2 Euclidean Geometry

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 came withthe publication of David Hilbert’s Grundlagen der Geometrie (Foundations of Geometry) in 1899.

2.1 Hilbert’s Axioms

We describe Hilbert’s axioms for plane geometry1 (next page). We will not follow Hilbert strictly, butrather 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, onand congruence.) At various places, definitions and notations are needed.

Definition 2.1. •←→AB denotes the line through distinct points A and B. This exists and is unique

by axioms I-1 and I-2.

• AB denotes the line segment consisting of distinct points A and B and all the points C on←→AB

such that C lies between A and B. This exists and is unique by axioms O-1, O-2 and O-3.

• The ray−→AB is the segment AB together with all points C on the line through A and B such that

B lies between A and C.

• An angle with vertex A consists of a point A together with its sides: two rays−→AB and

−→AC. We

denote the angle ∠BAC.

• 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.

• 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.

• Given a line ` and points A, B not on `, we say that the points lie on the same side of ` if thesegment AB does not intersect `. Otherwise A and B are on opposite sides of `.

In the usual model of plane geometry, a line extends infinitely in both directions, a segment 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

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

1

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 Continuity Suppose that the pointson ` are partitioned into two non-empty subsetsΣ1, Σ2 such that no point of Σ1 lies between twopoints of Σ2, and vice versa. Then there exists aunique point O lying on ` such that P1 ∗O ∗ P2if and only if O 6= P1, O 6= P2 and one of P1 or P2lies in Σ1 and the other in Σ2.

Definition: parallel lines

Playfair’s Axiom Given a line ` and a point Pnot on `, there exists exactly one line through Pparallel to `.

Axioms of Incidence

The axioms of incidence explain how the relation on interacts with points and lines. Hilbert’s pre-sentation differs from Euclid’s in that he states his axioms and definitions when needed. This meansthat results can be built up and models created of geometries that satisfy only the first few axioms,but none later. 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 yourselfthat 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

We can even prove some very simple theorems in incidence geometry: these only depend on axiomsI-1, I-2 and I-3!

Lemma 2.2. If two 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 2.3. Through any point there exists at least two lines.

Try to prove this one yourself.

There is not a lot more one can do simply with the incidence axioms, though there are many esotericexamples of incidence geometry, such as the Fano plane. Look them up if you are interested!

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

3

Of the order axioms, Pasch’s axiom (first published in 1882) is the mostimportant. It is necessary to properly define the concept of the side ofa line, which is used in several of Euclid’s arguments.

Definition 2.4. We say that distinct points A, B lie on the same side of aline ` if AB does not intersect `. The points lie on the opposite side of `if the segment intersects `.

`

A

B

C

Theorem 2.5 (Plane Separation). Let A, B, C be points not on a line `.

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

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

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

Otherwise said, every line 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 parts depend on Pasch’s axiom. We prove the first and leave the others as exercises.

Consider the triangle4ABC. If ` intersects AC then, by Pasch’s axiom, it would also intersect one ofthe sides AB or BC. It does neither of these things, whence A, C lie on the same side of `.The extreme case is where A, B, C are collinear: we omit this to avoid tediousness.

Armed with the above, one can properly define the notion of the inside/interior of a triangle as theintersection of three half-planes. One of Hilbert’s early results is to prove that every triangle has anon-empty interior.

Even though the concept of angle is not required until the congruence axioms, it is convenient toconsider the concept of interior point, so that we may properly consider the first of Euclid’s omissions.

Definition 2.6. A point I is interior to an 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

4

Theorem 2.7 (Crossbar Theorem). Suppose that I is interior to ∠BAC. Then the ray−→AI intersects BC.

Proof. Extend AB to a point D such that A lies between B and D (axiom O-2).

Since C is not on the line←→BD =

←→AB we have a triangle4BCD.

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

intersects one of the other edges at a point M.The point A is the unique intersection of three distinct lines:

←→AB,←→AC and

←→IM. However I and M are

on the same side2 of←→AB, thus

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

• M lies on the ray−→AI.

There are two cases which we show below: all we need to do is 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.

Corollary 2.8. If a line passes through one vertex of a triangle and aninterior point of the triangle, then it must pass through the side oppositethe vertex.

Euclid uses this concept several times, including in his con-struction of perpendiculars and of angle and segment bisectors(Theorems I. 9+10); we sketch the latter here.

Given an angle ∠BAC, construct E such that AB ∼= AE; theequilateral triangle (Thm I. 1) 4DBE on BE is seen to producethe bisector

−→AD after a couple of appeals to Side-Angle-Side.

The problem for Euclid is that he gives no argument for why−→AD should intersect BE, without which his construction goesnowhere!

Even with Pasch’s axiom and the crossbar theorem, it requiressome effort to repair Euclid’s proof: in particular, we’d need toshow that D is interior to the angle ∠BAC. . .

`

AB

C

A B

C

DE

2All interior points of the segments CE and BC lie on the same side of←→AB as C.

5

Axioms of Congruence

Hilbert uses the concept of congruence for several purposes, in particular:

1. To formalize Euclid’s pictorial reasoning (laying one triangle on top of another, etc.).

2. To tidy up Euclid’s liberal and confusing use of the word equal.

To Hilbert, segments or angles are equal only when they are precisely the same. Certainly equalsegments are congruent, as seen by the reflexivity part of the following result:

Lemma 2.9. Congruence of segments (angles) is an equivalence relation.

Proof. Reflexivity Let AB be given and apply C-1 to obtain a segment A′B′ such that AB ∼= A′B′.We sneakily use this twice and apply C-2:

AB ∼= A′B′ and AB ∼= A′B′ =⇒ AB ∼= AB

Symmetry Assume AB ∼= CD. By reflexivity, CD ∼= CD. C-2 now says that CD ∼= AB.

Transitivity If AB ∼= CD and CD ∼= EF, apply symmetry and C-2 to conclude AB ∼= EF.

Axioms C-4 and C-5 say the same thing for angles.

The most important congruence axiom is C-6 (Side-Angle-Side/SAS).3 This is Theorem I. 4 of theElements, which relies on Euclid’s unjustified procedure of laying one triangle on top on another. Itwas eventually realized that at least one triangle congruence theorem had to be an axiom. Hilbertassumes SAS and proceeds to prove the remaining congruence theorems. For example:

Theorem 2.10 (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 Gon 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

F

G

3Hilbert ultimately states a weaker version of this, merely that one pair of the remaining angles in the triangles arecongruent. The full SAS congruence is then a theorem which Hilbert proves just before proving ASA.

6

Geometry Without Circles

For reasons we’ll deal with shortly, Hilbert barely mentions circles, and yet constructions using circlesare at the heart of Euclid’s presentation. Hilbert therefore has alternative constructions of some ofEuclid’s basic results. We give sketches of a few.

Definition 2.11. A triangle is isosceles if two of its edges are congruent. The third edge is known asthe base.

Theorem 2.12 (Euclid I. 5). Isosceles triangles have congruent base angles.

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 areswapped: i.e.

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 use SAS!

The converse to this statement is also true (see the homework).

Dropping a Perpendicular Given a point C not on the line←→AB, use

axioms (C-4 and C-1) to transfer the angle ∠PAB to the other side ofthe line at A, thus creating a new point D.Since C and D lie on opposite sides of the line, it must intersect the lineat some point M.Otherwise, SAS shows that the triangles4MAC and4MAD are con-gruent whence the supplementary angles ∠AMC and ∠AMD are con-gruent also.Since these sum to a straight edge, they must be right angles.The only reason this construction might fail is if, by chance, M = A.But then the same argument applied to4BAC and4BAD shows thatwe have a right angle at A.

A B M

C

D

7

The SSS congruence This is done similarly to Euclid.Suppose 4ABC and 4DEF have sides congruent in pairs,then the triangles are congruent.Proof. Transfer the angle ∠EDF to A on the other side of ABfrom C to obtain the point G (axioms C-4 and C-1).SAS shows that4BAG ∼= 4EDF and so BG ∼= EF ∼= BC.Joining CG we now have two pairs of isosceles triangles withvertices at A and B and congruent base angles at C and G.Summing angles at C and G and applying SAS shows that4ABC ∼= 4ABG ∼= 4DEF as required.

A little work is necessary to properly define angle sum andto deal with the case where the sum is actually a subtraction,but we omit these details.

A B

C

G

D

E

F

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

Proof. Given4ABC, extend AB to D such that AB ∼= BD.If δ ∼= γ we can apply SAS to the triangles4ACB ∼= 4DBC which forces ε ∼= β.

Clearly D (on−→AB) lies on the same side of AC as B, but yet β and δ, and thus γ and ε, sum to a

straight edge. Contradiction. We conclude that δ 6= γ. Taking the vertical angle at B, we see that anexterior angle δ to a triangle cannot be congruent to either opposite interior angle α or β.

A B

C

D

α β

γ

δ

ε

A B

C

DE

α β

η

δ

ε

Step 1: δ � γ Step 2: δ 6< γ

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, which contradicts the above. We conclude the full exteriorangle theorem.

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.

Moreover, suppose that the above picture for the SSS congruence comprised two isosceles triangles: thefact that the base angles must now be less than a right angle proves that CG is in fact the perpendicularbisector of AB: in this manner, one may bisect both angles and segments in Hilbert’s geometry, anderect perpendiculars, without relying on circles. In particular, this fixes Euclid’s Theorems I. 9+10.

8

The SAA congruence If AB ∼= DE, ∠ABC ∼= ∠DEF, and∠BCA ∼= ∠EFD, then the triangles are congruent:

4ABC ∼= 4DEF

Proof. Choose the unique G on−→BC such that BG ∼= EF.

SAS says that4ABG ∼= 4DEF.

If G 6= C, then the triangle4AGC has an exterior angle and anopposite interior angle both congruent to ∠ABC, which con-tradicts the exterior angle theorem.

G

BA

C

A similar picture could be drawn when G lies on BC, though the argument is unchanged.

Is Euclid now fixed? At this point Hilbert has proved all of Euclid’s results in Book I of the Elementsprior to the application of the parallel postulate. Including Playfair’s Postulate allows the remainderof the results in Book I to be completed, including Pythagoras’ Theorem.

There is one remaining major problem with Euclid: his use of circles.

Comparing Lengths: Circles and Continuity

Neither Hilbert nor Euclid require an absolute notion of length, although additional axioms can beadded to accommodate this. However they both require a method of comparing lengths.

Definition 2.13. Let segments AB and CD be given. Usingaxiom C-1, construct the unique point E on

−→AB such that

AE ∼= CD. We write CD < AB if E lies between A and B.

Any two segments are then comparable: given AB andCD, precisely one of the following is true:

AB < CD, CD < AB, AB ∼= CD

A

B

CD

E

A similar approach can be done for angles using a discussion of interior points: indeed this wasalready necessary in order to prove some of the previous results.

We now proceed to define circles.

Definition 2.14. 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 segments are comparable, every point lies either inside, outside or on agiven circle.

O

RA

P

Q

9

The Axiom of Continuity Why does Hilbert barely mention circles? Part of his goal was to see whyand when each axiom is required; he wants to build as much geometry as possible without invokingcomplicated axioms. A major weakness of Euclid is that many of his proofs rely on the intersectionsof lines and circles. To use circles in this manner requires the axiom of continuity, undeniably the mostdifficult of Hilbert’s axioms.

The following statements can be proved from the Axiom of Continuity.

Theorem 2.15. 1. (Elementary Continuity Principle) If C is a circle such that P is inside and Q is out-side, then the segment PQ intersects C in exactly one point.

2. (Circular Continuity Principle) If C and D are circles such that D contains at least one point insideand one point outside C, then the circles intersect in exactly two points; these points lie on oppositesides of the line joining the circle centers.

The idea of the first is to partition the segment 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, whichguarantees the existence of the unique point O. Finally O is shown tolie on the circle itself.

P

QO

Σ1

Σ2

A full discussion of this (even a sketch of the second argument) lies beyond the difficulty of this class.What is perhaps more interesting is to think about a geometry in which the axiom of continuity isfalse. To do so, we can jump forward a little to analytic geometry and consider the set

Q2 = {(x, y) ∈ R2 : x, y ∈ Q}

of points in the plane with rational co-ordinates. A line in this geometry consists of all rational points(x, y) satisfying an equation of the form

ax + by = c where a, b, c ∈ Q

Almost all of Hilbert’s axioms hold for this geometry: here are two of the problems.

Axiom C-1 Consider the points A = (0, 0), B = (1, 0) and C = (1, 1).O = ( 1√

2, 1√

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

−→AC such

that AC ∼= AB. Clearly O is an irrational point and therefore notin the geometry: C-1 is false!

Continuity The same example shows the failure of the continuityaxiom, since the circle centered at A = (0, 0) with radius 1does not intersect the segment AC. More properly, the segmentAC = Σ1 ∪ Σ2 may be partitioned as shown and yet no point Oin the geometry separates Σ1, Σ2.

Σ1

Σ2

A B

CO

10

Equilateral triangles We are finally(!) in a position to prove the first proposition of the Elementsusing a corrected version of Euclid’s proof.

Definition 2.16. A triangle4ABC is equilateral if its three sides are congruent.

Theorem 2.17 (Euclid I.1). We may construct an equilateral triangle on a given segment AB.

Proof. Following Euclid, we start by constructing the cir-cles α and β centered at A and B, with radii AB.

Axiom O-2 says there exists a point D such that B lies be-tween A and D. By axiom C-1, let C be the point on the ray−→BD such that BC ∼= AB.

Clearly β contains A (inside α) and C (outside α).

By the Circular Continuity Principle, the circles intersectin precisely two points P, Q.

A B C D

P

Qα β

Since P lies on both circles (and is therefore distinct from A and B), it follows that AB ∼= AP ∼= BP.We conclude that4ABC is equilateral.

It is, in fact, possible to prove Euclid’s first theorem without using circles or assuming the continuityaxiom. Instead one requires several applications of Pythagoras’ Theorem which follows from Play-fair’s axiom on parallels (see the homework). 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!

Conclusions

There are several major errors in Book I of the Elements: each is addressed by Hilbert’s axioms:

1. Pasch’s axiom formalizes the concepts of inside and outside for triangles and guarantees that anyline entering a triangle must also leave it.

2. The congruence axioms, particularly SAS, replace Euclid’s reliance on pictorial reasoning.

3. The continuity axiom guarantees that lines and circles intersect and thus recovers the traditionalEuclidean constructions.

It is important to appreciate one of Hilbert’s main purposes; to understand why certain axioms arerequired. Thus he works as much as possible using early axioms, only introducing later ones whennecessary, akin to Euclid’s delayed use of the parallel postulate. This approach set the standard forthe modern axiomatic presentation of pure mathematics.

Euclid’s text is essentially practical. The Elements is presented more as a manual for how to constructgeometric figures. One can certainly recover the standard Euclidean constructions within Hilbert’sFoundations, but this was not his goal. The inspiration for 2200 years of work on Euclidean Geometrymight have been an attempt to fix the Elements, but the result was really a new way of looking atmathematics: Hilbert, more than Euclid, is arguably the father of axiomatics!

11

2.2 Measurement and Area in Euclidean Geometry

While it is difficult for us to work without an absolute notion of length, we put this off for a shortwhile to consider how Euclid dealt with area. The discussion was based on parallelograms.

Definition 2.18. A parallelogram consists of four points A, B, C, D, no three of which are collinear, andthe line segments AB, BC, CD, DA such that opposite sides are parallel.A rectangle is a parallelogram with all angles being right angles. A square is a rectangle, all of whosesides are congruent.

Theorem 2.19 (I. 34). Opposite sides of a parallelogram are congruent.

Proof. Draw a diagonal of the parallelogram, AC say. Since←→AC falls

on two pairs of parallel lines, the indicated angles in the picture arecongruent.

Now observe that 4DAC ∼= 4BCA via Theorem 2.10 (ASA),whence opposite sides are congruent.

A B

CD

We now make a loose definition of relative area.

Definition 2.20. Two figures have equal area if they are congruent or can be subdivided into mutuallycongruent pieces.A figure has double the area of another if the first can be subdivided into two subfigures, each of whichhas area equal to the latter. Other ratios may be defined similarly.

Figure in this context will almost always mean parallelogram or triangle. Euclid simply says that twofigures are ‘equal,’ or that one is ‘twice the other.’ The word area is never mentioned.

Example The parallelogram in Theorem 2.19 may be divided into two congruent triangles: theparallelogram thus has double the area of either triangle.

Theorem 2.21 (I. 35). If two parallelograms share the same base with theopposite sides on the same line, then the parallelograms have equal area.

Proof. If ABCD and ABEF are the parallelograms, the fact that←→CF

falls on two sets of parallel lines yields the angle congruences

∠FDA ∼= ∠ECB, ∠DFA ∼= ∠CEB

Since AD ∼= BC, Side-Angle-Angle forces

4ADF ∼= 4BCE

The first parallelogram ABCD may be divided into two pieces(4BCE and a quadrilateral ABED) which are congruent to a sub-division of the second parallelogram ABEF.

A second possible arrangement is shown. The proof is essentially thesame, though one might want to take some care regarding the threesmall triangles. . .

A B

CD EF

A B

CDEF

12

Corollary 2.22 (I. 41). If a parallelogram and triangle have the same base and height then the parallelogram istwice the triangle.

This is as close as Euclid comes to the formula A = 12 bh for the area of a triangle. Look back at

Euclid’s proof of Pythagoras’ Theorem (I. 47): the heart of the argument is the comparison of theareas of a square and a rectangle using the above corollary.

Aside: A Little HistoryYou should be familiar with the most common form of sloppiness in Euclidean Geometry: the con-flation of equality and congruence. To be precise,

AB = CD =⇒ AB ∼= CD and ∠ABC = ∠DEF =⇒ ∠ABC ∼= ∠DEF

however the converses are both false! One can be forgiven shomewhat, for the rot starts with Euclidhimself who uses equals for congruence of angles and segments, for areas and volumes, and even tocompare similar triangles!

A more important reason for this misuse is that modern intuition regarding length and angle tendsto be numerical. Given the way we are taught geometry in grade-school, it is extremely difficult for usto refer to segments and angles without mentioning their numerical length or (degree/radian) mea-sure. Indeed, colloquially, when we say “AB equals CD,” we usually mean that these segments haveequal length. While this is absolutely fine for real-world use, mathematics requires strict definitions.One of Hilbert’s legacies is to properly separate and distinguish Euclid’s multiple notions of equality.

Absolute measures of angle and length are not present in Euclid, nor are they mentioned in Hilbert’saxioms, though, as we saw previously, length and angle can be compared (recall Definition 2.13).Moreover, ratios of lengths are very much discussed in Euclid. For instance,

Definition 2.23. Suppose B lies between A and C in such a way that AB ∼= BC. Then B is the midpointof AC, and AC is double (the length of) AB.

More generally, if (say) five copies of segment AB laid end-to-end forms a segment congruent to seven copies of CD,then the segments are in the ratio

AB : CD = 7 : 5

A B

C D

Ratios of areas can be defined similarly, building on our discussion of parallelograms/triangles.

To the ancient Greeks, ‘length’ was physical while ‘number’ meant positive integer, and it made nosense to them to directly associate these concepts. If the only numbers are positive integers, thenat least fractions could be represented by ratios, but irrational numbers would be meaningless. ByEuclid’s time (300 BC), the Greeks were aware of the existence incommensurable lengths; i.e. segmentssuch that AB : CD 6= m : n for any m, n ∈ N, the side and diagonal of a square being such. Book Xof the Elements is largely devoted to the discussion of these. Even the acceptance of this was a hugeleap forward: there is some evidence that the Pythagoreans (500 BC) found the idea heretical!

The slow evolution of geometry from Euclid’s synthetic description to the modern analytic formula-tion (co-ordinates and numerical measurement) is one of the great stories of mathematical history.

13

Length and Angle Measure Axioms

To avoid continued frustration, it is now time to accept length and area as absolute numerical quan-tities. To do this rigorously requires a few more axioms. It is important to realize that none of thisis strictly necessary for pure Euclidean Geometry. Numerically measuring length and angle merelymakes the language seem more familiar and certain computations simpler. Of course, even a tinysimplification can have huge effect: results that would be very difficult to state suddenly becomemanageable, and some new mathematics becomes easier to discover. You should also be aware ofthe cost: relying on numerical calculation robs us of some intuition!

Length measure axioms

1. To every segment AB corresponds the distance |AB|, a positive number.

2. |AB| = |CD| ⇐⇒ AB ∼= CD.

3. |AB| < |CD| ⇐⇒ AB < CD (recall Definition 2.13)

4. If B lies between A and C, then |AB|+ |BC| = |AC|.Rather than attempt to memorize these axioms, you should simply observe that they reflect yourintuition! Here is a quick result proving the converse4 of axiom 4.

Lemma 2.24. If A, B, C are distinct and |AB|+ |BC| = |AC|, then B lies between A and C.

Proof. We exclude the other possibilities.

• Suppose A, B, C are non-collinear. Since |AB| < |AC|,axiom 3 says ∃D on AC such that |AD| = |AB|. Byaxiom 4, |DC| = |BC| and we have the picture withtwo isosceles triangles4BAD and4BCD. But α+ β isa straight edge, whence B lies on AC: contradiction. A

B

CD

α

α

β

β

• If B lies on the ray−→AC but not on AC, then C lies between A and B whence |AC|+ |BC| = |AB|

by axiom 4. Substituting |AB|+ |BC| = |AC| shows that 2 |BC| = 0: contradiction.B lying on

−→CA but not on AC is similar.

Angle measure axioms

1. To every angle ∠ABC corresponds a degree measure m∠ABC in the interval (0, 180).

2. m∠ABC = m∠DEF ⇐⇒ ∠ABC ∼= ∠DEF.

3. ∠ABC < ∠DEF ⇐⇒ m∠ABC < m∠DEF5

4. Suppose−→BP is interior to ∠ABC. Then m∠ABP + m∠PBC = m∠ABC.

5. Right angles measure 90°.4For an extra challenge, try proving the full4-inequality: for distinct A, B, C, |AB|+ |BC| ≥ |AC| with equality if and

only if B lies between A and C.5∠ABC < ∠DEF if there is a ray

−→EG interior to ∠DEF for which ∠ABC ∼= ∠DEG.

14

Theorem 2.25 (Uniqueness of measure). 1. Given a segment OP, there is a unique way to assign adistance to every segment in such a way that |OP| = 1.

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

This theorem is a consequence of the continuity axiom.

Area can now be computed numerically: a rectangle ABCD is defined6 to have area |AB| |CD|. Weimmediately have that the area of a triangle is half its base times height.

Aside: Cevians and triangle centersAfter Giovanni Ceva (1647–1734): a cevian is an line joining a vertex to the opposite side of the trian-gle. Certain results are much simpler in Euclidean rather than Analytic geometry. Here is an examplefrom the height of Euclidean geometry. Proving this using co-ordinates would involve a lot of nastyalgebra which would obscure the beauty of the result.

Theorem 2.26 (Ceva’s Theorem). Given a triangle4ABC and cevians AX, BY, CZ, then

|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. For instance,

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.

(⇒) This time define P as the intersection of BY and CZ. Theray−→AP meets BC at X′. We want to prove that X = X′. By the

assumption and the (⇐) direction, we know that

|BX′||X′C|

|CY||YA|

|AZ||ZB| = 1 =

|BX||XC|

|CY||YA|

|AZ||ZB|

=⇒ |BX′||X′C| =

|BX||XC|

A

B CX

YZ

P

A

B CXX′

YZ

P

whence X and X′ cut BC in the same ratio and are thus equal.

As a simple corollary, we easily see that the three bisectors of a triangle meet at a common point (thecentroid). This is immediate, since if X is the midpoint of |BC|, then |BX| = |XC|, etc.

6One can easily prove in Euclid that if two rectangles have the same height and bases in the ratio of p : q (wherep, q ∈ N), then their areas must also be in the ratio p : q. Our definition does two things: first we extend by continuity tothe situation where the ratio of lengths is irrational, and second we assert that a square with unit length sides has unit area.

15

2.3 Similar Triangles and Trigonometry

Definition 2.27. Two triangles are similar if corresponding pairs ofsides are in the same ratio (of length). We write4ABC ∼ 4XYZ.

While it is possible to work with similar triangles purely using ratiosof lengths in a relative manner, it is much more difficult. We’ll as-sume we have a well-defined length measure as defined previously. A B

C

XY

Z

Our primary result is the AAA theorem.

Theorem 2.28 (Angle-Angle-Angle). 1. Triangles are similar if and only if their angles come in mutu-ally congruent pairs.

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

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

Lemma 2.29. Suppose ` intersects sides AB and AC of 4ABCat D and E as in the picture. The following are equivalent:

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)

Now drop perpendiculars from B, C to ` and note that4BDE and4CDE share a 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 each side of |BD|

|AD| =|CE||AE| 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 from part 1: d1 = d2 is the common height of4BCE and4BDE.We conclude that4ABC ∼ 4ADE.

16

(3 ⇒ 1) Use Playfair’s Axiom to construct the unique par-allel m to BC through D. This intersects AC at G. Since1⇒ 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

There are two places in the Lemma where we used the parallel postulate/Playfair: in the proof that3 =⇒ 1 and in the observation7 that d1 = d2 ⇐⇒ ` is parallel to BC. We should not thereforeexpect AAA to hold in geometries without unique parallels. Indeed it will be seen that AAA is atheorem for congruent triangles in Hyperbolic Geometry!

Trigonometric Functions

We can use similar triangles to define the concepts of sine and cosine.

Definition 2.30. Given an acute angle ∠ABC, construct D on−→BC so that4ADB is right-angled at D.

We may then define

sin∠ABC :=|AD||AB| cos∠ABC :=

|BD||AB|

Theorem 2.31. Sine and cosine are well-defined.

We need the Theorem because we made a choice inthe Definition: we had to choose an explicit triangle4ABC, not merely an angle ∠ABC, the latter of whichdepends only on rays. We therefore need to check thatif ∠A′B′C′ is congruent to ∠ABC, then their sines (andcosines) are equal.

A

B CD

A′

B′

C′

D′

Proof. By construction, triangles4ABD and4A′B′D′ have two pairs of mutually congruent angles.Since the sum of the angles in a triangle is a straight edge, it follows that the third pair of angles arecongruent. AAA applies, whence the triangles are similar and we have equal ratios

|AD||AB| =

|A′D′||A′B′|

|BD||AB| =

|B′D′||A′B′|

In particular, sin∠ABC = sin∠A′B′C′ and cos∠ABC = cos∠A′B′C′.

This is very different to how ancient mathematicians defined the forerunners of sine and cosine. Theyused chords of circles rather than triangles. The word trigonometry (literally triangle measure) wasn’tcoined until 1595!

7This is really Euclid’s discussion about triangles which share a base having equal area if and only if their heights areequal; this certainly requires the parallel postulate.

17

2.4 Circle Geometry

We conclude our survey of Euclidean Geometry with a few results about circles. These are the pri-mary topic of Book III of the Elements. Since you should now be quite comfortable with EuclideanGeometry, several of the details are left as exercises.

Definition 2.32. With regard 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 chordpoints (major or minor by length).

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

A

B

C

P

O

Theorem 2.33. The angle at the center of a circle is twice the inscribed angle.

Proof. Join O to each of A, B, P, thus breaking 4ABP into three isosceles triangles. Now count an-gles. . .

The details are an exercise: to be thorough, you should check that the argument works if the chordAB is a diameter, when O lies outside 4ABP, and even (with suitably careful definition) when thecentral angle is larger than 180°.

Corollary 2.34. 1. If two triangles in a circle share the same arc, then the angles opposite that arc arecongruent.

2. Angles in a semi-circle are right angles.

3. A quadrilateral whose points lie on a circle has its opposite angles supplementary.

Theorem 2.35. Given three non-collinear points, there exists aunique circle through them.

Proof. Let AB, AC be sides of the triangle. Construct their per-pendicular bisectors. The intersection point O is the center ofthe required circle.

Convince yourself of this. In particular, explain why no othercircle can pass through the given points.

A

B

C

O

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Tangency

Definition 2.36. A line is tangent to a circle if they intersect exactly once.

Theorem 2.37. A line is tangent to a circle if and only if it is perpendicular to the radius at the point oftangency.

Proof. Suppose a line ` through T is perpendicular to the radiusOT, as in the picture.

Let P 6= T lie on `. But then,

m∠OPT < 90° = m∠OTP =⇒ |OT| < |OP| (∗)

Thus P lies outside the circle and the line intersects only at T: theline is tangent.

For the converse, if ` is tangent at T but not perpendicular to OT,construct the perpendicular to ` through O: why do we get a con-tradiction?

T

O

P

`

(∗) is Euclid’s Theorem I. 19, and an exercise in the homework.

Theorem 2.38. Given a point outside a circle, there are exactly two lines through this point which are tangentto the circle.

Proof. Given a circle centered at O and a point P outside the circle, draw the circle centered at themidpoint of OP passing through O and P. The intersections of these circles are the points of tangency.

Draw the picture and explain why!

19