B ∗ C, then A, B and C are distinctpoints on the same line and C ∗ B ∗ A.

O-2 For any distinct points A and B, there is atleast one point C such that A ∗ B ∗ C.

O-3 If A, B, C are distinct points on the sameline, exactly one lies between the othertwo.

Definitions: segment AB and triangle4ABC

O-4 (Pasch’s Axiom) Let 4ABC be a triangleand ` a line not containing any of A, B, C. If` contains a point of the segment AB, thenit also contains a point of either AC or BC.

Axioms of Congruence

Definition: ray−→AB

C-1 If A, B are distinct points and A′ is a point,then for each ray r from A′ there is aunique point B′ on r such that AB ∼= A′B′.Moreover AB ∼= BA.

C-2 If AB ∼= EF and CD ∼= EF, then AB ∼= CD.

C-3 If A ∗ B ∗ C, A′ ∗ B′ ∗ C’, AB ∼= A′B′ andBC ∼= B′C′, then AC ∼= A′C′.

Definitions: angle ∠ABC, side of line←→AB

C-4 Given ∠BAC and a ray−−→A′B′, there is a

unique ray−−→A′C′ on a given side of

←−→A′B′

such that ∠BAC ∼= ∠B′A′C′.

C-5 If ∠ABC ∼= ∠GHI and ∠DEF ∼= ∠GHI,then ∠ABC ∼= ∠DEF. Moreover, ∠ABC ∼=∠CBA.

C-6 (Side-Angle-Side) Given triangles 4ABCand 4A′B′C′, if AB ∼= A′B′, AC ∼= A′C′,and ∠BAC ∼= ∠B′A′C′, then the trianglesare congruent.

Axiom of ContinuitySuppose that the points on ` are partitioned intotwo non-empty subsets Σ1, Σ2 such that no pointof Σ1 lies between two points of Σ2, and viceversa. Then there exists a unique point O lyingon ` such that P1 ∗O ∗ P2 if and only if O 6= P1,O 6= P2 and one of P1 or P2 lies in Σ1 and theother in Σ2.

Playfair’s AxiomDefinition: parallel linesGiven a line ` and a point P not on `, there existsexactly one line through P parallel to `.

Axioms of Incidence

The axioms of incidence explain how the relation on interacts with points and lines. Hilbert’s presen-tation differs from Euclid’s in that he states his axioms and definitions only when required. Modelsof geometries that satisfy only the first few axioms can be provided as we go. For example, there isessentially only one model satisfying the incidence axioms (I-1, I-2, I-3) with exactly three points, andtwo with exactly four points. You should convince yourself that this is the case.

` = {A, B}

m = {A, C}

n = {B, C}

A

B

C

A B

C

D A

B

C

D

3 points, 3 lines 4 points, 6 lines 4 points, 4 lines

By I-3, the 3-point geometry is the smallest possible incidence geometry. We can even prove somevery simple theorems in incidence geometry: these only depend on axioms I-1, I-2 and I-3!

Lemma 2.13. If distinct lines intersect, then they do so in exactly one point.

Proof. Suppose A, B are distinct points of intersection. By axiom I-2, there is exactly one line throughA and B. Contradiction.

Lemma 2.14. Through any point there exists at least two lines.

This is left as an exercise. There are other esoteric examples of incidence geometry, such as the Fanoplane: look some up if you’re interested! Since our main goal is to understand Euclidean Geometry,we move on to the next set of axioms.

Axioms of Order: Sides of a Line and Pasch’s Axiom

The axioms of order explain the use of the ternary relation between for points. Their inclusion inHilbert’s axioms is in no small part due to the work of Moritz Pasch, after whom Pasch’s axiom isnamed (c.1882).

Once you allow the order axioms in addition to the incidence axioms,geometries with only finitely many points bacome impossible! To seethis note that axioms O-1 and O-2 generate a new point C on the line←→AB such that A ∗ B ∗ C. Repeat the exercise to obtain D such thatB ∗ C ∗ D, etc. You should convince yourself that D 6= A so that wegenuinely obtain a new point. In this fashion we see that any line con-tains unboundedly many points.

AB C

DE

10

Of the order axioms, Pasch’s axiom O-4 (first published in 1882) is themost important. It is necessary to properly define the concept of theside of a line (Definition 2.12, part 7), which is used in several of Eu-clid’s arguments. Note how the Definition refers to the same and oppo-site side of a line: this almost presupposes that every line has preciselytwo sides: such an assertion certainly needs a proof.

`

A

B

C

Theorem 2.15 (Plane Separation). We consider the sidedness of points A, B, C not on a line `: thereare essentially three cases.

1. If A, B lie on the same side and B, C lie on the same side, then A, C lie on the same side.

2. If A, B lie on opposite sides and B, C lie on opposite sides, then A, C lie on the same side.

3. If A, B lie on opposite sides and B, C lie on the same side, then A, C lie on opposite sides.

Otherwise said, ` separates the plane into exactly two half-planes; the two sides of `.

`

A

B

C

Case 1

`

A

B

C

Case 2

`

A

BC

Case 3

Proof. All three cases depend on Pasch’s axiom. We prove the first and leave the others as exercises.

We prove the contrapositive. If AC intersects `, then ` intersects one of the sides of4ABC. By Pasch’saxiom, it also intersects one of the other sides AB or BC.Three further extreme cases are when A, B, C are collinear: we omit these to avoid tediousness.

Armed with this, Hilbert defines the inside/interior of a triangle as the intersection of three half-planesand proves that every triangle has a non-empty interior.

Though angles are not required until Hilbert’s congruence axioms, it is convenient to consider theconcept of interior point so that we may properly consider one of Euclid’s early omissions.

Definition 2.16. A point I is interior to angle ∠BAC if:

• I lies on the same side of←→AB as C, and,

• I lies on the same side of←→AC as B.

Otherwise said, I lies in the intersection of two half-planes.It is now possible to compare angles: for instance ∠BAI < ∠BAC. A B

C

I

11

Theorem 2.17 (Crossbar Theorem).

• Suppose I is interior to ∠BAC. Then−→AI intersects BC.

• More specifically, if a line passes through a vertex and aninterior point of a triangle, then it must pass through theside opposite the vertex.

`

AB

C

Proof. Extend AB to a point D such that A lies between B and D (O-2). Since C is not on←→BD =

←→AB

we have a triangle4BCD.←→AI intersects one edge of4BCD at A and does not cross any vertices. By Pasch’s axiom, it intersectsone of the other edges (BC or CD) at a point M. Observe:

• I lies on the same side of←→AB =

←→BD as C, since it is interior to ∠BAC;

• M lies on the same side of←→AB =

←→BD as C, since it is an interior point of either BC or CD.

We conclude (plane separation) that I and M lie on the same side of←→AB, thus

• The segment IM does not intersect←→AB and so A does not lie on IM.

• M lies on−→AI.

There are two generic cases: we need to show that the latter is impossible.

A B

C

I

D

M

Correct arrangementA B

C

I

D

M

The impossible case

If M lies on CD, then I and M must lie on opposite sides of←→AC. But then IM intersects

←→AC, which it

must do at A. This contradicts the fact that A does not lie on IM.

Euclid regularly uses the crossbar theorem without justification,including in his construction of perpendiculars and of angle andsegment bisectors (Theorems I. 9+10); we sketch this here.

Given∠BAC, construct E such that AB ∼= AE. Construct D (equi-lateral triangle Thm I. 1) and appeal to Side-Angle-Side (Thm I. 4)to see that

−→AD bisects BE.

The problem is that Euclid gives no argument for why−→AD should

intersect BE, without which the construction goes nowhere!A B

C

DE

12

Even with Pasch’s axiom and the crossbar theorem, it requires some effort to repair Euclid’s proof: inparticular, we’d need to show that D is interior to the angle ∠BAC: No matter, we shall provide analternative construction of the bisector once we’ve considered congruence more properly.

Exercises. 2.2.1. Prove Lemma 2.14.

2.2.2. Give a model for each of the 5-point incidence geometries: how many are there?(Hint: remember that order doesn’t matter, so the only issue is how many points lie on each line!)

2.2.3. In the discussion on page 10, explain why D 6= A.

2.2.4. We prove part 2 of the plane separation theorem. Suppose that A, B lie on oppoisite side of `,as do B, C. Suppose, for a contradiction, that A, C also lie on opposite sides of `. This meansthat ` intersects all three sides of4ABC!

(a) Label the intersections of ` with each side D ∈ BC, E ∈ AC and F ∈ AB. WLOG E liesbetween D and F: draw a picture of this arrangement, it will look very strange!

(b) By applying Pasch’s axiom to4DBF and←→AC, obtain a contradiction.

2.2.5. Suppose you are given distinct points A, B. Using only axioms up to and including O-4 andthe theorem that the intersection of distinct lines is unique (follows from I-2), show the ex-istence of each of the points C, D, E, F in the picture in alphabetical order. Hence conclude theexistence of a point F lying between A and B.

(a) Explain why D does not lie on←→AB.

(b) Explain why E does not lie on4ABD.

(c) Explain why E 6= C (so that←→CE exists).

(d) Explain why F lies on AB and not on BD. A B

C

D

E

F

2.2.6. Use the previous exercise to prove that the interior of a triangle is non-empty.(Hint: construct a point E relative to4ABC using the previous exercise, then prove that it lies on thesame side of

←→AB as C, etc. . . )

2.2.7. Consider the proof of the Crossbar Theorem.

(a) Explain how we know that−→AI does not pass through any of the vertices of4BCD.

(b) There are really three further possible cases for the arrangement of I, M with respect to4BCD. What are they?

13

2.3 Hilbert’s Axioms, part II: Congruence

The congruence axioms formalize some of Euclid’s constructions and pictorial reasoning, and replacehis confusing use of the word equal. Segments/angles are equal only when they are precisely thesame, the reflexivity part of the following , which depends only on axioms C-1, C-2, C-4 and C-5.

Lemma 2.18. Congruence of segments/angles is an equivalence relation.

Segment/Angle Transfer and Comparison Neither Hilbert nor Euclid require an absolute notion oflength, though both require a method of comparing lengths.

Definition 2.19. Let segments AB and CD be given.

Using axiom C-1, construct the unique point E on−→CD such

that CE ∼= AB: we have now transferred AB onto−→CD.

We write AB < CD if E lies between C and D, etc.C

D

AB

E

By O-3, any two segments are now comparable: given AB, CD, precisely one of the following holds:

AB < CD, CD < AB, AB ∼= CD

C-3 says that congruence respects the addition of congruent segments. Angle transfer, comparisonand addition can be considered similarly using interior points (Definition 2.16).

Side-Angle Side/SAS The final congruence axiom is the SAS congruence property. This is Euclid’sTheorem I. 4 which he ‘proves’ by the unjustified procedure of laying one triangle on top of another.It was ultimately realized that at least one of the four triangle congruence results (SAS, ASA, SSS andSAA) had to be an axiom. Hilbert assumes SAS and proceeds to prove the remainder. For example:

Theorem 2.20 (Angle-Side-Angle/ASA, Euclid I. 26, case I). Suppose4ABC and4DEF satisfy

∠ABC ∼= ∠DEF, AB ∼= DE, ∠BAC ∼= ∠EDF

Then the triangles are congruent.

Proof. Given the assumptions, axiom C-1 gives a unique point G on the ray−→EF such that EG ∼= BC.

Axiom C-6 (SAS) says that ∠BAC ∼= ∠EDG which, by assump-tion, is congruent to ∠EDF.

Since F and G lie on the same side of←→DE, axiom C-4 says that they

lie on the same ray through D.

But then F and G both lie on two distinct lines (←→EF and

←→DF): we

conclude that F = G.

By SAS we conclude that4ABC ∼= 4DEF.

A B

C

D E

FG

14

Geometry Without Circles

Circles are are the heart of Euclid’s constructions yet, for reasons we’ll deal with shortly, Hilbert es-sentially ignores them. Hilbert therefore has alternative approaches to some of Euclid’s basic results.We sketch a few.

Theorem 2.21 (Euclid I. 5). An isosceles trianglea has congruent base angles.

aIsoscles means equal legs: i.e. two sides of the triangle are congruent. The remaining side is the base.

Euclid’s argument is famously unpleasant, relying on a complicated construction. Hilbert does thingsfar more speedily.

Proof. Suppose WLOG that AB ∼= AC so that4ABC is isosceles. Con-sider a ‘new’ triangle 4A′B′C′ = 4ACB where the base points areswitched:

A′ = A, B′ = C, C′ = B

We have:

• ∠BAC ∼= ∠CAB (axiom C-5) =⇒ ∠BAC ∼= ∠B′A′C′.

• AB ∼= AC =⇒ AB ∼= A′B′ and AC ∼= A′C′.

SAS says that ∠ABC ∼= ∠A′B′C′ ∼= ∠ACB.

A = A′

B = C′ C = B′

The proof is very sneaky: just relabel the original triangle and apply SAS!

The converse to this statement is in the Exercises.

Dropping a Perpendicular Suppose C does not lie on `. By axiomI-3, ` =

←→AB for some points A, B.

Use axioms C-4 and C-1 to transfer ∠BAC to the other side of the lineat A, creating a new point D.

Since C and D lie on opposite sides of←→AB, it must intersect CD at some

point M.SAS applied to 4MAC and 4MAD shows that ∠AMC ∼= ∠AMD:since these sum to a straight edge, they must be right angles.

• The construction still works if∠BAC is greater than a right angle:∠MAC is now supplementary to ∠BAC rather than equal.

• In the extreme case M = A, we have no triangles and SAS can-not be applied. Instead follow the same argument starting with∠ABC. . .

A B M

C

D

A generalization of this construction facilitates a corrected argument for the SSS congruence: Euclidagain resorted to his flawed ‘laying on top of’ reasoning.

15

Theorem 2.22 (Side-Side-Side/SSS, Euclid I. 8). If two triangles have sides congruent in pairs,then the triangles are congruent.

Proof. Suppose 4ABC and 4DEF have sides congruent inpairs as shown.Transfer∠EDF to A on the other side of AB from C to obtainG (axioms C-4 and C-1).SAS shows that4BAG ∼= 4EDF and so BG ∼= EF ∼= BC.Joining CG we now have two isosceles triangles with apicesat A and B and congruent base angles at C and G.Summing angles at C and G and applying SAS shows that4ACB ∼= 4AGB ∼= 4DFE as required.

As with the perpendicular argument above, we should alsocarefully deal with the situtations where the sum is a sub-traction or the triangle is right-angled at A or B.

A B

C

G

D

E

F

The Exterior Angle Theorem (Euclid I. 16) Euclid’s approach requires a bisector which he obtainsfrom circles. Hilbert does things a little differently.

Proof. Given4ABC, extend AB to D such that AB ∼= BD.In the first picture below, assume δ ∼= γ. Apply SAS to see that4ACB ∼= 4DBC: in particular ε ∼= β.

Clearly A and D lie on opposite sides of←→BC: but then ε + γ ∼= β + δ is a straight edge and so A, D are

distinct points lying on two lines! Contradiction.We conclude that δ � γ. By taking the vertical angle to δ at B, we see that δ is not congruent to αeither.

A B

C

D

α β

γ

δ

ε

A B

C

DE

α β

η

δ

ε

Step 1: δ ∼= γ is a contradiction Step 2: δ < γ is a contradiction

In the second picture, assume δ < γ. Transfer the angle to C as shown where η ∼= δ: by the crossbartheorem, we obtain an intersection point E. But now δ is an exterior angle of 4EBC and congruentto an interior angle η of the same triangle: contradiction.

The only non-contradictory possibility is that δ > γ (similarly δ > α), which concludes the proof.

The Exterior Angle Theorem also proves that the sum of any two angles in a triangle is strictly lessthan a straight edge: α + β < δ + β.

16

Is Euclid now fixed? Almost! In the exercises we show how the following may be achieved inHilbert’s system:

• Construction of an isosceles triangle on a segment AB. With this one can construct segmentand angle bisectors (Euclid I. 9+10).

• SAA congruence (Euclid I. 26, case II): the last remaining triangle congruence theorem.

At this point we’ve recovered almost all of Book I prior to the application of the parallel postulate.Including Playfair’s axiom allows the remainder of Book I to be completed, including Pythagoras’Theorem, all without circles!

Exercises. Except for the first and last questions, all exercises should be answered in neutralgeometry without reference to the continuity axiom: you should not use circles, or anythingregarding uniqueness of parallels such as Playfair’s axiom or the angle sum of a triangle being180°. All you need are the results in sections 2.2 and 2.3.

2.3.1. Draw a picture to suggest why Angle-Angle-Angle (AAA) and Side-Side-Angle (SSA) are nottriangle congruence theorems in Euclidean geometry.

2.3.2. Use Hilbert’s axioms C-4 and C-5 to prove that congruence of angles is an equivalence relation.

2.3.3. (a) Use ASA to prove the converse to Theorem 2.21: if the base angles are congruent then atriangle is isosceles.

(b) Find an alternative proof that relies on the Exterior Angle Theorem.(c) Explain why the base angles of an isosceles triangle are acute (less than a right angle).

2.3.4. Suppose AB is given. By axiom I-3, ∃C 6∈ ←→AB.If4ABC is not isosceles, then WLOG assume ∠ABC < ∠BAC.By C-1 and C-4, transfer ∠ABC to A to produce D on the sameside of

←→AB as C and for which

∠ABC ∼= ∠BAD, BC ∼= AD

A B

C D

M

(a) Explain why rays−→AD and

−→BC intersect at a point M.

(b) Why is4MAB isosceles?(c) Explain how to produce the perpendicular bisector of AB.(d) Explain how to construct an angle bisector using the above discussion.

2.3.5. We prove Theorems I. 18, 19 and 20, on comparisonsof angles and sides in a triangle. For clarity, suppose4ABC has sides and angles labelled as in the picture.

c

b a

A B

C

α β

γ

(a) (I. 18) Assume a < c. Prove that α < γ.(Hint: let D on AB satisfy BD ∼= a)

(b) (I. 19) Prove the converse: if α < γ, then a < c.(Hint: Show that a ≥ c is impossible)

(c) (I. 20) Prove the triangle-inequality. For any triangle, a + b > c.(Hint: Let E lie on

−→BC such that CE ∼= b and apply the previous part)

17

2.3.6. Prove the SAA congruence. If4ABC and4DEF satisfy

AB ∼= DE, ∠ABC ∼= ∠DEF and ∠BCA ∼= ∠EFD

then the triangles are congruent4ABC ∼= 4DEF.(Hint: Let G on

−→BC be such that BG ∼= EF, then apply SAS

and the Exterior Angle Theorem)

G

BA

C

2.3.7. Hilbert’s published SAS axiom is actually weaker than we stated.

Given triangles 4ABC and 4A′B′C′, if AB ∼= A′B′, AC ∼= A′C′, and ∠BAC ∼=∠B′A′C′, then ∠ABC ∼= ∠A′B′C′.

Use this to prove the full SAS congruence theorem (axiom C-6 as we’ve stated it).(Hint: try a trick similar to that in the proof of ASA)

2.3.8. Consider the picture constructed as follows.

• BF is the perpendicular bisector of AC

• BD ∼= AB

• BE ∼= AD

• BF ∼= AE

Use Pythagoras’ Theorem to prove that4ACF is equilateral. A B C

DE

F

(Since Pythagoras’ requires Playfair’s axiom, this construction is false in hyperbolic geometry)

2.4 Circles and Continuity

Definition 2.23. Let O and R be distinct points. The circle C with center Oand radius OR is the collection of points A such that OA ∼= OR.A point P lies inside the circle C if P = O or OP < OR.A point Q lies outside if OR < OQ.Since all segments are comparable, any point lies inside, outside or on a givencircle.

O

RA

P

Q

A major weakness of Euclid is that many of his proofs rely on circle intersections. To use circles inthis manner requires the Axiom of Continuity, from which the following statements can be proved.

Theorem 2.24. Suppose C and D are circles.

1. (Elementary Continuity Principle) If P lies inside and Q outside C, then PQ intersects C inexactly one point.

2. (Circular Continuity Principle) If D contains a point inside and another outside C, then thecircles intersect in exactly two points; these lie on opposite sides of the line joining the circlecenters.

18

Since the Axiom of Continuity is much more technical than the previous axioms, Hilbert barely men-tioned circles: he wanted to build as much geometry as possible using only the simplest axioms.

The idea of the first principle is to partition PQ into two pieces:

Σ1 consists of the points lying inside or on C,

Σ2 consists of the points lying outside C.

One shows that Σ1 and Σ2 satisfy the assumptions of the Axiom: theunique point O then exists and is shown to lie on C itself.

P

QO

Σ1

Σ2

C

A full discussion of this (even a sketch of the second argument) requires significant Analysis andtherefore lies beyond the scope of this course. What is perhaps more interesting is to consider ageometry in which the axiom of continuity is false.

Example 2.25. The geometry Q2 = {(x, y) ∈ R2 : x, y ∈ Q} of points in the plane with rationalco-ordinates satisfies almost all of Hilbert’s axioms, however C-1 and continuity are false.

Axiom C-1 Given points A = (0, 0), B = (1, 0) and C = (1, 1), we seethat O = ( 1√

2, 1√

2) is the unique point (in R2) on the ray r =

−→AC

such that AC ∼= AB. ClearlyO is an irrational point and thereforenot in the geometry.

Continuity The circle centered at A = (0, 0) with radius 1 does notintersect the segment AC. More properly, the segment AC =Σ1 ∪ Σ2 may be partitioned as shown and yet no point O in thegeometry separates Σ1, Σ2.

Σ1

Σ2

A B

CO

Equilateral triangles We can finally correct Euclid’s proof of the first proposition of the Elements!

Definition 2.26. A triangle4ABC is equilateral if its three sides are congruent.

Theorem 2.27 (Euclid I.1). An equilateral triangle many be constructed on a given segment AB.

Proof. Following Euclid, take the circles α and β centered at A and B, with radii AB.Axiom O-2: ∃D such that A ∗ B ∗ D.Axiom C-1: let C ∈ −→BD be such that BC ∼= AB.Circular continuity principle: β contains A (inside α) andC (outside α) whence the circles intersect in precisely twopoints P, Q.Since P lies on both circles (and is therefore distinct fromA and B), it follows that AB ∼= AP ∼= BP and that 4ABCis equilateral.

A B C D

P

Qα β

If one allows Playfair’s axiom on unique parallels, Euclid’s result can be proved without using circlesor the continuity axiom (see Exercise 2.3.8.). Nonetheless, we are finally able to say that every resultin Book I of Euclid is correct, even if his original axioms and arguments are insufficient!

19

Basic Circle Geometry

We continue our survey of Euclidean Geometry with a few results about circles, many of whichare found in Book III of the Elements. As such, from this point onwards we assume all of Hilbert’sAxioms including Playfair and Continuity: indeed these results often rely on their consequences,namely angle-sums in triangles and the circular continuity priniple.

Definition 2.28. With reference to the picture:

• A chord AB is a segment joining two points on a circle.

• A diameter BC is a chord passing through the center O.

• An arc is part of the circular edge between the chord points(major or minor by length).

• ∠AOB is a central angle and ∠APB an inscribed angle.

• 4ABP is inscribed in the circle.

A

B

C

P

O

Since you should by now be quite comfortable with Euclidean Geometry, most of the details are leftas exercises.

Theorem 2.29 (III. 20). The central angle is twice the inscribed angle: ∠AOB = 2∠APB.

Simply join O to A, B, P, breaking4ABP into three isosceles triangles and count angle sums. . .

Corollary 2.30. 1. (III. 21) If inscribed triangles share a side, the opposite angles are congruent.

2. (III. 31) A triangle in a semi-circle is right-angled (Thales).

3. (III. 22) A quadrilateral inscribed in a circle has its opposite angles supplementary (summing toa straight edge).

Theorem 2.31. Given three non-collinear points, there existsa unique circle through them.

This is similar to Thm III. 1: just construct the perpendiularbisectors of two sides.

A

B

C

O

Definition 2.32. A line is tangent to a circle if it intersects the circle exactly once.

Theorem 2.33 (III. 18, 19 (in part)). A line is tangent to a circle if and only if it is perpendicular tothe radius at the point of tangency.

20

Proof. Suppose ` through T is perpendicular to the radius OT.Let P be another point on `. But then (Exercise 2.3.5.),

∠OPT < ∠OTP =⇒ OT < OP

Thus P lies outside the circle, which ` intersects only at T: the lineis therefore tangent.

The converse is an exercise.

T

O

P

`

Theorem 2.34. Through a point outside a circle, exactly two lines are tangent to the circle.

Exercises. 2.4.1. Give formal proofs of all parts of Corollary 2.30.

2.4.2. Prove Theorem 2.31.

2.4.3. Complete the proof of Theorem 2.33 by showing the converse.(Hints: suppose ` isn’t tangent and drop a perpendicular to O. Use the result of Exercise 2.4.5.)

2.4.4. Given a circle centered at O and a point P outside the circle, draw the circle centered at themidpoint of OP. Explain why the intersections of these circles are the points of tangencyrequired in Theorem 2.34, and hence complete its proof.

2.4.5. (Hard) If a line contains a point inside a circle, show that it intersects the circle in two points.(Hint: first show that some point on the line lies outside the circle)

2.5 Similar Triangles, Length and Trigonometry

To avoid continued frustration, it is time to modernize and accept length and area as absolute nu-merical quantities. It is important to realize that none of this is necessary. Measuring length and anglemerely makes the language more familiar and certain computations simpler.

Axioms 2.35 (Length and Angle Measure).

L1 To each AB corresponds its length |AB|, a positive real number

L2 |AB| = |CD| ⇐⇒ AB ∼= CD

L3 |AB| < |CD| ⇐⇒ AB < CD (recall Definition 2.19)

L4 If B lies between A and C, then |AB|+ |BC| = |AC|A1 To each ∠ABC corresponds its degree measure m∠ABC, a real number in the interval (0, 180)

A2 m∠ABC = m∠DEF ⇐⇒ ∠ABC ∼= ∠DEF

A3 m∠ABC < m∠DEF ⇐⇒ ∠ABC < ∠DEF (recall Definition 2.16)

A4 If P is interior to ∠ABC, then m∠ABP + m∠PBC = m∠ABC

A5 Right angles measure 90°

Don’t memorize these! Just read and observe how they fit your intuition for how measurement

21

of length and angle behave. The axioms for length and angle are very similar, the only signficantdifference being that angle measure is normalized to a fixed scale by fixing the measure of a rightangle at 90°. To do the same for length requires only a choice of a reference segment of length 1.Indeed the following is a consequence of the continuity axiom.

Theorem 2.36 (Uniqueness of measure).

1. Given a segment OP, there is a unique way to assign a length to every segment in such a waythat |OP| = 1.

2. There is a unique way to assign a degree measure to every angle.

While the uniqueness of measure requires the continuity axiom, including Playfair means that rect-angles can be defined, which allows the easy numerical measurement of area.

Definition 2.37. The area (measure) of a rectangle is the product of the lengths of its base and height.

Given a fixed length measure, a square with side length 1 necessarily has area 1. By Euclid’s discus-sion towards the end of Book I, we immediately see that parallelograms also have area measuringbase times height and triangles half this.

The AAA Similarity Theorem

Similar triangles are partly the concern of Book VI of the Elements.

Definition 2.38. Triangles are similar, written4ABC ∼ 4XYZ,if their sides are in the same length ratio

|AB||XY| =

|BC||YZ| =

|CA||ZX|

AB

C

XY

Z

We could work with similar triangles as did Euclid, without a numerical length measure and usingratios of lengths in a relative manner, AB : XY = BC : YZ, etc. This approach is unnecessarilyconfusing to modern readers.

Our primary result is the following, essentially Theorems VI. 2–5.

Theorem 2.39 (Angle-Angle-Angle).

1. Triangles are similar if and only if their angles come in mutually congruent pairs.

2. Suppose a line intersects two sides of a triangle. The smaller triangle so created is similar to theoriginal if and only if the line is parallel to the third side of the triangle.

The proof is an irritating exercise which follows from a tickly lemma.

22

Lemma 2.40. Suppose ` intersects sides AB and AC of4ABC at D and E as in the picture. The following areequivalent:

1. ` is parallel to BC

2.|BD||AD| =

|CE||AE|

3. 4ABC ∼ 4ADE

`

A

B C

D E

hk

d1 d2

Proof. (1 ⇐⇒ 2) Note first that

|BD||AD| =

area(BDE)area(ADE)

(4BDE,4ADE have same height h, whence area ∝ base)

|CE||AE| =

area(CDE)area(ADE)

(triangles have the same height k; again area ∝ base)

Drop perpendiculars from B and C to ` and note that4BDE and4CDE share the base DE. Thus

` is parallel to BC ⇐⇒ d1 = d2 ⇐⇒ area(BDE) = area(CDE) ⇐⇒ |BD||AD| =

|CE||AE|

(1 + 2⇒ 3) Add |AD||AD| =

|AE||AE| to both sides of part 2 to obtain |AB|

|AD| =|AC||AE| . Now observe

|AB||BD| =

area(ABE)area(BDE)

|CE||AE| =

area(BCE)area(ABE)

Apply part 2 and the above to see that

|AB||AD| =

|AB||BD|

|CE||AE| =

area(BCE)area(BDE)

=|BC||DE|

where the last equality follows since d1 = d2 is the common height of4BCE and4BDE.

(3 ⇒ 1) By Playfair, let m be the unique parallel to BCthrough D. This intersects AC at G. Since 1⇒ 3, we have

4ABC ∼ 4ADG

The transitivity of ∼ forces4ADE ∼ 4ADG: the similarityratio is 1, whence G = E and m = `.

`

m

A

B CD

E

G

Given the liberal use of Playfair’s axiom,5 we should not expect AAA similarity in non-Euclideangeometry. Indeed we shall later see that AAA is a theorem for congruent triangles in hyperbolicgeometry!

5d1 = d2 ⇐⇒ ` parallel to BC is also Playfair: compare Exercise 2.1.2. (Thm I. 46).

23

Trigonometric Functions We can use similar triangles to define the sine and cosine of an angle.

Definition 2.41. Given an acute angle ∠ABC, drop a perpendicular from A to−→BC at D so that

∠ADB is a right angle. Now define

sin∠ABC :=|AD||AB| cos∠ABC :=

|BD||AB|

Theorem 2.42. Angles have the same sine (cosine) if and only if they are congruent.a

aIn essence, this says that SSA is a congruence theorem, provided the angle is 90°!

Proof. Consider the picture: assume ∠ABC ∼= ∠A′B′C′ and drop perpendiculars to D, D′.Since 4ABD and 4A′B′D′ have two pairs of mutu-ally congruent angles, the third pair is congruent also.AAA applies, whence the triangles are similar and wehave equal ratios

|AD||AB| =

|A′D′||A′B′|

|BD||AB| =

|B′D′||A′B′|

A

B CD

A′

B′

C′

D′

In particular, sin∠ABC = sin∠A′B′C′ and cos∠ABC = cos∠A′B′C′.The converse is an exercise.

This is very different to how the ancient forerunners of sine and cosine were defined using chords ofcircles rather than triangles. The word trigonometry (lit. triangle measure) wasn’t coined until 1595!

Cevians: non-examinable After Giovanni Ceva (1647–1734), a cevian is a segment joining a vertexto the opposite side of a triangle. Here is a beautiful result from the height of Euclidean geometry.Good luck trying to prove this using co-ordinates (analytic geometry)!

Theorem 2.43 (Ceva’s Theorem). Given a triangle4ABC and cevians AX, BY, CZ,

|BX||XC|

|CY||YA|

|AZ||ZB| = 1 ⇐⇒ the cevians meet at a common point P

Proof. (⇐) We have pairs of triangles with the same heights, the areas of which are proportional tothe lengths of their bases:

area(ABX)

area(AXC)=|BX||XC| =

area(PBX)

area(PXC)

=⇒ area(ABX)− area(PBX)

area(AXC)− area(PXC)=

area(ABP)area(APC)

=|BX||XC|

Repeat for the other ratios and multiply to get 1.The converse is an exercise.

A

B CX

YZ

P

24

Exercises. 2.5.1. Let4ABC have a right-angle at C. Drop a perpendicular from C to←→AB at D.

(a) Prove that D lies between A and B.

(b) Prove that you have three similar triangles

4ACB ∼ 4ADC ∼ 4CDB

(c) Use these facts to prove Pythagoras’ Theorem.(Use the picture, where a, b, c, x, y are lengths)

ab

c = x + y

x y

A B

C

D

2.5.2. Suppose AD and BC are chords of a circle which intersect at P. Use similar triangles to showthat |AP|

|BP| =|PC||PD| .

2.5.3. Prove a simplified version of the SAS similarity theorem:

|AB||AG| =

|AC||AH| ⇐⇒ 4ABC ∼ 4AGH

(Hint: construct BJ parallel to GH and appeal to AAA)

A

BC

G

H2.5.4. By excluding the other possibilities, prove the converse of length axiom L4:

If A, B, C are distinct and |AB|+ |BC| = |AC|, then B lies between A and C.

2.5.5. Use Pythagoras’ to prove that sin 45° = cos 45° = 1√2, that sin 60° =

√3

2 and cos 60° = 12 .

2.5.6. Prove the converse of Theorem 2.42: if sin∠ABC = sin∠A′B′C′, then ∠ABC ∼= ∠A′B′C′.(Hint: create right-triangles and prove they are similar. To make things easier, label the side lengthso, a, h, etc.)

2.5.7. (Hard) Explain how to prove the full AAA Theorem using Lemma 2.40.

The last two questions rely on the optional discussion of Cevians.

2.5.8. We prove the (⇒) direction of Ceva’s Theorem.

Assume X, Y, Z satisfy the formula: following the pic-ture, define P as the intersection of BY and CZ and letthe ray

−→AP meet BC at X′.

Use the (⇐) direction of Ceva’s Theorem to prove thatX′ = X.

A

B CXX′

YZ

P

2.5.9. (a) A median of a triangle is a segment from a vertex to the midpoint of the opposite side.Use Ceva’s Theorem to prove that the medians of a triangle meet at a point (the centroid).

(b) The medians split the triangle into six sub-triangles. Prove that all have the same area.

(c) Prove that the centroid is exactly 2/3 of the distance along each median.

25