2 Euclidean Geometry
2.1 Euclid’s Postulates and Book I of the Elements
Euclid’s Elements (c. 300 BC) formed a core part of European and Arabic curricula until the mid 20
th
century. Several examples are shown below.
Earliest Fragment c. AD 100 Full copy, Vatican, 9
th
C Pop-up edition, 1500s
Latin translation, 1572 Color edition, 1847 Textbook, 1903
Many of Euclid’s arguments can be found online, and you can read Byrne’s 1847 edition here: the
cover is Euclid’s proof of Pythagoras’. We present an overview of Book I.
Undefined Terms E.g., point, line, etc.
4
Axioms/Postulates
5
A1 If two objects equal a third, then the objects are equal (= is transitive)
A2 If equals are added to equals, the results are equal (a = c & b = d =⇒ a + b = c + d)
A3 If equals are subtracted from equals, the results are equal
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.
4
In fact Euclid attempted to define these: ‘A point is that which has no part,’ and ‘A line has length but no breadth.’
5
In Euclid, an axiom is considered somewhat more general than a postulate. Here the postulates contain the geometry.
8
The first three postulates describe the intuitive ruler and compass constructions. P4 allows Euclid to
compare angles at different locations. P5 is usually known as the parallel postulate.
Euclid’s system doesn’t quite fit the modern standard. Some axioms are vague (what are ‘things’?)
and we’ll consider several more-serious shortcomings later. For now we clarify two issues and intro-
duce some notation.
Segments To Euclid, a line had finite extent—we call such a (line) segment. The segment joining points
A, B is denoted AB. In modern geometry, a line extends as far as is permitted.
Congruence Euclid uses equal where modern mathematicians say congruent. We’ll express, say, con-
gruent angles as ∠ABC
∼
=
∠DEF rather than ∠ABC = ∠DEF.
Basic Theorems `a la Euclid
Theorems were typically presented as a problem. Euclid first provides a construction (P1–P3) before
proving that his construction solves the problem.
Theorem 2.1 (I. 1). Problem: to construct an equilateral triangle on a given segment.
The labelling I. 1 indicates Book I, Theorem 1.
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.
We claim that △ABC is equilateral.
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.
A
B
C
Euclid proceeds to develop several well-known constructions and properties of triangles.
• (I. 4) Side-angle-side (SAS) congruence: if two triangles have two pairs of congruent sides and
the angles between these are congruent, then the remaining sides and angles are congruent in
pairs.
AB
∼
=
DE
∠ABC
∼
=
∠DEF
BC
∼
=
EF
=⇒
AC
∼
=
DF
∠BCA
∼
=
∠EFD
∠CAB
∼
=
∠FDE
• (I. 5) An isosceles triangle has congruent base angles.
• (I. 9) To bisect an angle.
• (I. 10) To find the midpoint of a segment.
• (I. 15) If two lines/segments cut one another, opposite angles are congruent.
Have a look at some of Euclid’s arguments online. These are worth reading despite there being logical
issues with Euclid’s presentation. We’ll revisit these results in the Exercises and next two sections.
9
Parallel Lines: Construction & Existence
Definition 2.2. Lines are parallel if they do not intersect. Segments are parallel if no extensions of
them intersect.
In Euclid, a line is not parallel to itself. The next result is one of the most important in Euclidean
geometry, in that it describes how to create a parallel line through a given point.
Theorem 2.3 (I. 16 Exterior Angle Theorem). If one side of a
triangle is extended, then the exterior angle is larger than either
of the opposite interior angles.
In the picture, we have δ > α and δ > β.
α
β
δ
Euclid did not quantify angles numerically: δ > α means that α is congruent to some angle inside δ.
Proof. Construct the bisector BM of AC (I. 10).
Extend BM to E such that BM
∼
=
ME (I. 2) and connect CE (P1).
The opposite angles at M are congruent (I. 15).
SAS (I. 4) applied to △AMB and △CME says ∠BAM
∼
=
∠EMC,
which is clearly smaller than the exterior angle at C.
A
B
C
M
E
Bisect BC and repeat the argument to see that β < δ.
The proof in fact constructs a parallel (CE) to AB through C, as the next result shows.
Theorem 2.4 (I. 27). If a line falls on two other lines such that the
alternate angles (α, β) are congruent, then the two lines are parallel.
α
β
The alternate angles in the exterior angle theorem are those at A and C: CE really is parallel to AB.
Proof. If the lines were not parallel, they would meet on one
side. WLOG suppose they meet on the right side at C.
The angle β at B, being exterior to △ABC, must be greater than
the angle α at A (I. 16): contradiction.
α
β
A
B
C
Euclid combines this with the vertical angles theorem (I. 15) to finish the first half of Book I.
Corollary 2.5 (I. 28). If a line falling on two other lines makes con-
gruent angles, then the two lines are parallel.
Thus far, Euclid uses only postulates P1–P4. In any model in which these hold:
Given a line ℓ and a point C not on ℓ, there exists a parallel to ℓ through C
10
Parallel Lines: Uniqueness, Angle-sums & Playfair’s Postulate
Euclid finally invokes the parallel postulate to prove the converse of I. 27, showing that the congruent
alternate angle approach is the only way to have 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 ℓ, m meet on the left side of
the picture, whence ℓ and m are not parallel.
m
ℓ
α
γ
β
n
The most well-known result about triangles is now in our grasp, that the interior angles sum to a
straight edge. Euclid words this slightly differently.
Theorem 2.7 (I. 32). If one side of a triangle is extended, the exterior angle is congruent to the sum
of the opposite interior angles.
This is not a numerical sum, though for familiarity’s sake we’ll
often write 180° for a straight edge and 90° for a right-angle.
In the picture we’ve labelled angles as Greek letters for clarity.
The result amounts to showing that
e
α +
e
β
∼
=
α + β.
A
B
C
D
E
β
˜
β
α
˜
α
Proof. Construct CE parallel to BA as in I. 16, so that
e
α
∼
=
α.
BD falls on parallel lines AB and CE, whence
e
β
∼
=
β (Corollary of I. 29).
Axiom A2 shows that ∠ACD =
e
α +
e
β
∼
=
α + β.
The parallel postulate is stated in the negative (angles don’t sum to a straight edge, therefore lines are
not parallel). Though we cannot be sure, Euclid possibly chose this formulation in order to facilitate
proofs by contradiction. Unfortunately the effect is to obscure the meaning of the parallel postulate.
Here is a more modern interpretation.
Axiom 2.8 (Playfair’s Postulate). Given a line ℓ and a point C
not on ℓ , at most one parallel line m to ℓ passes through C.
m
ℓ
C
Our discussion up to now shows that the parallel postulate
implies Playfair.
• Let A, B ∈ ℓ and construct the triangle △ABC.
• The exterior angle theorem constructs E and thus a par-
allel m to ℓ by I. 27.
• I. 29 invokes the parallel postulate to prove that this is
the only such parallel.
ℓ
m
B
A
C
E
11
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 (P5) are equivalent.
Proof. (P5 ⇒ Playfair) We proved this above.
(Playfair ⇒ P5) We prove the contrapositive. Assume postulates P1–P4 are true and that P5 is false.
Using quantifiers, and with reference to the picture in I. 29, we restate the parallel postulate:
P5: ∀ pairs of lines ℓ, m and ∀ crossing lines n, β + γ < 180° =⇒ ℓ, m not parallel.
Its negation (P5 false) is therefore:
∃ parallel lines ℓ , m and a crossing line n for
which β + γ < 180°
This is without loss of generality: if β + γ > 180°, consider
the angles on the other side of n.
By the the exterior angle theorem/I. 28, we may build a
parallel line
ˆ
ℓ to ℓ through the intersection C of m and n
(in the picture,
ˆ
β
∼
=
β). Crucially, this only requires postu-
lates P1–P4!
ℓ
m
γ
β
n
ˆ
ℓ
ℓ
m
C
ˆ
β
γ
β
Observe that
ˆ
ℓ and m are distinct since
ˆ
β + γ
∼
=
β + γ < 180°. We therefore have a line ℓ and a
point C not on ℓ, though which pass (at least) two parallels to ℓ: Playfair’s postulate is false.
Non-Euclidean Geometry
That Euclid waited so long before invoking the uniqueness of parallels suggests he was trying to
establish as much as he could about triangles and basic geometry in its absence. By contrast, every-
thing from I. 29 onwards relies on the parallel postulate, including the proof that the angle sum in a
triangle is 180°. For centuries, many mathematicians believed, though none could prove it, that such
a fundamental fact about triangles must be true independent of the parallel postulate.
Loosely speaking, a non-Euclidean geometry is a model for which a parallel through an off-line point
either doesn’t exist or is non-unique. It wasn’t until the 17–1800s and the development of hyperbolic
geometry (Chapter 4) that a model was found in which Euclid’s first four postulates hold but for which
the parallel postulate is false.
6
We shall eventually see that every triangle in hyperbolic geometry has angle
sum less than 180°, though this will require a lot of work! For a more eas-
ily visualized non-Euclidean geometry consider the sphere. A rubber band
stretched between three points on its surface describes a spherical triangle: an
example with angle sum 270° is drawn. A similar game can be played on a
saddle-shaped surface: as in hyperbolic geometry, ‘triangles’ will have angle
sum less than 180°.
6
This shows that the parallel postulate is independent; in fact all Euclid’s postulates are independent. They are also
consistent (the ‘usual’ points and lines in the plane are a model), but incomplete: a sample undecidable is in Exercise 5.
12
Pythagoras’ Theorem
Following his discussion of parallels, Euclid shows that parallelograms with the same base and
height are equal (in area) (I. 33–41), before providing constructions of parallelograms and squares
(I. 42–46). Some of this is in Exercise 2. Immediately afterwards comes the capstone of Book I.
Theorem 2.10 (I. 47 Pythagoras’ Theorem). 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.
1. Construct squares on each side of △ABC (I. 46) and a
parallel AL to BD (I. 16).
2. AB
∼
=
FB and BD
∼
=
BC since sides of squares are con-
gruent. Moreover ∠ABD
∼
=
∠FBC since both contain
∠ABC and a right-angle.
3. Side-angle-side (I. 4) says that △ABD and △FBC are
congruent (identical up to rotation by 90°).
4. I. 41 compares areas of parallelograms and triangles
with the same base and height:
Area(□ABFG) = 2 Area(△FBC)
= 2 Area(△ABD)
= Area
BOLD
5. Similarly Area(□ACKH) = Area
OCEL
.
A
B
CO
LD E
F
G
H
K
6. Sum the rectangles to obtain □BCED and complete the proof.
Euclid finishes Book I with the converse, which we state without proof. Euclid’s argument is very
sneaky—look it up!
Theorem 2.11 (I. 48). If the (areas of the) squares on two sides of a triangle equal the (area of the)
square on the third side, then the triangle has a right-angle opposite the third side.
The Elements contains thirteen books. Much of the remaining twelve discuss further geometric con-
structions, including in three dimensions. There is also a healthy dose of basic number theory includ-
ing what is now known as the Euclidean algorithm.
While undoubtedly a masterpiece of logical reasoning, Euclid’s presentation has several flaws. Most
problematic is his reliance on pictorial reasoning: for instance, he ‘proves’ the SAS and SSS congru-
ence theorems (I. 4 & 8) by laying one triangle on top of another, a process not justified by his axioms
(look it up online or Byrne). In a modern sense, Euclid’s approach is part axiomatic system and part
model: his reasoning requires a visual/physical representation of lines, circles, etc. Because of these
issues, we now turn to a more modern description of Euclidean geometry, courtesy of David Hilbert.
13
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. To help prove Pythagoras’, Euclid makes use of the following results. Prove them as best as
you can. Full rigor is tricky, but the pictures should help!
(a) (I. 11) At a given point on a line, to construct a perpendicular.
(b) (I. 46) To construct a square on a given segment.
(c) (I. 35) Parallelograms on the same base and with the same height have equal area.
(d) (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
C
D
A
B
C
D EF
Theorem I. 11 Theorem I. 46 Theorem I. 35
3. Consider spherical geometry (page 12), where lines are paths of shortest distance (great circles).
(a) Which of Euclid’s postulates P1–P5 are satisfied by this geometry?
(b) (Hard) Where does the proof of the exterior angle theorem fail in spherical geometry?
4. (a) State the negation of Playfair’s postulate.
(b) 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 perpen-
dicular to the other.
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 Q
2
= {(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.
6. The standard proof of the converse of Pythagoras’ theorem (I. 48) is, in fact, a corollary of the
original! Look it up and explain the argument as best you can.
14
2.2 Hilbert’s Axioms I: Incidence and Order
The long process of identifying and correcting the errors and omissions in Euclid’s Elements culmi-
nated in the 1899 publication of David Hilbert’s Grundlagen der Geometrie (Foundations of Geometry).
In the next two sections we consider some of the details of Hilbert’s approach, providing a modern
and logically superior description of Euclidean geometry.
Hilbert’s axioms for plane geometry
7
are listed on the next page. The undefined terms consist of two
types of object (points and lines), and three relations (between ∗, on ∈ and congruence
∼
=
). For brevity
we’ll often use/abuse set notation, viewing a line as a set of points, though this is not necessary. At
various places, definitions and notations are required.
Definition 2.12. Throughout, A, B, C denote points and ℓ , m lines.
Line:
←→
AB denotes the line through distinct A, B. This exists and is unique by axioms I-1 and I-2.
Segment: AB := {A, B} ∪ {C : A ∗C ∗ B} consists of distinct endpoints A, B and all interior points C
lying between them.
Ray:
−→
AB := AB ∪ {C : A ∗ B ∗ C} is a ray with vertex A. In essence we extend AB beyond B.
Triangle: △ABC := AB ∪ BC ∪ CA where A, B, C are non-collinear. Triangles are congruent if their
sides and angles are congruent in pairs.
Sidedness: Distinct A, B, not on ℓ, lie on the same side of ℓ if AB ∩ ℓ = ∅. Otherwise A and B lie on
opposite sides of ℓ .
Angle: ∠BAC :=
−→
AB ∪
−→
AC has vertex A and sides
−→
AB and
−→
AC.
Parallelism: Lines ℓ and m intersect if there exists a point lying on both: ∃A ∈ ℓ ∩m. Lines are parallel
if they do not intersect. Segments/rays are parallel when the corresponding lines are parallel.
The pictures represent these notions in the usual model of Cartesian geometry.
A
B
A
B
A
B
A
B
C
Line
←→
AB Segment AB Ray
−→
AB Triangle △ABC
ℓ
A
B
ℓ
A
B
A
B
C
A
ℓ
m
Same side Opposite sides Angle ∠BAC Intersection A ∈ ℓ ∩ m
7
Like Euclid, Hilbert also covered 3D geometry—we only give the axioms for plane geometry. With regard to our
desired properties (Definition 1.6), his system is about as good as can be hoped. Essentially one only one model exists,
which is almost the same thing as completeness. In the absence of the continuity axiom, the axioms are consistent; in
line with G
¨
odel’s theorems (1.8), consistency cannot be proved once continuity is included. As stated, the axioms are not
quite independent, though this can be remedied: O-3 does not require existence (follows from Pasch’s axiom), C-1 does
not require uniqueness (follows from uniqueness in C-4) and C-6 can be weakened slightly.
15
Hilbert’s Axioms for Plane Geometry
Undefined terms
1. Points: use capital letters, A, B, C . . .
2. Lines: use lower case letters, ℓ, m, n, . . .
3. On: A ∈ ℓ is read ‘A lies on ℓ’
4. Between: A ∗B ∗C is read ‘B lies between
A and C’
5. Congruence:
∼
=
is a binary relation on
segments or angles
Axioms of Incidence
I-1 For any distinct A, B there exists a line ℓ
on which lie A, B.
I-2 There is at most one line through distinct
A, 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 three
points not all on the same line.
Axioms of Order
O-1 If A ∗ B ∗ C, then A, B, C are distinct
points on the same line and C ∗ B ∗ A.
O-2 Given distinct A, B, there is at least one
point C such that A ∗ B ∗ C.
O-3 If A, B, C are distinct points on the same
line, exactly one lies between the others.
Definitions: segment AB and triangle △ABC
O-4 (Pasch’s Axiom) Let △ABC be a triangle
and ℓ a line not containing any of A, B, C.
If ℓ contains a point of the segment AB,
then it also contains a point of either AC
or BC.
Definitions: sides of line
←→
AB and ray
−→
AB
Axioms of Congruence
C-1 (Segment transference) Let A, B be distinct
and r a ray based at A
′
. Then there exists a
unique point B
′
∈ r for which 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
′
and
BC
∼
=
B
′
C
′
, then AC
∼
=
A
′
C
′
.
Definitions: angle ∠ABC
C-4 (Angle transference) Given ∠BAC and
−−→
A
′
B
′
, there exists a unique ray
−−→
A
′
C
′
on
a given side of
←−→
A
′
B
′
for which ∠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 △ABC
and △A
′
B
′
C
′
, if AB
∼
=
A
′
B
′
, AC
∼
=
A
′
C
′
,
and ∠BAC
∼
=
∠B
′
A
′
C
′
, then the triangles
are congruent.
8
Axiom of Continuity
Suppose a line/segment is partitioned into non-
empty subsets Σ
1
, Σ
2
such that no point of Σ
1
lies
between two points of Σ
2
and vice versa.
Then there exists a unique point O satisfying
P
1
∗ O ∗ P
2
, if and only if O = P
1
, O = P
2
and
one of P
1
or P
2
lies in Σ
1
and the other in Σ
2
.
Playfair’s Axiom
Definition: parallel lines
Given a line ℓ and a point P /∈ ℓ, at most one line
through P is parallel to ℓ.
8
Its sides/angles are congruent in pairs. We extend congruence to other geometric objects similarly.
16
Axioms of Incidence: Finite Geometries
The axioms of incidence describe the relation on. An incidence geometry is any model satisfying axioms
I-1, I-2, I-3. Perhaps surprisingly, there exist incidence geometries with finitely many points!
Examples 2.13. By I-3, an incidence geometry requires at least three points.
A 3-point geometry exists, and is unique up to relabelling:
I-3 says the points A, B, C must be non-collinear. By I-1 and
I-2, each pair lies on a unique line, whence there are precisely
three lines
ℓ = {A, B}, m = {A, C}, n = {B, C}
Up to relabelling, there are two incidence geometries with four
points: one is drawn; how many lines has the other?
ℓ
m
n
A
B
C
A
B
C
D
3 points, 3 lines 4 points, 6 lines
The final picture is a seven-point incidence geometry called the Fano
plane, which finds many applications particularly in combinatorics. Each
point lies on precisely three lines and each line contains precisely three
points—each dot is colored to indicate the lines to which it belongs.
Don’t be fooled by the black line looking ‘curved’ and seeming to cross
the blue line near the top, for the line really only contains three points!
We can even prove some simple theorems in incidence geometry.
Lemma 2.14. 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 at most one line through
A and B. Contradiction.
Lemma 2.15. Given any point, there exist at least two lines on which it lies.
The proof is an exercise. While incidence geometry is fun, our main goal is to understand Euclidean
geometry, so we move on to the next set of axioms.
Axioms of Order: Sides of a Line, Pasch’s Axiom & the Crossbar Theorem
The axioms of order describe the ternary relation between. Their inclusion in Hilbert’s axioms is due
in no small part to the work of Moritz Pasch, after whom Pasch’s axiom (O-4, c. 1882) is named. This
axiom is very powerful; in particular, it permits us to define the interiors of several geometric objects,
and to see that these are non-empty.
Lemma 2.16. Every segment contains an interior point.
We leave the proof to Exercise 5. By inducting on the Lemma, every segment contains infinitely many
points, whence the above finite geometries are not valid models once the order axioms are included.
17
To get much further, it is necessary to establish that a line has precisely two sides (Definition 2.12). This
concept lies behind several of Euclid’s arguments, without being properly defined in the Elements.
Theorem 2.17 (Plane Separation). A line ℓ separates all points not on ℓ into two half-planes: the two
sides of ℓ . To be explicit, suppose none of the points A, B, C lie on ℓ, then:
1. If A, B lie on the same side of ℓ 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.
ℓ
A
B
C
ℓ
A
B
C
ℓ
A
B
C
Case 1 Case 2 Case 3
Proof. We prove the contrapositive of case 1. Suppose A, B, C are non-collinear. If AC intersects ℓ,
then ℓ intersects one side of △ABC. By Pasch’s axiom, it also intersects either AB or BC.
The other cases are exercises, and we omit the tedious collinear possibilities.
Plane separation/sidedness allows us to properly define interiors of angles and triangles.
Definition 2.18. 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.
A
B
C
I
A point I is interior to triangle △ABC if it is interior to all three of its angles ∠ABC, ∠BAC and
∠ACB. Otherwise said, I lies in the triple intersection of three of the half-planes defined by the
triangle’s sides.
Interior points permit us to compare angles: if I is interior to ∠BAC, then ∠BAI < ∠BAC has obvious
meaning without resorting to numerical angle measure.
Corollary 2.19. Every angle has an interior point.
Proof. Given ∠BAC, consider any interior point I of the segment BC. This plainly lies on the same
side of
←→
AB as C and on the same side of
←→
AC as B.
In Exercise 8, we check that the interior of a triangle is non-empty.
Pasch’s axiom could be paraphrased: If a line enters a triangle, it must come out. We haven’t quite
established this crucial fact, however. What if the line passes through a vertex?
18
Theorem 2.20 (Crossbar Theorem). Suppose I is interior to ∠BAC.
Then
−→
AI intersects BC.
In particular, if a line passes through a vertex and an interior point of
a triangle, then it intersects the side opposite the vertex.
ℓ
A
B
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 triangle △BCD. Since
←→
AI intersects one edge of △BCD at A and does not cross any vertices
(think about why. . . ), Pasch says it intersects one of the other edges (BC or CD) at some point M.
The result follows from applying plane separation to the lines
←→
AB =
←→
BD and
←→
AC. First observe:
Since I, M lie on the same side of
←→
AB =
←→
BD as C, it follows that IM does not intersect
←→
AB.
Since A, I, M are collinear and A ∈
←→
AB, it follows that A /∈ I M.
If M ∈ BC, we are done. Our goal is to show that M ∈ CD is a contradiction.
A
B
C
I
D
M
A
B
C
I
D
M
correct arrangement
9
contradiction
Suppose, for contradiction, that M ∈ CD. Relative to
←→
AC:
• I and B lie on the same side since I is interior to ∠BAC;
• B and D lie on opposite sides, since B ∗ A ∗ D and
←→
AC =
←→
BD = AB;
• D and M lie on the same side since M ∈ CD and
←→
CD =
←→
AC.
By plane separation, I, M lie on opposite sides of
←→
AC. The collinearity of A, I, M then forces the
contradiction A ∈ IM.
Euclid repeatedly uses the crossbar theorem without justification,
including in his construction of perpendiculars and angle/segment
bisectors (Theorems I. 9+10). We sketch the latter here.
Given ∠BAC, construct E such that AB
∼
=
AE. Construct D using
an equilateral triangle (I. 1). SSS (I. 8) shows that ∠BAC is bisected,
and SAS (I. 4) that
−→
AD bisects BE.
Quite apart from Euclid’s arguments for SAS and SSS being suspect
(we’ll deal with these in the next section), he gives no argument for
why D is interior to ∠BAC or why
−→
AD should intersect BE!
A
B
C
D
E
Even with Pasch’s axiom and the crossbar theorem, it requires some effort to repair Euclid’s proof. No
matter, we’ll provide an alternative construction of the bisector once we’ve considered congruence.
9
The pictures could be modified: e.g., I = M and A ∗ I ∗ M are also correct arrangements (M ∈ BC).
19
Exercises 2.2. 1. Label the vertices in the Fano plane 1 through 7 (any way you like). As we did in
Example 2.13 for the 3-point geometry, describe each line in terms of its points.
2. Prove Lemma 2.15.
3. 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)
4. Consider the proof of the crossbar theorem. Explain how we know that
←→
AI does not contain
any of the vertices of △BCD.
5. You are given distinct points A, B. Using the axioms of incidence and order and Lemma 2.14
(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 (Lemma 2.16).
During your construction, address the following issues:
(a) Explain why D does not lie on
←→
AB.
(b) Explain why E does not lie on △ABD.
(c) Explain why E = C (whence
←→
CE exists).
(d) Explain why F lies on AB and not on BD.
A
B
C
D
E
F
6. We complete the proof of the plane separation theorem (2.17).
(a) Prove part 3 (it is almost a verbatim application of Pasch’s axiom).
(b) Suppose a line ℓ intersects all three sides of △ABC but no vertices.
This results in a very strange picture (we’ve labelled the intersec-
tions D, E, F and WLOG chosen D ∗ E ∗ F).
Apply Pasch’s axiom to △DBF and
←→
AC to obtain a contradiction.
Hence establish part 2 of the plane separation theorem.
ℓ
A
B
C
D
E
F
7. Suppose A, B, C are distinct points on a line ℓ.
(a) Explain why there exists a line m = ℓ such that B ∈ m.
(b) Prove that A ∗ B ∗C ⇐⇒ A and C lie on opposite sides of m.
(c) Suppose A ∗ B ∗ C. Use part (b) to prove the following:
i. B is the only point common to the rays
−→
BA and
−→
BC.
ii. If D ∈ ℓ is any point other than B, prove that D lies in precisely one of
−→
BA or
−→
BC.
8. Prove that the interior of a triangle is non-empty.
(Hint: use Exercise 5 to construct a suitable I, then prove that it lies on the correct side of each edge)
20
2.3 Hilbert’s Axioms II: Congruence
Hilbert’s congruence axioms address two primary issues in Euclid.
1. Euclid’s use of equal is confusing. In Hilbert, segments/angles are now equal only when they
are precisely the same (this amounts to the reflexivity part of the next result).
2. Euclid’s frequent and unjustified use of pictorial reasoning. We previously discussed Euclid’s
erroneous approach to the SAS and SSS triangle congruence theorems. It was eventually real-
ized that one of the triangle congruences has to be an axiom: SAS is Hilbert’s C-6.
We start with a small piece of bookkeeping.
Lemma 2.21. Congruence of segments/angles is an equivalence relation.
Proof. (Reflexivity) Let AB be given. Apply C-1 to obtain A
′
B
′
such that AB
∼
=
A
′
B
′
. We sneakily use
this twice and apply C-2 to obtain
AB
∼
=
A
′
B
′
and AB
∼
=
A
′
B
′
=⇒ AB
∼
=
AB
(Symmetry) Assume AB
∼
=
CD. By reflexivity, CD
∼
=
CD. By C-2 we have CD
∼
=
AB.
(Transitivity) Suppose AB
∼
=
CD and CD
∼
=
EF. By symmetry, EF
∼
=
CD. Axiom C-2 now shows that
AB
∼
=
EF.
Axioms C-4 and C-5 say essentially the same thing for angles (see Exercise 2).
Segment/Angle Transfer and Comparison
Hilbert’s axioms of segment and angle transference are crucial for comparing non-collinear segments
and angles with distinct vertices.
Definition 2.22. Let segments AB and CD be given.
By axiom C-1, let E be the unique point on
−→
CD such that CE
∼
=
AB:
we have transferred AB onto
−→
CD.
We write AB < CD if E lies between C and D, etc.
C
D
A
B
E
By O-3, any two segments are 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 adjacent congruent segments. Unique angle trans-
fer, comparison and addition follow similarly from axiom C-4 and Definition 2.18 (interior points).
Neither Hilbert nor Euclid use or require an absolute notion of length/angle-measure: the compari-
son AB < CD does not indicate a relationship between numerical quantities (lengths). Introducing
numerical length requires the inclusion of the real numbers (and thus far more axioms)—for purity
reasons, we postpone this until Section 2.5.
21
The Triangle Congruence Theorems: SAS, ASA, SSS & SAA
Hilbert assumes side-angle-side (SAS) and proceeds to prove the remainder. Here is the first of these;
we’ll cover SSS momentarily and SAA in Exercise 6.
Theorem 2.23 (Angle-Side-Angle/ASA, Euclid I. 26, case I). Suppose △ABC and △DEF satisfy
∠ABC
∼
=
∠DEF, AB
∼
=
DE, ∠BAC
∼
=
∠EDF
Then the triangles are congruent (∠ACB
∼
=
∠DFE, AC
∼
=
DF and BC
∼
=
EF).
Hilbert’s approach modifies Euclid’s: instead of laying △ABC on top of △DEF, he creates a new
triangle △DEG
∼
=
△ABC and proves that G = F.
Proof. Segment transfer provides the unique point G ∈
−→
EF such that EG
∼
=
BC.
SAS applied to AB
∼
=
DE, ∠ABC
∼
=
∠DEG (= ∠DEF), BC
∼
=
EG,
says ∠BAC
∼
=
∠EDG (
∼
=
∠EDF) (this last is by assumption).
Since F and G lie on the same side of
←→
DE, angle transfer (C-4) says
they lie on the same ray through D.
But then F and G both lie on two distinct lines (
←→
EF =
←→
EG and
←→
DF =
←→
DG). We conclude that F = G.
By SAS we conclude that △ABC
∼
=
△DEF.
A
B
C
D E
F
G
Geometry Without Circles
Circles are at the heart of Euclid’s constructions. Yet, for reason we’ll address in Section 2.4, Hilbert
essentially ignores them. We sketch a few of his alternative approaches to Euclid’s basic results.
Theorem 2.24 (Euclid I. 5). An isosceles triangle has congruent base angles.
Isosceles means equal legs: two sides of the triangle are congruent. The remaining side is the base.
Euclid’s argument relies on a famously complicated construction (look it up!). Hilbert does things
more speedily and sneakily, by relabelling the original triangle and applying SAS.
Proof. Suppose △ABC is isosceles where AB
∼
=
AC. Consider a ‘new’
triangle △A
′
B
′
C
′
= △ACB where the base points are switched:
A
′
:= A, B
′
:= C, C
′
:= B
Observe:
• ∠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
′
22
Dropping a Perpendicular As with the majority of Book I, Euclid accomplishes this using circle
intersections.
10
Hilbert instead uses segment/angle transference and the concept of sidedness.
Suppose we are given a line ℓ and a point P not on ℓ. Our goal is to
construct a point M ∈ ℓ such that PM intersects ℓ in a right-angle.
Let A, B be distinct points on ℓ (axiom I-3) so that ℓ =
←→
AB.
By axioms C-4 and C-1, we may transfer AP to the other side of ℓ at
A, creating a new point Q.
Since P and Q lie on opposite sides of ℓ, the line intersects PQ at some
point M. There are two cases to consider.
• In the generic case M = A (pictured), SAS applied to △MAP
and △MAQ shows that ∠AMP
∼
=
∠AMQ. Since these angles
sum to a straight edge (PQ), they are both right-angles.
• In the extreme case M = A, there are no triangles and SAS
cannot be applied. Instead, observe that B does not lie on PQ
(which axioms/results make this clear?!) and apply the above
argument with B instead of A.
ℓ
A
B
M
P
Q
A generalization of this construction facilitates a corrected argument for the SSS triangle congruence.
Theorem 2.25 (Side-Side-Side/SSS, Euclid I. 8). Suppose △ABC and △DEF have sides congruent
in pairs:
AB
∼
=
DE, BC
∼
=
EF, AC
∼
=
DF
Then the triangles are congruent (∠ABC
∼
=
∠DEF, ∠BCA
∼
=
∠EFD, ∠CAB
∼
=
∠FDE).
The strategy is similar to the proof of ASA. Hilbert creates a new triangle △DEG
∼
=
△ABC, though
this time with G on the opposite side of
←→
DE to F.
Proof. Transfer ∠BAC to D on the other side of
←→
DE from F to
obtain G (axioms C-4 and C-1).
SAS (AB
∼
=
DE, ∠BAC
∼
=
∠EDG, AC
∼
=
DG) shows that
EG
∼
=
BC
∼
=
EF. Otherwise said, △DEG
∼
=
△ABC.
Join FG to produce isosceles triangles △FDG and △FEG
with base FG, both with congruent angles at F and G.
Sum angles at F and G and apply SAS (DF
∼
=
DG, ∠DFE
∼
=
∠DGE, EF
∼
=
EG) to see that △DEF
∼
=
△DEG.
We conclude that △ABC
∼
=
△DEG
∼
=
△DEF, as required.
To be completely formal, we should also carefully deal with
the situations where the sum is a subtraction or the triangle
is right-angled at A or B.
D E
F
G
A
B
C
10
Consider the picture for Thm. I. 11 in Exercise 2.2.2.
23
Exterior Angle Theorem (Thm. 2.3, I. 16) Euclid’s approach uses a bisector which he obtains from
circles. Hilbert does things a little differently.
Proof. Given △ABC, extend AB to D such that AC
∼
=
BD. For clarity, we label angles with Greek let-
ters as in the first picture below. We show that γ < δ by proving that the alternatives are impossible.
1. (δ ≇ γ) Assume δ
∼
=
γ. By SAS, △ACB
∼
=
△DBC; in particular ϵ
∼
=
β. Since A and D lie on
opposite sides of
←→
BC, we see that ϵ + γ
∼
=
β + δ is a straight edge. But then A, D are distinct
points lying on two lines! Contradiction.
2. (δ < γ) Assume δ < γ. Transfer δ to C as shown to obtain η
∼
=
δ. By the crossbar theorem, we
obtain an intersection point E. But now δ is an exterior angle of △EBC congruent to an interior
angle η of the same triangle, contradicting part 1.
A
B
C
D
α
β
γ
δ
ϵ
A
B
C
D
E
α
β
η
δ
ϵ
Step 1: δ
∼
=
γ is a contradiction Step 2: δ < γ is a contradiction
Take the vertical angle to δ at B and repeat the argument to see that α < δ.
The proof also shows that the sum of any two angles in a triangle is strictly less than a straight edge:
α + β < δ + β = 180°.
Is Euclid now fixed? Almost! In the exercises we show how the following may be achieved:
• 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.
We’ve now recovered almost all of Book I prior to the application of the parallel postulate. Including
Playfair’s axiom completes the remainder, including Pythagoras’, all without circles!
Exercises 2.3. Except for question 8, answer everything without reference to the continuity axiom,
circles, or the uniqueness of parallels (e.g., Playfair’s axiom, (tri)angle sum = 180°).
1. Draw pictures to suggest why you don’t expect Angle-Angle-Angle (AAA) and Side-Side-
Angle (SSA) to be triangle congruence theorems.
2. Use Hilbert’s axioms C-4 and C-5 to prove that congruence of angles is an equivalence relation.
3. (a) Use ASA to prove that if the base angles are congruent then a triangle is isosceles.
(b) Find an alternative argument that relies the exterior angle theorem.
(Hint: this is essentially the same as the proof of Exercise 5 (a))
(c) Explain why the base angles of an isosceles triangle are acute (less than a right-angle).
24
4. Given AB, axiom I-3 says ∃C ∈
←→
AB.
If △ABC is not isosceles, then WLOG assume ∠ABC < ∠BAC.
Transfer ∠ABC to A to produce D on the same side of
←→
AB as C with
∠ABC
∼
=
∠BAD, BC
∼
=
AD
A
B
C
D
M
(a) Explain why rays
−→
AD and
−→
BC intersect (at some point M).
(b) Why is △MAB isosceles?
(c) Describe how to produce the perpendicular bisector of AB.
(d) Explain how to construct an angle bisector using the above discussion.
5. We prove Theorems I. 18, 19 and 20 on comparisons
of angles and sides in a triangle. For clarity, suppose
△ABC 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: converse to I. 18) α < γ =⇒ a < c.
(Hint: Prove the contrapositive)
(c) (I. 20: triangle inequality) a + b > c.
(Hint: Let E lie on
−→
BC such that CE
∼
=
b and apply I. 19)
6. Prove the SAA congruence. If △ABC and △DEF satisfy
AB
∼
=
DE, ∠ABC
∼
=
∠DEF and ∠BCA
∼
=
∠EFD
then the triangles are congruent: △ABC
∼
=
△DEF.
(Hint: Let G ∈
−→
BC be such that BG
∼
=
EF, apply SAS and the exterior
angle theorem)
G
B
A
C
7. Hilbert’s published SAS axiom is weaker than we’ve stated.
Given triangles △ABC and △A
′
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 the proof of ASA)
8. Construct the picture on the right, where BF is the perpendicular
bisector of AC, and
BD
∼
=
AB, BE
∼
=
AD, BF
∼
=
AE
Use Pythagoras’ Theorem to prove that △ACF is equilateral.
This construction requires Playfair’s axiom and thus unique parallels. It
does not require circle intersections (continuity) like Theorem 2.1 (I. 1).
A
B
C
D
E
F
25
2.4 Circles and Continuity
Definition 2.26. Let O and R be distinct points. The circle C with center O and
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 given
circle.
O
R
A
P
Q
A major weakness of Euclid is that many of his proofs rely on circle intersections, rather than lines.
To use circles in this manner requires the Axiom of Continuity. This much more technical than the
other axioms. It is likely for this reason that Hilbert barely mentions circles, instead wanting to build
as much geometry as possible using only the simplest axioms.
Here are the facts necessary before Euclid’s approach can be followed.
Theorem 2.27. Suppose C and D are circles.
1. (Elementary/line-circle Continuity Principle) If P lies inside and Q outside C, then PQ inter-
sects C in exactly one point.
2. (Circular Continuity Principle) If D contains a point inside and another outside C, then they
intersect in exactly two points. These lie on opposite sides of the line joining the circle centers.
The idea of the first principle is to partition PQ into two pieces:
Σ
1
consists of the points lying on or inside C
Σ
2
consists of the points lying outside C
One shows that Σ
1
and Σ
2
satisfy the assumptions of the axiom. The
unique point O then exists and is shown to lie on C itself. Some of this are
in Exercise 6. The circular continuity principle is harder.
P
Q
O
Σ
1
Σ
2
C
What is perhaps more interesting is to consider a geometry in which the axiom of continuity is false.
Example 2.28. The geometry Q
2
= {(x, y) ∈ R
2
: 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
that O = (
1
√
2
,
1
√
2
) is the unique point (in R
2
) on the ray r =
−→
AC
such that AC
∼
=
AB. Clearly O is an irrational point and therefore
not in the geometry.
Continuity The circle centered at A = (0, 0) with radius 1 does not
intersect the segment AC. More properly, AC = Σ
1
∪ Σ
2
may be
partitioned as shown and yet no point O in the geometry separates
Σ
1
, Σ
2
.
Σ
1
Σ
2
A
B
C
O
26
Equilateral triangles We can finally correct Euclid’s proof of the first proposition of the Elements!
Theorem 2.29 (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) and
C (outside α) so the circles intersect in two points P, Q.
Since P lies on both circles (and is therefore distinct from
A and B), we have AB
∼
=
AP
∼
=
BP whence △ABC is
equilateral.
A
B
C
D
P
Q
α
β
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). Nevertheless, 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!
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. From this point on, we assume all Hilbert’s axioms including
Playfair and continuity. Indeed what follows often relies on their consequences, particularly angle-
sums in triangles and the circular continuity principle.
Definition 2.30. 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
)
AB is part of the circular edge between chord points
(major or minor by length).
• ∠AOB is a central angle and ∠APB an inscribed angle.
• △ABP is inscribed in its circumcircle.
A
B
C
P
O
Since these ideas shouldn’t be new, most of the details are left as exercises.
Theorem 2.31 (III. 20). The central angle is twice the inscribed angle: ∠AOB = 2∠APB.
For a sketch proof, join OP, breaking △ABP into three isosceles triangles and count angle sums.
Corollary 2.32. 1. (III. 21) If inscribed triangles share a side, the opposite angles are congruent.
2. (III. 22) An inscribed quadrilateral has opposite angles supplementary (summing to 180°).
3. (Thales’ Theorem III. 31) A triangle in a semi-circle is right-angled.
27
Theorem 2.33. Any triangle has a unique circumcircle.
This is similar to III. 1: construct the perpendicular bisectors of
two sides as in the picture.
A
B
C
O
Definition 2.34. A line is tangent to a circle if it intersects the circle exactly once.
Theorem 2.35 (III. 18, 19 (part)). A line is tangent to a circle if and only if it is perpendicular to the
radius at an intersection point.
Proof. (⇐) Suppose ℓ through T is perpendicular to the radius OT.
Let P be any another point on ℓ. But then (Exercise 2.3.5),
∠OPT < 90° = ∠OTP =⇒ OT < OP
thus P lies outside the circle. Every point on ℓ except T lies outside
the circle, so T is the unique intersection, and ℓ is therefore tangent.
The converse is an exercise.
T
O
P
ℓ
Theorem 2.36. 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 2.32.
2. Prove Theorem 2.33.
3. Complete the proof of Theorem 2.35 by showing the (⇒) direction.
(Hint: suppose T is an intersection and that the angle isn’t 90°, and drop a perpendicular to ℓ. . . )
4. Given a circle centered at O and a point P outside the circle, draw the circle centered at the
midpoint of OP passing through O and P. Explain why the intersections of these circles are the
points of tangency required in Theorem 2.36, and hence complete its proof.
5. (a) Prove Theorem 2.31 when O is an interior point to △ABP.
(b) Prove Theorem 2.31 when O is an exterior point to △ABP.
6. Suppose A ∗C ∗B and that O /∈
←→
AB. Use Exercise 2.3.5 to show that
OC < max
OA, OB
If A, B are interior to a circle centered at O, conclude that C is also.
(This is part of what’s needed to demonstrate the elementary continuity principle:
no point of Σ
2
lies between two points of Σ
1
. Can you prove the other condition?)
A
B
C
O
7. If a line contains a point inside a circle, show that it intersects the circle in two points.
(Hint: first construct two points on the line lying outside the circle)
28
2.5 Similar Triangles, Length and Trigonometry
In the geometry of Euclid & Hilbert, there are no numerical measures of length or angle. Relative
measure is built in (Definition 2.22), and we’ve denoted right-angles and straight edges by 90° & 180°
purely for convenience. To avoid continued frustration it is time we introduced explicit numerical
measure, though to do so properly requires more axioms!
Axioms 2.37 (Length and Angle Measure).
L1 To each segment AB corresponds a unique length
|
AB
|
, a positive real number
L2
|
AB
|
=
|
CD
|
⇐⇒ AB
∼
=
CD
L3
|
AB
|
<
|
CD
|
⇐⇒ AB < CD (Definition 2.22)
L4 If A ∗ B ∗ C, then
|
AB
|
+
|
BC
|
=
|
AC
|
A1 To each ∠ABC corresponds a unique degree measure m∠ABC, a real number between 0 and 180
A2 m∠ABC = m∠DEF ⇐⇒ ∠ABC
∼
=
∠DEF
A3 m∠ABC < m∠DEF ⇐⇒ ∠ABC < ∠DEF (Definition 2.18)
A4 If P is interior to ∠ABC, then m∠ABP + m∠PBC = m∠ABC
A5 Right-angles measure 90°
Don’t memorize these axioms, just observe how they fit your intuition. Angle measure in Euclidean
geometry has two notable differences from what you might expect:
• (A1) All angles measure strictly between 0° and 180°. A straight edge isn’t an angle, though
such is commonly denoted 180°, and there are no reflex angles (> 180°).
• (A2) Angles are non-oriented, measuring the same in reverse (m∠ABC = m∠CBA).
The axioms for length and angle follow the same pattern except for us explicitly fixing the scale of
angle measure (A5). To do the same for length requires only a choice of a reference segment of length
1. The following is a consequence of the continuity axiom.
Theorem 2.38 (Uniqueness of measure).
1. Given OP, there is a unique way to assign a length to every segment such that
|
OP
|
= 1.
2. There is a unique way to assign a degree measure to every angle.
The segment OP in part 1 provides a length-scale
for a ruler. We measure the length of any segment
by moving this rule on top of the desired segment
(congruence!).
0
1
2
3
4
5
1
O
P
0
1
2
3
4
5
|QR| = 4.6
Q
R
29
Area Measure If we also include Playfair’s axiom, then the discussion at the end of Book I of Euclid
becomes valid, and rectangles can be defined (see e.g., Exercise 2.1.2).
Definition 2.39. The area (measure) of a rectangle is the product of its base and height (measures).
Given a length measure, a square with side length 1 necessarily has area 1. Relative to a base segment,
the height of a triangle is the length of the perpendicular dropped from the vertex.
Since every rectangle is a parallelogram and a triangle half a parallelogram, Eu-
clid’s discussion (Thm. I. 35) amounts to the familiar area formulæ:
area(parallelogram) = bh, area(triangle) =
1
2
bh
While these expressions are nice to have, they are not necessary. Indeed every-
thing that follows depends only on area ratios: e.g.,
1
2
bh
1
1
2
bh
2
=
h
1
h
2
.
h
b
Lemma 2.40. If triangles have congruent bases, then their areas are in the same ratio as their heights.
The same holds with the roles of heights and bases reversed.
Similarity and the AAA Theorem Similar triangles are the concern of Book VI of the Elements.
Definition 2.41. Triangles are similar, written △ABC ∼ △XYZ, if
their sides are in the same length ratio
|
AB
|
|
XY
|
=
|
BC
|
|
YZ
|
=
|
CA
|
|
ZX
|
A
B
C
X
Y
Z
Euclid discusses these using non-numerical ratios of segments (e.g., AB : XY = BC : YZ). This is
unnecessarily confusing for modern readers, indeed some of the most difficult parts of the Elements
are where he describes what this should mean, particularly for irrational ratios (Books V & X).
Our primary result comes in two versions, where the second (which we’ll prove) is a special case of
the first.
Theorem 2.42 (Angle-Angle-Angle/AAA, Euclid VI. 2–5).
1. Triangles are similar if and only if their angles come in mutually con-
gruent 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.
ℓ
A
B
C
The picture should convince you that 1 ⇒ 2 follows from the uniqueness of parallels (Playfair’s
axiom, Corollary 2.5 & Theorem 2.6). This reliance is crucial! We should not expect AAA similarity
in non-Euclidean geometry, and indeed shall see later that it is false in hyperbolic geometry (Chapter
4), where AAA is a theorem for congruent triangles! The converse (2 ⇒ 1) is left to Exercise 10.
30
Proof (AAA similarity, part 2). Suppose ℓ intersects △ABC at
points D, E as shown. Drop perpendiculars to create distances
h, k, d
1
, d
2
as indicated. We prove:
ℓ is parallel to BC ⇐⇒ △ABC ∼ △ADE
( ⇒) Suppose ℓ is parallel to BC. Playfair
11
tells us that d
1
= d
2
.
By Lemma 2.40, triangles with the same height have areas
proportional to their bases:
d
1
d
2
ℓ
h
k
A
B
C
D
E
|
BD
|
|
AD
|
=
area(BDE)
area(ADE)
(△BDE, △ADE have same height h)
|
CE
|
|
AE
|
=
area(CDE)
area(ADE)
(△CDE, △ADE have same height k)
Since △BDE and △CDE share base DE, we see that
ℓ is parallel to BC ⇐⇒ d
1
= d
2
⇐⇒ area(BDE) = area(CDE)
⇐⇒
|
BD
|
|
AD
|
=
|
CE
|
|
AE
|
(∗)
Add 1 =
|
AD
|
|
AD
|
=
|
AE
|
|
AE
|
to both sides to obtain one part of the similarity ratio
|
AB
|
|
AD
|
=
|
AD
|
+
|
BD
|
|
AD
|
=
|
AE
|
+
|
CE
|
|
AE
|
=
|
AC
|
|
AE
|
It remains to see that this ratio equals
|
BC
|
|
DE
|
. Again using common heights (h, k) of triangles,
|
AB
|
|
BD
|
=
area(ABE)
area(BDE)
|
CE
|
|
AE
|
=
area(BCE)
area(ABE)
(†)
=⇒
|
AB
|
|
AD
|
(∗)
=
|
AB
|
|
BD
|
|
CE
|
|
AE
|
(†)
=
area(BCE)
area(BDE)
=
|
BC
|
|
DE
|
where the last equality follows since △BCE and △BDE have common height d
1
= d
2
.
( ⇐) Suppose △ABC ∼ △ADE. By Playfair, let m be the
unique parallel to BC through D. This intersects AC at a
point G. We must prove that G = E (consequently m = ℓ).
By the (⇒) direction above,
△ABC ∼ △ADG
d
1
d
2
ℓ
m
A
B
C
D
E
G
However, ∼ is plainly transitive (it is an equivalence relation), whence △ADE ∼ △ADG. The
similarity ratio is 1 =
|
AD
|
|
AD
|
, whence
|
AE
|
|
AG
|
= 1 =⇒
|
AE
|
=
|
AG
|
=⇒ E = G
11
d
1
= d
2
⇐⇒ ℓ parallel to BC is Playfair! Compare Exercise 2.1.2 (Thm I. 46) on the construction of a square...
31
Applications of Similarity: Trigonometric Functions, Cevians and the Butterfly Theorem
We finish with several applications of similarity which hopefully give an idea of what can be done
without co-ordinates. None of these ideas were known to Euclid.
Definition 2.43. Given an acute angle ∠ABC (m∠ABC < 90°), drop a perpendicular from A to
−→
BC
at D so that ∠ADB is a right-angle. Define
sin ∠ABC :=
|
AD
|
|
AB
|
cos ∠ABC :=
|
BD
|
|
AB
|
Early trigonometry dates to a few hundred years after Euclid, though the approach was different.
12
Theorem 2.44. Angles have the same sine (cosine) if and only if they are congruent.
Proof. Assume ∠ABC
∼
=
∠A
′
B
′
C
′
as in the picture, and drop perpendiculars to D, D
′
.
Since △ABD and △A
′
B
′
D
′
have two pairs of mutually
congruent angles, the third pair is congruent also. AAA
applies, the triangles are similar and
|
AD
|
|
AB
|
=
|
A
′
D
′
|
|
A
′
B
′
|
|
BD
|
|
AB
|
=
|
B
′
D
′
|
|
A
′
B
′
|
A
B
C
D
A
′
B
′
C
′
D
′
In particular, sin ∠ABC = sin ∠A
′
B
′
C
′
and cos ∠ABC = cos ∠A
′
B
′
C
′
.
The converse is an exercise.
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
it using co-ordinates!
Theorem 2.45 (Ceva’s Theorem). Given △ABC and cevians AX, BY, CZ,
|
BX
|
|
XC
|
|
CY
|
|
YA
|
|
AZ
|
|
ZB
|
= 1 ⇐⇒ the cevians meet at a common point P
Proof. (⇐) This is simply a repeated application of Lemma 2.40.
area(ABX)
area(AXC)
=
|
BX
|
|
XC
|
=
area(PBX)
area(PXC)
=⇒
|
BX
|
|
XC
|
(∗)
=
area(ABX) −area(PBX)
area(AXC) −area(PXC)
=
area(ABP)
area(APC)
Repeat for the other ratios and multiply to get 1.
A simple justification of (∗) and the converse are an exercise.
A
B
C
X
Y
Z
P
12
The ancient forerunners of sine and cosine were defined using chords of circles rather than triangles. The word
trigonometry (literally triangle measure) wasn’t coined until 1595.
32
Theorem 2.46 (Butterfly Theorem). We are given the following data
as in the picture:
• PQ is a chord of a circle with midpoint M.
• AC and BD are chords meeting at M.
• X, Y are the chord-intersections as shown.
Then M is the midpoint of XY.
A
B
C
D
M
P
Q
X
Y
This beautiful result dates to 1803-5 and has several proofs. We present an argument relying on
similar triangles.
Proof. For convenience we introduce several numerical
lengths:
• z =
|
PM
|
=
|
MQ
|
, x =
|
XM
|
and y =
|
YM
|
.
• Drop perpendiculars from X, Y to chords AD, BC,
and label the lengths x
1
, x
2
, y
1
, y
2
as shown.
The four colored pairs of angles are congruent: vertical an-
gles at M, and inscribed angles at A, B, C, D.
We compare sides of several similar triangles:
•
x
x
1
=
y
y
1
and
x
x
2
=
y
y
2
=⇒
x
y
=
x
1
y
1
=
x
2
y
2
•
|
AX
|
|
BY
|
=
x
1
y
2
and
|
XD
|
|
YC
|
=
x
2
y
1
x
y
x
1
x
2
y
1
y
2
A
B
C
D
M
P
Q
X
Y
The result follows by putting these observations together and applying Exercise 1 twice:
13
x
2
y
2
=
x
1
x
2
y
1
y
2
=
|
AX
||
XD
|
|
BY
||
YC
|
=
|
PX
||
XQ
|
|
PY
||
YQ
|
=
( z − x)(z + x)
( z + y)(z −y)
=
z
2
− x
2
z
2
−y
2
=⇒ x
2
( z
2
−y
2
) = y
2
( z
2
− x
2
)
=⇒ x
2
= y
2
=⇒ x = y
as required.
Exercises 2.5. 1. Let AD and PQ be chords of a circle which intersect at X. Use similar triangles to
prove that
|
AX
||
XD
|
=
|
PX
||
XQ
|
13
The denominators are equal by applying the Exercise to the chords BC and PQ.
33
2. Let △ABC 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
△ACB ∼ △ADC ∼ △CDB
(c) Use these facts to prove Pythagoras’ Theorem.
(Use the picture, where a, b, c, x, y are lengths)
a
b
c = x + y
x
y
A
B
C
D
3. Prove a simplified version of the SAS similarity theorem:
|
AB
|
|
AG
|
=
|
AC
|
|
AH
|
⇐⇒ △ABC ∼ △AGH
(Hint: construct BJ parallel to GH and appeal to AAA)
A
B
C
G
H
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.
5. Use Pythagoras’ to prove that sin 45° = cos 45° =
1
√
2
, that sin 60° =
√
3
2
and cos 60° =
1
2
.
6. Prove the converse of Theorem 2.44: if sin ∠ABC = sin ∠A
′
B
′
C
′
, then ∠ABC
∼
=
∠A
′
B
′
C
′
.
(Hint: create right-triangles and prove they are similar. Label the side lengths o, a, h, etc.)
7. We complete the proof of Ceva’s theorem.
(a) If p, q, r, s are non-zero real numbers, verify that α =
p
q
=
r
s
=⇒ α =
p−r
q−s
.
(b) Assume X, Y, Z satisfy Ceva’s formula.
Define P as the intersection of BY and CZ and let
−→
AP
meet BC at X
′
.
Prove the (⇒) direction of Ceva’s theorem by using
the (⇐) direction to show that X
′
= X.
A
B
C
X
X
′
Y
Z
P
8. (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) (Hard) Medians split a 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.
9. Prove that similarity of triangles is an equivalence relation.
(Don’t use AAA since its proof requires this fact!)
10. (Hard) Explain how to prove (2 ⇒ 1) in the AAA Theorem (2.42).
34
