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 Islamic 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 the Pythagorean Theorem. 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 somewhat more general than a postulate. Here the postulates contain the geometry.
8
The first three postulates describe ruler and compass constructions. P4 allows Euclid to compare angles
at different locations. P5 is usually called 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 permitted, often infinitely.
Congruence Euclid uses equal where modern mathematicians say congruent. We’ll express congruent
segments and angles in the modern fashion, e.g., ∠ABC
∼
=
∠DEF. Equal angles/segments
must be genuinely the same object (same location, etc.).
Basic Theorems `a la Euclid
Theorems were typically presented as a problem. Euclid provides a constructive solution (P1–P3)
before proving that his construction really does solve 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. A triangle is equilateral if its three sides are congruent.
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 also 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, say in Byrne’s edition. These are worth reading despite
the logical issues in Euclid’s presentation. We’ll revisit these basic results in the Exercises and in the
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, for 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 within our grasp: 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 the sake of familiarity we’ll
often write 180° for a straight-edge and 90° for a right-angle.
In the picture we’ve labelled angles with 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 the lines
are not parallel). Euclid possibly chose this formulation to facilitate proofs by contradiction, though
the unfortunate 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 m to ℓ passes through C.
m
ℓ
C
Our discussion thus far shows that the parallel postulate im-
plies 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 which share a vertex: 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 /∈ IM.
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).
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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. (a) 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)
(b) It is possible for there to be a 6-point incidence geometry so that each line contains pre-
cisely three points? Why/why not?
4. Consider the proof of the crossbar theorem. How can we be certain 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)
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