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’s proof 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 right angles 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 little vague (what are a, b?!), and there are several more serious shortcomings. We’ll consider some of these later, 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 and denote 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 physically produced via simple tools. P4 allows Euclid to compare angles located at different locations. P5 is usually known as the parallel postulate.

Basic Theorems `ala Euclid Many theorems were phrased as a problem for which Euclid ﬁrst provides a construction (postulates P1–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: C 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. Then ABC is equilateral. 4 A B To prove this, observe that AB and AC are radii of the circle centered at A, while AB and BC are radii of the circle centered at B. By Axiom A1, the three sides of ABC are congruent. 4 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 with the angles between these being congruent also, then the remaining sides and angles are congruent in pairs. Formulaically, this might be written:   AB = DE AC = DF  ∼  ∼ ABC = DEF = BCA = EFD ∠ ∼ ∠ ⇒ ∠ ∼ ∠   BC ∼= EF ∠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 of them 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 a parallel line through a point which is not on any extension of that line.

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

Proof. Draw the bisector BM of AC (Thm I. 10). A E Extend BM the same distance beyond M to E (Thm I. 2). Connect CE (P1). M The opposite angles at M are congruent (Thm I. 15). SAS (Thm I. 4) applied to AMB and CME says that 4 4 ∠BAM ∼= ∠EMC, which is clearly less than the exterior angle. B C 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. It remains to prove that this line really is parallel.

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

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

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

3 Theorem 2.5 (I. 28). If a line falling on two other lines makes 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 postulates are true, it is therefore possible to create parallel lines using the construction of the Exterior Angle Theorem. Euclid now proves the converse to Theorem I. 27, invoking the parallel postulate for the 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.

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

Angles in a triangle Euclid is ﬁnally in a position to prove the most well-known result about triangles: 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 of the interior opposite angles.

In the language of the picture, the result claims that A E ∠ACD ∼= ∠BAC + ∠ABC. For clarity we’ll label the angles with Greek letters. α

Proof. Construct CE parallel to BA as in Theorem I. 16, so α˜ ˜ that ∠ACE ∼= ∠CAB. β β BD falls on two parallel lines, whence ∠ABC ∼= ∠ECD B C D (Corollary of I. 29). But then ∠ACD ∼= ∠ACE + ∠ECD ∼= ∠CAB + ∠ABC.

Non-Euclidean Geometry Euclid’s proof that the angle sum in a triangle is 180° relies on the parallel postulate. That he waited so long suggests that he was trying to establish as much as he could about triangles and basic geometry in its absence. For centuries, many mathematicians assumed that such 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’s ﬁrst four postulates hold but for which the parallel postulate is false.2 We shall eventually see that every 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 consistent (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 a more easily visualized non-Euclidean geometry3 consider elliptic geometry, whose simplest example is the round sphere. Imagine stretching a rubber band between three points on a sphere: the result is a spherical triangle. In the picture, one corner of an equilateral triangle is placed at the top of the sphere and the other two corners lie on the equator: the sum of the angles is clearly 270°! Elliptic geometry is not enough to show that the parallel postulate is independent in Euclidean geometry since it does not satisfy all of the ﬁrst 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 will have 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 lines are not parallel). Euclid possibly chose this formulation to facilitate proofs by contradiction, but the effect 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 a m P point P not on `, there exists exactly one parallel line m through P. `

It should immediately be clear that the parallel postulate P m Q implies Playfair’s postulate.

• Let A, B ` and construct the triangle ABP. ∈ 4 • The Exterior Angle Theorem (I. 16) constructs Q and thus a parallel m to ` by Theorem I. 27. • Theorem I. 29 invokes the parallel postulate to prove that this is the only such parallel. B ` A In fact the postulates are equivalent:

Theorem 2.9. In the presence of Euclid’s ﬁrst four postulates, Playfair’s postulate and the parallel postulate 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 is false. We therefore need to consider the negation of the parallel postulate. Using quantiﬁers, and with reference 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: n m parallel lines `, m and a crossing line n, such ∃ that β + γ < straight edge γ β The assumption is without loss of generality: if β + γ were ` greater than a straight edge, consider the angles on the other side of n. ˆ m β P Use the construction of the Exterior Angle Theorem (I. 16) `ˆ to build a parallel line `ˆ to ` through the intersection P of γ ˆ m and n: in the picture, β ∼= β. This only requires results β prior to I. 29 and so is perfectly legitimate in this context. `

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

Pythagoras’ Theorem Following his discussion of parallels, Euclid shows that parallelograms with the same base and height are equal (in area) (Thms I. 33–41), before providing explicit constructions of parallelograms and squares (Thms I. 42–46). Some of this is covered in the exercises. Immediately afterwards, Euclid presents 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 triangle ABC is assumed to have a right angle at A. 4 1. Construct squares on each side of ABC (Thm. I. 46) H 4 and a parallel AL to BD (e.g. Thm. I. 16). G 2. We have AB ∼= FB and BD ∼= BC (sides of squares K are congruent). We also have ∠ABD ∼= ∠FBC since A both contain ABC and a right angle. ∠ F 3. By Side-Angle-Side (Thm. I. 4), ABD and FBC 4 4 are congruent (identical up to rotation by 90°). B CO 4. By Thm I. 41,

Area( ABFG) = 2 Area( FBC) 4 = 2 Area( ABD) 4 = Area L BOLD DE 5. Similarly Area(ACKH) = Area OCEL .

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, the triangle 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 Euclidean algorithm. This is enough pure-Euclid for us however; we now turn to a more modern description of Euclidean geometry, courtesy of David Hilbert.

Exercises. 2.1.1. (a) Prove the ‘vertical angle theorem’ (I. 15): if two lines cut one another, opposite angles 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 parallelograms and 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. D C DEF C D

C A B

A B A B 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 parallel. 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 is perpendicular 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 a { ∈ } sensible deﬁnition of line and circle, show that the line-circle continuity property is undecidable 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, his approach contains several errors and omissions. Over the centuries, mathematicians identified these and worked towards a correct axiomatic system for Euclidean Geometry. The culmination of this work 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 follow Hilbert strictly, but rather to explore how his system can address some of the difficulties/omissions in Euclid. The undefined terms consist of two types of objects (points and lines), and three relations (between, on and 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 and B. 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 ABC consists of segments AB, BC and CA where A, B, C are non-collinear. Two 4 triangles 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 not intersect. 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, a segment is bounded, and a ray extends inﬁnitely in one direction.

C B B B

A A A A B Line ←→AB Segment AB Ray −→AB Angle ∠BAC

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

8 Hilbert’s Axioms for Plane Geometry

Undefined terms Axioms of Congruence 1. Points (use capital letters, e.g., A, B, C) Definition: ray −→AB C-1 If A, B are distinct points and A is a point, 2. Lines (use lower case letters, e.g., `) 0 then for each ray r from A0 there is a 3. On (a point A lies on a line `) unique point B0 on r such that AB ∼= A0B0. Moreover AB ∼= BA. 4. Between (a point B lies between points A, C, written A B C) C-2 If AB = EF and CD = EF, then AB = CD. ∗ ∗ ∼ ∼ ∼ 5. Congruence = (of segments/angles) C-3 If A B C, A0 B0 C’, AB = A B and ∼ ∗ ∗ ∗ ∗ ∼ 0 0 BC ∼= B0C0, then AC ∼= A0C0. Axioms of Incidence Definitions: angle ∠ABC, side of line ←→AB I-1 For any distinct A, B there exists a line ` on which lie A, B. C-4 Given ∠BAC and a ray −−→A0B0, there is a unique ray −−→A C on a given side of ←−→A B I-2 There is at most one line through distinct 0 0 0 0 such that ∠BAC = ∠B0 A0C0. A, B (A and B both on the line). ∼ C-5 If ABC = GHI and DEF = GHI, Notation: line ←→AB through A and B ∠ ∼ ∠ ∠ ∼ ∠ then ∠ABC ∼= ∠DEF. Moreover, ∠ABC ∼= I-3 On every line there exist at least two dis- ∠CBA. tinct points. There exist at least three C-6 (Side-Angle-Side) Given triangles ABC points not all on the same line. 4 and A B C , if AB = A B , AC = A C , 4 0 0 0 ∼ 0 0 ∼ 0 0 and ∠BAC ∼= ∠B0 A0C0, then the triangles Axioms of Order/Betweenness are congruent. O-1 If A B C, then A, B and C are distinct ∗ ∗ points on the same line and C B A. Axiom of Continuity ∗ ∗ Suppose that the points on ` are partitioned into O-2 For any distinct points A and B, there is at two non-empty subsets Σ1, Σ2 such that no point least one point C such that A B C. ∗ ∗ of Σ1 lies between two points of Σ2, and vice O-3 If A, B, C are distinct points on the same versa. Then there exists a unique point O lying on ` such that P O P if and only if O = P , line, exactly one lies between the other 1 ∗ ∗ 2 6 1 O = P and one of P or P lies in Σ and the two. 6 2 1 2 1 other in Σ2. Definitions: segment AB and triangle ABC 4 O-4 (Pasch’s Axiom) Let ABC be a triangle Playfair’s Axiom 4 and ` a line not containing any of A, B, C. If Definition: parallel lines ` contains a point of the segment AB, then Given a line ` and a point P not on `, there exists it also contains a point of either AC or BC. exactly 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. Models of geometries that satisfy only the first few axioms can be provided as we go. For example, there is essentially only one model satisfying the incidence axioms (I-1, I-2, I-3) with exactly three points, and two with exactly four points. You should convince yourself that this is the case.

C C D

n = B, C { } m = A, C { } C B D A ` = A, B { } A A B B 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 some very 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 through A 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 Fano plane: 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 in Hilbert’s axioms is in no small part due to the work of Moritz Pasch, after whom Pasch’s axiom is named (c.1882). Once you allow the order axioms in addition to the incidence axioms, geometries with only ﬁnitely many points bacome impossible! To see E this note that axioms O-1 and O-2 generate a new point C on the line D B C ←→AB such that A B C. Repeat the exercise to obtain D such that ∗ ∗ A B C D, etc. You should convince yourself that D = A so that we ∗ ∗ 6 genuinely obtain a new point. In this fashion we see that any line contains unboundedly many points.

10 Of the order axioms, Pasch’s axiom O-4 (first published in 1882) is the C most important. It is necessary to properly define the concept of the side 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 precisely B two sides: such an assertion certainly needs a proof. A

Theorem 2.15 (Plane Separation). We consider the sidedness of points A, B, C not on a line `: there are 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 `.

B A A

A ` ` C ` C B B C Case 1 Case 2 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 of ABC. By Pasch’s 4 axiom, 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-planes and proves that every triangle has a non-empty interior. Though angles are not required until Hilbert’s congruence axioms, it is convenient to consider the concept 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: C • I lies on the same side of ←→AB as C, and, • I lies on the same side of ←→AC as B. I Otherwise said, I lies in the intersection of two half-planes. It is now possible to compare angles: for instance ∠BAI < ∠BAC. A B

11 Theorem 2.17 (Crossbar Theorem). C • Suppose I is interior to ∠BAC. Then −→AI intersects BC. • More speciﬁcally, if a line passes through a vertex and an ` interior point of a triangle, then it must pass through the B side opposite the vertex. A

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 triangle BCD. 4 ←→AI intersects one edge of BCD at A and does not cross any vertices. By Pasch’s axiom, it intersects 4 one 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. C C

I M I M D A B D A B Correct arrangement 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, C including in his construction of perpendiculars and of angle and segment bisectors (Theorems I. 9+10); we sketch this here. E D Given ∠BAC, construct E such that AB ∼= AE. Construct D (equilateral 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 A B intersect BE, without which the construction goes nowhere!

12 Even with Pasch’s axiom and the crossbar theorem, it requires some effort to repair Euclid’s proof: in particular, we’d need to show that D is interior to the angle ∠BAC: No matter, we shall provide an alternative 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 = A. 6 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 means that ` intersects all three sides of ABC! 4 (a) Label the intersections of ` with each side D BC, E AC and F AB. WLOG E lies ∈ ∈ ∈ between D and F: draw a picture of this arrangement, it will look very strange! (b) By applying Pasch’s axiom to DBF and ←→AC, obtain a contradiction. 4 2.2.5. Suppose you are given distinct points A, B. Using only axioms up to and including O-4 and the theorem that the intersection of distinct lines is unique (follows from I-2), show the existence of each of the points C, D, E, F in the picture in alphabetical order. Hence conclude the existence of a point F lying between A and B.

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

(b) Explain why E does not lie on ABD. 4 C (c) Explain why E = C (so that CE←→ exists). 6 (d) Explain why F lies on AB and not on BD. A F B

E 2.2.6. Use the previous exercise to prove that the interior of a triangle is non-empty. (Hint: construct a point E relative to ABC using the previous exercise, then prove that it lies on the 4 same 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 of BCD. 4 (b) There are really three further possible cases for the arrangement of I, M with respect to BCD. What are they? 4

13 2.3 Hilbert’s Axioms, part II: Congruence The congruence axioms formalize some of Euclid’s constructions and pictorial reasoning, and replace his confusing use of the word equal. Segments/angles are equal only when they are precisely the same, 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 of length, though both require a method of comparing lengths.

Deﬁnition 2.19. Let segments AB and CD be given. D Using axiom C-1, construct the unique point E on CD−→ such E that CE = AB: we have now transferred AB onto CD−→. ∼ C We write < if lies between and , etc. A AB CD E C D B

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, comparison and 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’s Theorem 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 and SAA) 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). Suppose ABC and DEF satisfy 4 4 ∠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. C Axiom C-6 (SAS) says that ∠BAC ∼= ∠EDG which, by assumption, 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. A G B But then F and G both lie on two distinct lines (←→EF and ←→DF): we F conclude that F = G. By SAS we conclude that ABC = DEF. 4 ∼ 4 DE

14 Geometry Without Circles Circles are are the heart of Euclid’s constructions yet, for reasons we’ll deal with shortly, Hilbert essentially 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 things far more speedily.

Proof. Suppose WLOG that AB = AC so that ABC is isosceles. Con- A = A ∼ 4 0 sider a ‘new’ triangle A B C = ACB where the base points are 4 0 0 0 4 switched:

A0 = A, B0 = C, C0 = B

We have:

• BAC = CAB (axiom C-5) = BAC = B A C . ∠ ∼ ∠ ⇒ ∠ ∼ ∠ 0 0 0 • AB = AC = AB = A0B0 and AC = A0C0. ∼ ⇒ ∼ ∼ B = C0 C = B0 SAS says that ∠ABC ∼= ∠A0B0C0 ∼= ∠ACB.

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 axiom C I-3, ` = ←→AB for some points A, B. Use axioms C-4 and C-1 to transfer ∠BAC to the other side of the line at 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 MAC and MAD shows that AMC = AMD: A B M 4 4 ∠ ∼ ∠ 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 cannot be applied. Instead follow the same argument starting with D ∠ABC... A generalization of this construction facilitates a corrected argument for the SSS congruence: Euclid again 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 ABC and DEF have sides congruent in 4 4 E pairs as shown. Transfer ∠EDF to A on the other side of AB from C to obtain G (axioms C-4 and C-1). C F SAS shows that BAG = EDF and so BG = EF = BC. 4 ∼ 4 ∼ ∼ Joining CG we now have two isosceles triangles with apices at A and B and congruent base angles at C and G. D Summing angles at C and G and applying SAS shows that A B ACB = AGB = DFE as required. 4 ∼ 4 ∼ 4 As with the perpendicular argument above, we should also carefully deal with the situtations where the sum is a sub- traction or the triangle is right-angled at A or B. G

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

Proof. Given ABC, extend AB to D such that AB = BD. 4 ∼ In the ﬁrst picture below, assume δ = γ. Apply SAS to see that ACB = DBC: in particular e = β. ∼ 4 ∼ 4 ∼ Clearly A and D lie on opposite sides of ←→BC: but then e + 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 α either. C C γ e η e

α β δ α β δ A B D A E B D 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 crossbar theorem, we obtain an intersection point E. But now δ is an exterior angle of EBC and congruent 4 to 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 less than a straight edge: α + β < δ + β.

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

• Construction of an isosceles triangle on a segment AB. With this one can construct segment and 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 neutral geometry without reference to the continuity axiom: you should not use circles, or anything regarding uniqueness of parallels such as Playfair’s axiom or the angle sum of a triangle being 180°. 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 not triangle 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 a triangle 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 ←→AB. C D ∃ 6∈ If ABC is not isosceles, then WLOG assume ABC < BAC. 4 ∠ ∠ By C-1 and C-4, transfer ∠ABC to A to produce D on the same M side of ←→AB as C and for which

∠ABC ∼= ∠BAD, BC ∼= AD (a) Explain why rays −→AD and −→BC intersect at a point M. A B (b) Why is MAB isosceles? 4 (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 comparisons C of angles and sides in a triangle. For clarity, suppose ABC has sides and angles labelled as in the picture. γ 4 b a (a) (I. 18) Assume a < c. Prove that α < γ. (Hint: let D on AB satisfy BD = a) ∼ α β (b) (I. 19) Prove the converse: if α < γ, then a < c. (Hint: Show that a c is impossible) A c B ≥ (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. If ABC and DEF satisfy 4 4 G C AB ∼= DE, ∠ABC ∼= ∠DEF and ∠BCA ∼= ∠EFD then the triangles are congruent ABC = DEF. 4 ∼ 4 (Hint: Let G on −→BC be such that BG ∼= EF, then apply SAS and the Exterior Angle Theorem) A B 2.3.7. Hilbert’s published SAS axiom is actually weaker than we stated. Given triangles ABC and A B C , if AB = A B , AC = A C , and BAC = 4 4 0 0 0 ∼ 0 0 ∼ 0 0 ∠ ∼ ∠B0 A0C0, then ∠ABC ∼= ∠A0B0C0. 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) F 2.3.8. Consider the picture constructed as follows. E • BF is the perpendicular bisector of AC D • BD ∼= AB • BE ∼= AD • BF ∼= AE A B C Use Pythagoras’ Theorem to prove that ACF is equilateral. 4 (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 with center O C A R and radius OR is the collection of points A such that OA ∼= OR. A point P lies inside the circle if P = O or OP < OR. C A point Q lies outside if OR < OQ. P O Since all segments are comparable, any point lies inside, outside or on a given circle. Q

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

Theorem 2.24. Suppose and are circles. C D 1. (Elementary Continuity Principle) If P lies inside and Q outside , then PQ intersects in C C exactly one point. 2. (Circular Continuity Principle) If contains a point inside and another outside , then the D C circles intersect in exactly two points; these lie on opposite sides of the line joining the circle centers.

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: Q Σ2 Σ1 consists of the points lying inside or on , Σ1 O C P Σ consists of the points lying outside . 2 C One shows that Σ1 and Σ2 satisfy the assumptions of the Axiom: the C unique point then exists and is shown to lie on itself. O C A full discussion of this (even a sketch of the second argument) requires signiﬁcant Analysis and therefore lies beyond the scope of this course. What is perhaps more interesting is to consider a geometry 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 rational { ∈ ∈ } co-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 see C that = ( 1 , 1 ) is the unique point (in R2) on the ray r = −→AC O Σ2 O √2 √2 such that AC = AB. Clearly is an irrational point and therefore ∼ O Σ1 not in the geometry. A B Continuity The circle centered at A = (0, 0) with radius 1 does not intersect the segment AC. More properly, the segment AC = Σ Σ may be partitioned as shown and yet no point in the 1 ∪ 2 O geometry separates Σ1, Σ2.

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

Deﬁnition 2.26. A triangle ABC is equilateral if its three sides are congruent. 4

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. ∃ ∗ ∗ P Axiom C-1: let C −→BD be such that BC = AB. ∈ ∼ Circular continuity principle: β contains A (inside α) and C (outside α) whence the circles intersect in precisely two points P, Q. A B C D Since P lies on both circles (and is therefore distinct from α β A and B), it follows that AB = AP = BP and that ABC ∼ ∼ 4 is equilateral. Q

If one allows Playfair’s axiom on unique parallels, Euclid’s result can be proved without using circles or the continuity axiom (see Exercise 2.3.8.). Nonetheless, we are ﬁnally able to say that every result in 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 which are found in Book III of the Elements. As such, from this point onwards we assume all of Hilbert’s Axioms 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

•A chord AB is a segment joining two points on a circle. B •A diameter BC is a chord passing through the center O.

• An arc is part of the circular edge between the chord points O (major or minor by length). C • ∠AOB is a central angle and ∠APB an inscribed angle. • ABP is inscribed in the circle. P 4 Since you should by now be quite comfortable with Euclidean Geometry, most of the details are left as exercises.

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

Simply join O to A, B, P, breaking ABP into three isosceles triangles and count angle sums. . . 4 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 to a straight edge).

Theorem 2.31. Given three non-collinear points, there exists A a unique circle through them. C

This is similar to Thm III. 1: just construct the perpendiular bisectors of two sides. B 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 to the radius at the point of tangency.

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

∠OPT < ∠OTP = OT < OP ⇒ T Thus P lies outside the circle, which ` intersects only at T: the line is therefore tangent. O The converse is an exercise. `

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 the midpoint of OP. Explain why the intersections of these circles are the points of tangency required 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 numerical quantities. It is important to realize that none of this is necessary. Measuring length and angle merely 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 signficant difference being that angle measure is normalized to a fixed scale by fixing the measure of a right angle 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 way that 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 rectangles 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 discussion towards the end of Book I, we immediately see that parallelograms also have area measuring base 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, written ABC XYZ, 4 ∼ 4 C if their sides are in the same length ratio Z AB BC CA | | = | | = | | Y XY YZ ZX | | | | | | X A B

We could work with similar triangles as did Euclid, without a numerical length measure and using ratios of lengths in a relative manner, AB : XY = BC : YZ, etc. This approach is unnecessarily confusing 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 the original 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.

AB AB CE area(BCE) BC | | = | | | | = = | | AD BD AE area(BDE) DE | | | | | | | | where the last equality follows since d = d is the common height of BCE and BDE. 1 2 4 4 (3 1) By Playfair, let m be the unique parallel to BC A ⇒ through D. This intersects AC at G. Since 1 3, we have ⇒ ABC ADG E ` 4 ∼ 4 m The transitivity of forces ADE ADG: the similarity D G ∼ 4 ∼ 4 ratio is 1, whence G = E and m = `. B C

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

5d = d ` parallel to BC is also Playfair: compare Exercise 2.1.2. (Thm I. 46). 1 2 ⇐⇒

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 AD BD sin ∠ABC := | | cos ∠ABC := | | AB 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 ∼= ∠A0B0C0 and drop perpendiculars to D, D0. Since ABD and A B D have two pairs of mutu- 4 4 0 0 0 A0 ally congruent angles, the third pair is congruent also. A AAA applies, whence the triangles are similar and we have equal ratios B0 C AD A D BD B D 0 | | = | 0 0| | | = | 0 0| B D C AB A B AB A B D0 | | | 0 0| | | | 0 0| In particular, sin ∠ABC = sin ∠A0B0C0 and cos ∠ABC = cos ∠A0B0C0. The converse is an exercise.

This is very different to how the ancient forerunners of sine and cosine were deﬁned using chords of circles 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 vertex to 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 triangle ABC and cevians AX, BY, CZ, 4 BX CY AZ | | | | | | = 1 the cevians meet at a common point P XC YA ZB ⇐⇒ | | | | | | Proof. ( ) We have pairs of triangles with the same heights, the areas of which are proportional to ⇐ the lengths of their bases: area(ABX) BX area(PBX) A = | | = Y area(AXC) XC area(PXC) | | Z area(ABX) area(PBX) area(ABP) BX = − = = | | ⇒ area(AXC) area(PXC) area(APC) XC P − | | Repeat for the other ratios and multiply to get 1. B X C The converse is an exercise.

24 Exercises. 2.5.1. Let ABC have a right-angle at C. Drop a perpendicular from C to ←→AB at D. 4 (a) Prove that D lies between A and B. C (b) Prove that you have three similar triangles b a ACB ADC CDB 4 ∼ 4 ∼ 4 x y (c) Use these facts to prove Pythagoras’ Theorem. A D B (Use the picture, where a, b, c, x, y are lengths) c = x + y 2.5.2. Suppose AD and BC are chords of a circle which intersect at P. Use similar triangles to show AP PC that |BP| = |PD| . | | | | 2.5.3. Prove a simpliﬁed version of the SAS similarity theorem: A AB AC | | = | | ABC AGH AG AH ⇐⇒ 4 ∼ 4 B | | | | C (Hint: construct BJ parallel to GH and appeal to AAA) G H 2.5.4. 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 , that sin 60° = √3 and cos 60° = 1 . √2 2 2 2.5.6. Prove the converse of Theorem 2.42: if sin ∠ABC = sin ∠A0B0C0, then ∠ABC ∼= ∠A0B0C0. (Hint: create right-triangles and prove they are similar. To make things easier, label the side lengths o, 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. A ⇒ Y Assume X, Y, Z satisfy the formula: following the pic- Z ture, deﬁne P as the intersection of BY and CZ and let the ray −→AP meet BC at X0. P Use the ( ) direction of Ceva’s Theorem to prove that ⇐ X0 = X. B X0 X C 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.