A MINIMAL VERSION OF HILBERT’S AXIOMS FOR PLANE
GEOMETRY
WILLIAM RICHTER
Abstract. This note is intended to be useful to good high school students
wanting a rigorous treatment of Hilbert’s axioms for plane geometry.
1. Introduction
Euclid’s Elements [6] is very creative geometry, but it contains errors involving
angle addition. These errors were found and fixed by Hilbert [12], by adding axioms
involving betweenness of points. However it seems that Euclid’s errors are made in
every widely used high school Geometry text in the US today. A good high school
student should be able to read Venema’s college geometry text [16], which has both
rigorous and elegant proofs. This note is intended to help such students. Moise’s
college geometry text [15], which influenced Venema’s, may be another good choice.
Venema lists the minimal version of Hilbert’s axioms [16, App. B1], but he
uses stronger axioms, which obviate §3 and §4. The geometry texts for graduate
students [7, 10, 13] also use stronger Hilbert’s axioms. We prove that the minimal
Hilbert’s axioms suffice. This result was proved by Hilbert, Moore, Wylie, and
Greenberg [12, 9, 17, 7], but our proofs seem elegant, and are collected in one source.
Venema also uses Birkhoff’s axioms [4] that the real line R measures segments and
angles, but he does not explain Birkhoff’s work. In §9 we discuss Birkhoff’s work,
explained by MacLane [14], which gives an alternative to Hilbert’s work by using
directed angles and taking the crossbar theorem and its converse as an axiom.
Venema discusses a widely used high school Geometry text [2], but not its lack
of rigor. A high school student reading Venema’s text might ask why they should
not read [2] instead. In §10 we discuss Venema’s proof of the triangle sum theorem
(Lemma 10.1) and contrast it with the inadequate proof of [2]. There is a fairly
rigorous proof in a old high school Geometry text [1] Venema mentions, which
adopts the plane separation axiom (Prop. 4.14) missing from [2], correctly defines
the interior of a angle (Definition 3.3), but does not prove the Crossbar Theorem 3.8.
Hilbert’s axiom [12] are actually for 3-dimensional space, and in §8 we discuss a
minimal version of Hilbert’s entire set of axioms, and prove various 3-dimensional
results of Euclid [6], profiting from the rigorous solid geometry proofs in [1].
Greenberg’s text [7] seems too difficult for high school students, but it seems to
be the fundamental text on Hilbert’s work, and it was a big influence on our paper
(and [10, 13]) and and we cite his results freely. We reprove some of his results.
Greenberg’s survey [8] gives reasons to use Hilbert’s axioms instead of, following
Birkhoff, assuming R, and contains interesting mathematical logic.
In §2 we explain Hilbert’s simplest axioms, e.g. that two points determine a line,
and explain how we, using set theory, do not give strictly axiomatic proofs.
1
2
WILLIAM RICHTER
In §3 we give Hilbert’s axiom for betweenness of points and show that a line l
defines an equivalence relation ∼l . Our proof of the crossbar theorem (Thm. 3.8), a
tool that shows two lines intersect, is simpler than the proofs of Greenberg, Moise
and Venema [7, 15, 16], as we do not use their plane separation axiom (Prop. 4.14).
In §4 we show a line separates the plane into two half-planes ([7, 10, 15, 16]
take this as an axiom). First we give in Prop. 4.1 Wylie’s proof [17, §4] that
Greenberg’s strengthened version of Hilbert’s axiom II.5 follows from Hilbert’s (our
axiom B4). Next we reprove (Prop. 4.10) Moore’s result [9] that Hilbert’s axiom
II.4 is redundant, with essentially Moore’s proof. (Wylie [17, §6] gave an interesting
and different proof of Moore’s result.) These two results imply Prop. 4.14.
In §5, we begin proving the surprisingly difficult results (Prop. 7.9 and 10.2) that
a quadrilateral with opposite sides (or angles) congruent is a parallelogram.
In §6 we give Hilbert’s axioms for congruence of both segments and angles. We
deduce Greenberg’s angle addition result from an angle subtraction result and also
an angle ordering result which Greenberg deduces from the angle addition result.
The SAS axiom in [7, 10, 16] is stronger than Hilbert’s, and we give Hilbert’s proof
(Lemma 6.1) that the weaker SAS axiom C6 suffices.
Hartshorne [10] gives an excellent exposition of Euclid’s Elements [6] together
with the Hilbert fixes needed [10, Thm. 10.4]. We explain how his rigorization of
Euclid’s Prop. I.10 shows that Hilbert’s axiom II.2 follows from our weaker axioms.
However Hartshorne’s work contains gaps that should trouble a high school student.
In §7 we fill in needed details to his proofs of Euclid’s Propositions I.7, I.21 and
I.27, and work his overly difficult parallelogram exercise [10, Ex. 10.10]. In §7 we
adopt the Euclidean parallel postulate for the remainder of the paper.
Thanks to Bjørn Jahren, who recommended [7, 10], found [17] and discovered
the proof of Prop. 4.1 independently. Thanks to Benjamin Kordesh, the high school
Geometry student “test pilot” for the paper, Miguel Lerma, who found [9], Takuo
Matsuoka, and Stephen Wilson, who recommended [16], for helpful conversations.
2. Hilbert’s simplest axioms
Following Venema, we shall give a set-theoretic version of Hilbert’s axioms. We
are given a set which is called a plane. Elements of this set are called points. We are
given certain subsets of a plane called lines. There are axioms that a plane and its
subset lines must satisfy, explained here and also §3 and §6. In §8 we will consider
many different planes, but for now we discuss one plane, referred to as ‘the plane’.
A line l is then the set of all point P in the plane so that P ∈ l. Two lines l and
m are then equal if P ∈ l iff P ∈ m, for all points P . As is usual we will write ‘iff’
for ‘if and only if’. We have as usual the intersection l ∩ m of two lines l and m,
the set of points P in the plane such that P ∈ l and P ∈ m. We will call points
collinear if there is a line in the plane containing them. Hilbert’s simplest axioms
(I for incidence) are
I1. Given distinct points A and B in the plane, there is a unique line l in the
plane such that A ∈ l and B ∈ l.
I2. Given a line l in the plane, there exists two distinct points A and B in the
plane such that A ∈ l and B ∈ l.
I3. There are at least three non-collinear points in the plane.
We will call AB the unique line in the plane containing A and B. We will need
A MINIMAL VERSION OF HILBERT’S AXIOMS FOR PLANE GEOMETRY
3
Lemma 2.1. Given two distinct lines l and m, if X ∈ l ∩ m, then l ∩ m = {X}.
Proof. Suppose that A 6= X and A ∈ l ∩ m. By axiom I1, l = AB = m. This
contradicts our assumption, so there can not exist any such A.
We are not working in a purely axiomatic framework, where proofs are given
using only the axioms and logic. The plane is what logicians call a model of Hilbert’s
plane geometry axioms, and we use Hilbert’s axioms together with set theory to give
proofs in the model. The advantage of a model is that we can use our set theory
skills, and this simplifies proofs, and even the statement of axioms. In a purely
axiomatic framework, Lemma 2.1 would be written in this hard to read fashion.
• For all lines l and m with l 6= m, and for every point X, if X ∈ l and
X ∈ m, then for every point Y with Y 6= X, either Y 6∈ l or Y 6∈ m.
So it’s very convenient to use sets freely and not write purely axiomatic proofs.
Later we will define an angle as a set of two elements, which we could not define in
a purely axiomatic framework, and more circumlocution would be required.
So we will use set-theoretic constructions like l ∩ m, {X}, and complements. So
for any line l, the complement of l is the set of points X in the plane not on l.
Given two lines l and m, the complement l − m is the set of points P with P ∈ l
and P 6∈ m. Set theory allows us to make arguments like the following. Given two
sets L and M with L ⊂ M and M ⊂ L, it follows that L = M . By using set theory,
we make our proofs easier and practice an important mathematical skill.
There is a trade-off in our free use of set theory, though. Axiomatic geometry is
important is for artificial intelligence (AI), because a robot supplied with the Hilbert
axioms can prove everything about points, lines and planes that we can prove using
Hilbert’s axioms. This follows from what is called Goedel’s Completeness Theorem
in First Order Logic (FOL). But to achieve this this AI advantage, we would have
to phrase our proofs entirely in FOL, and not freely use set theory. It would be
a good future project to learn enough mathematical logic to understand Goedel’s
result, and how to rewrite our proofs in FOL, to better understand AI.
3. The betweenness relation ∗ and the crossbar theorem
We have an undefined betweenness relation ∗. For all points A, B and C we have
the statement A ∗ B ∗ C, which is supposed to capture the idea that B lies between
A and C on some line. The betweenness axioms are
B1. If A ∗ B ∗ C, then C ∗ B ∗ A, and A, B and C are distinct points on a line.
B2. Given two distinct points A and B, there exists a point C such that A∗B∗C.
B3. If A, B and C are distinct points on a line, then exactly one of the statements A ∗ B ∗ C, A ∗ C ∗ B and B ∗ A ∗ C is true.
B4. Let A, B and C be non-collinear points not on line l. If there exists D ∈ l
so that A ∗ D ∗ C, there exists an X ∈ l such that A ∗ X ∗ B or B ∗ X ∗ C.
Using ∗, we define the open interval (A, B) as the subset of the plane
(A, B) = {C | A ∗ C ∗ B} ⊂ AB.
Then A ∗ C ∗ B iff C ∈ (A, B). Note that (A, A) = ∅. All we will ever use of B1 is
B1’. For any points A and C, there is a subset (A, C) of the plane. If A 6= C,
then (A, C) ⊂ AC − {A, C} and (A, C) = (C, A). Furthermore (A, A) = ∅.
Using open intervals, we have equivalent versions of axioms B2 and B3.
4
WILLIAM RICHTER
B2’. Given distinct points A and B, there exists a point C such that C ∈ (A, B).
B3’. Given distinct collinear points A, B and C, exactly one of the statements
A ∈ (B, C), B ∈ (A, C) and C ∈ (A, B) is true.
Axiom B4 is Hilbert’s axiom II.5. Hilbert draws a picture of a triangle with
vertices A, B and C with a line l intersecting the open intervals (A, C) and (A, B).
Take a line l. We define the relation ∼l on the complement of l. Take two points
A, B 6∈ l. We write A ∼l B if (A, B) ∩ l = ∅. We write A l B if (A, B) ∩ l 6= ∅.
We have two equivalent versions of B4.
B4’. Take a line l and three non-collinear points A, B and C not on l. If A l C,
then A l B or B l C.
B4”. Take a line l and three non-collinear points A, B and C not on l. If A ∼l B
and B ∼l C, then A ∼l C.
In Proposition 4.14 we prove a stronger version of axiom B4” that Greenberg [7,
p. 64] essentially takes as an axiom. A simple result using the relation ∼l is
Lemma 3.1. Take a point on a line X ∈ m and a line l with m ∩ l = {X}. Take
points A, B ∈ m − {X}, and suppose that X 6∈ (A, B). Then A, B 6∈ l and A ∼l B.
Proof. A 6∈ l, since A 6= X and m ∩ l = {X}. Similarly B 6∈ l. Assume (for a
contradiction) that A l B. Then there exists a point G ∈ (A, B) ∩ l. By axiom
B1’, (A, B) ⊂ m. So G ∈ m ∩ l = {X}. Thus G = X. But this contradicts
X 6∈ (A, B). Hence A l B is false, so A ∼l B.
A relation is an equivalence relation if it is reflexive, symmetric and transitive.
Lemma 3.2. Take a line l. Then the following statements are true.
(1) The relation ∼l is an equivalence relation on the complement of l.
(2) Take points A, B and C not on l. If A ∼l B and B ∼l C, then A ∼l C.
Proof. ∼l is obviously reflexive and symmetric, as (A, A) = ∅ and (A, B) = (B, A).
Transitivity is the second assertion (2), which we now prove. So take points A, B
and C not on l with A ∼l B and B ∼l C. We must prove A ∼l C.
By axiom B4”, A ∼l C if A, B and C are not collinear. Since ∼l is reflexive,
A ∼l C if A, B and C are not distinct. So we can assume that A, B and C are
distinct collinear points.
Let m = AC. If the line m does not intersect l, then clearly A ∼l C. So assume
that l and m intersect. The lines l and m are distinct, because our three points
belong to m but not l. Therefore, m ∩ l = {X}, for some point X, by Lemma 2.1.
By axiom I2, there exists a point E 6= X on l. Thus E 6∈ m. By axiom B2’ there
exists a point B 0 with B ∈ (E, B 0 ). Let n = BB 0 . By Lemma 2.1, n ∩ l = {E},
because B 6∈ l. Therefore B 0 6∈ l, so B 0 is not on l. By axiom B3’, E 6∈ (B, B 0 ).
So B ∼l B 0 , by Lemma 3.1. The lines m and n are distinct, because E ∈ n. Thus
by Lemma 2.1, m ∩ n = {B}. Thus A 6∈ n, and so A, B and B 0 are not collinear.
Since A ∼l B and B ∼l B 0 , axiom B4” implies that A ∼l B 0 .
By an argument similar to the previous paragraph, C, B and B 0 are not collinear,
and C ∼l B 0 . Since ∼l is symmetric, B 0 ∼l C.
But A, B 0 and C are also not collinear, since m ∩ n = {B}. Thus by axiom B4”,
A ∼l C. This proves (2). Since this shows ∼l is transitive, this proves (1).
We use Moise’s definition of a ray [15, p. 55], which is simpler than Greenberg’s.
A MINIMAL VERSION OF HILBERT’S AXIOMS FOR PLANE GEOMETRY
5
−−→
Definition 3.3. For distinct points O and P , the ray OP is the subset of the plane
−−→
(1)
OP = {X ∈ OP | O 6∈ (X, P )}.
Let O, A and B be non-collinear points, and let a = OA and b = OB. Define
the angle ]BAC as the (unordered) set consisting of the two rays
−→ −−→
]AOB = {OA, OB}.
We define the interior of the angle ]AOB as the subset of the plane
int ]AOB = {P | P 6∈ a, P 6∈ b, P ∼a B and P ∼b A}.
We prove a result of Greenberg’s [7, Prop. 3.7] that is in a sense a converse to
the Crossbar Theorem 3.8 we prove below.
Lemma 3.4. Take non-collinear points O, A and B in the plane and a point
G ∈ (A, B). Then G ∈ int ]AOB.
Proof. Let a = OA, b = OB, and l = AB. Then a ∩ l = {A}, by Lemma 2.1,
since B 6∈ a. Hence G 6∈ a, as G ∈ l and G 6= A, by axiom B1’. A 6∈ (G, B), by
axiom B3’, since G ∈ (A, B). Thus G ∼a B, by Lemma 3.1, since a ∩ l = {A} and
A 6∈ (G, B). Similarly G 6∈ b and G ∼b A, as B 6∈ (G, A). Thus G ∈ int ]AOB. We prove Greenberg’s result [7, Prop. 3.8(a)].
−−→
Lemma 3.5. If X ∈ int ]AOB, then OX − {O} ⊂ int ]AOB.
Proof. Let a = OA, b = OX and x = OX. x ∩ a = {O}, by Lemma 2.1, since
X 6∈ a. Similarly x ∩ b = {O}. Since X ∈ int ]AOB, X ∼a B and X ∼b A. Take
−−→
P ∈ OX − {O}. Then P 6∈ a, since P 6= O and x ∩ a = {O}. Thus P ∼a X, by
Lemma 3.1, since O 6∈ (P, X). Similarly P ∼b X. By Lemma 3.2, P ∼a B and
P ∼b A. Thus P ∈ int ]AOB.
We prove an easy trichotomy law for interiors of angles, inspired by [7, Prop. 3.21].
Lemma 3.6. Take distinct points O and A. Let a = OA. Take points P, Q 6∈ a with
P ∼a Q and P , Q and O non-collinear. Then P ∈ int ]QOA or Q ∈ int ]P OA.
Proof. Let p = OP and q = OQ. Then q ∩ a = {O}, by Lemma 2.1, since Q 6∈ a.
Thus A 6∈ q, since A 6= O. Then p ∩ q = {O}, by Lemma 2.1, since P 6∈ q.
Assume P 6∈ int ]QOA. Then P q A, since P 6∈ q, P 6∈ a and P ∼a Q. Take a
point G ∈ (P, A) ∩ q. Then G ∈ int ]P OA by Lemma 3.4, so G ∼a P , G 6∈ a, and
G 6= O. Since P ∼a Q, Lemma 3.2 implies Q ∼a G. Thus O 6∈ (Q, G), and then
−−→
Q ∈ OG − {O}. Hence Q ∈ int ]P OA, by Lemma 3.5.
We give a simpler proof of Greenberg’s result [7, Prop. 3.8(c)].
Lemma 3.7. Take non-collinear points O, A and B, a point D ∈ int ]AOB and
a point A0 with O ∈ (A, A0 ). Then B ∈ int ]DOA0 .
Proof. Let a = OA and b = OB. By axiom I1, a = OA0 . Then a ∩ b = {O}, by
Lemma 2.1, since B 6∈ a. Then A0 6∈ b and A b A0 , since A0 ∈ a − {O}. Since
D ∈ int ]AOB, D 6∈ a, D 6∈ b, D ∼a B, and D ∼b A. Thus D b A0 , by Lemma 3.2
and a short proof by contradiction. Hence D 6∈ int ]BOA0 . Thus B ∈ int ]DOA0 ,
by Lemma 3.6, since B, D 6∈ a, D ∼a B, and D 6∈ b.
We give a simpler proof of Greenberg’s crossbar theorem [7, p. 69].
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WILLIAM RICHTER
Theorem 3.8. Given non-collinear points O, A and B, and a point D ∈ int ]AOB,
−−→
the ray OD intersects the interval (A, B).
Proof. Let a = OA, b = OB and l = OD. By axiom B2’ there is a point A0 ∈ a and
so that O ∈ (A, A0 ). Thus B ∈ int ]DOA0 , by Lemma 3.7, and A l A0 . Hence
B ∼l A0 . Thus B l A, by Lemma 3.2 and a short proof by contradiction. So there
exists a point G ∈ (A, B) ∩ l. Then G ∈ int ]AOB by Lemma 3.4, so G ∼a B. But
D ∼a B since D ∈ int ]AOB. Thus G ∼a D by Lemma 3.2. Thus O 6∈ (G, D), so
−−→
−−→
G ∈ OD. Thus G ∈ (A, B) ∩ OD, and we are done.
Our proof of the Crossbar Theorem 3.8 is simpler than the proofs of Greenberg [7,
p. 69], Venema [16, Thm. 5.7.15] and Moise [15, p. 69], which use Proposition 4.14,
but only really need Lemma 4.11, based on Lemmas 4.4–4.8 which order four points.
4. plane separation
Wylie’s argument [17, §4] proves what Greenberg calls Pasch’s theorem [7, p. 67],
which Greenberg deduces from a stronger axiom than axiom B4.
Proposition 4.1. For A, B and C non-collinear points in the complement of a
line l, at least one of the statements A ∼l B, A ∼l C and B ∼l C is true.
Proof. Suppose none of the statements are true, so A l B, A l C and B l C.
Then there are points X, Y, Z ∈ l so that X ∈ (A, B), Y ∈ (A, C) and Z ∈ (B, C).
By axiom B3, one of X, Y, Z ∈ l lies between the other two. We may assume that
Y ∈ (X, Z). The points B, X and Z are non-collinear because A, B and C are.
Let m = AC. By axiom B3, A 6∈ (B, X), and AB ∩ m = {A}, by Lemma 2.1, since
B 6∈ m. Thus X ∼m B by Lemma 3.1. Similarly B ∼m Z. Thus X ∼m Z by axiom
B4”. But Y ∈ (X, Z) ∩ m, so X m Z, and we have reached a contradiction.
Now we prove, in a series of exercises, the collinear case of Proposition 4.1.
We specialize Lemma 3.2 to a fixed line.
Lemma 4.2. Take a point on a line O ∈ m and points P, Q, R ∈ m − {O}. If
O 6∈ (P, Q) and O 6∈ (Q, R), then O 6∈ (P, R).
Proof. By axiom I3, there exists E 6∈ m. Let l = OE. Then m ∩ l = {O}, by
Lemma 2.1, because E 6∈ m. By Lemma 3.1 and the hypothesis, P ∼l Q and
Q ∼l R. Thus P ∼l R, by Lemma 3.2. Therefore O 6∈ (P, R), by Lemma 3.1.
−−→
We now show that the ray OP is in a sense independent of P .
−−→
−−→ −−→
Lemma 4.3. Take points O, P and Q 6= O, with P ∈ OQ − {O}. Then OP = OQ.
−−→ −−→
If R ∈ (O, Q), then OR = OQ.
−−→
Proof. Take X ∈ OP . Then O 6∈ (X, Q), by Lemma 4.2, since O 6∈ (X, P ) and
−−→
−−→ −−→
O 6∈ (P, Q). Hence X ∈ OQ. Thus OP ⊂ OQ. But O 6∈ (Q, P ), by axiom B1’, so
−−→
−−→ −−→
−−→ −−→
Q ∈ OP . The same argument shows that OQ ⊂ OP . Thus OP = OQ.
Now assume R ∈ (O, Q). Then O, R and Q are distinct collinear points, and
−−→
−−→ −−→
O 6∈ (R, Q), by axiom B3’. Thus R ∈ OQ − {O}. By the first part, OR = OQ. We will prove Hilbert’s axiom II.4 is redundant using only the following result.
A MINIMAL VERSION OF HILBERT’S AXIOMS FOR PLANE GEOMETRY
7
−→ −−→
Lemma 4.4. Given points A, O and B with O ∈ (A, B), we have OA∩ OB = {O}.
−−→
−→
If X ∈ OB − {O}, then X 6∈ OA.
−→ −−→
Proof. O ∈ OA ∩ OB. So the second statement implies the first statement. So take
−−→
X ∈ OB − {O}. A, O and B are distinct by B1’. Let m = AB. Then m = OB by
axiom I1. Then A, X, B ∈ m − {O}. Since O 6∈ (X, B) and O ∈ (A, B), we have
−→
O ∈ (A, X), by Lemma 4.2 and a simple proof by contradiction. Thus X 6∈ OA. Given distinct collinear points A, B, C and D, we will say that the 4-tuple
hA, B, C, Di is an ordered sequence if B ∈ (A, C), C ∈ (B, D), B ∈ (A, D) and
C ∈ (A, D). Note that by symmetry, if hA, B, C, Di is an ordered sequence, then
hD, A, B, Ci is also an ordered sequence. An an easy corollary of Lemma 4.4 is
Lemma 4.5. Let A, B, C and D be distinct points on a line, and suppose that
B ∈ (A, C) and C ∈ (B, D). Then hA, B, C, Di is an ordered sequence.
−−→
Proof. D 6∈ (B, C), by axiom B3’. So B ∈ (A, C) and D ∈ BC − {B}. Thus
−−→
D 6∈ BA, by Lemma 4.4. Thus B ∈ (D, A), so B ∈ (A, D) by B1’ symmetry. To
prove C ∈ (A, D), we apply B1’ symmetry to the result we have just proved. Since
C ∈ (D, B) and B ∈ (C, A), we have C ∈ (D, A), so C ∈ (A, D).
With Venema’s plane separation axiom [16, Axiom 5.5.2] in mind, we now prove
Lemma 4.6. Take points B ∈ (A, C). Then (A, B) ⊂ (A, C).
Proof. Take X ∈ (A, B). Then X 6= B, by B1’, and B 6∈ (X, A), by axiom B3’,
−−→
−−→
so X ∈ BA − {B}. Since B ∈ (C, A), by B1’ symmetry, X 6∈ BC, by Lemma 4.4.
Thus B ∈ (X, C). Thus hA, X, B, Ci is an ordered sequence, by Lemma 4.5, and
so X ∈ (A, C). Thus (A, B) ⊂ (A, C).
The proof of the above result proves also that
Lemma 4.7. Let A, X, B and C be distinct points on a line, and suppose that
X ∈ (A, B) and B ∈ (A, C). Then hA, X, B, Ci is an ordered sequence.
This implies
Lemma 4.8. Let A, B, C and X be distinct collinear points. Then at least one of
the statements X 6∈ (A, B), X 6∈ (B, C) and X 6∈ (A, C) is true.
Proof. By axiom B3’, one point is between the other two. So assume B ∈ (A, C).
If X 6∈ (A, B), we are done, so assume X ∈ (A, B). Then hA, X, B, Ci is an ordered
sequence by Lemma 4.7, so B ∈ (X, C). Thus X 6∈ (B, C), by axiom B3’.
Now we have the 4-tuple ordered sequence result we have been aiming for.
Lemma 4.9. Let A, B, C and X be distinct collinear points, with B ∈ (A, C).
(1) Either A ∈ (X, B) or X ∈ (A, B) or X ∈ (B, C) or C ∈ (B, X).
(2) One of the 4-tuples hX, A, B, Ci, hA, X, B, Ci, hA, B, X, Ci or hA, B, C, Xi
is an ordered sequence.
Proof. By axiom B3’, A ∈ (X, B) or X ∈ (A, B) or B ∈ (A, X). Assume B ∈
(A, X). Since B ∈ (A, C), we have B 6∈ (C, X) by Lemma 4.8. Then by axiom B3’,
X ∈ (B, C) or C ∈ (B, X). This proves (1).
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WILLIAM RICHTER
To prove (2), we consider four cases. If A ∈ (X, B), then hX, A, B, Ci is an ordered sequence by Lemma 4.5, since B ∈ (A, C). If X ∈ (A, B), then hA, X, B, Ci
is an ordered sequence by Lemma 4.7. If X ∈ (B, C), we have X ∈ (C, B) and
B ∈ (C, A). Thus hC, X, B, Ai is an ordered sequence by Lemma 4.7, and so by symmetry, hA, B, X, Ci is also an ordered sequence. If C ∈ (B, X), then hA, B, C, Xi
is an ordered sequence by Lemma 4.5.
Hilbert’s axiom II.4 is now an easy corollary of Lemma 4.9.
Proposition 4.10 (Moore). Four distinct collinear points A, B, C and D can be
renamed P1 , P2 , P3 and P4 so that hP1 , P2 , P3 , P4 i is an ordered sequence.
Proof. By axiom B3’, one of the points A, B and C must be between the other
two. Since we are allowed to rename the points, we can assume that B ∈ (A, C).
By Lemma 4.9, one of the 4-tuples hD, A, B, Ci, hA, D, B, Ci, hA, D, X, Ci or
hA, D, C, Xi is an ordered sequence. In all four case we can rename the points
P1 , P2 , P3 and P4 to obtain our ordered sequence hP1 , P2 , P3 , P4 i.
Lemma 4.8 also implies a result suggested by Lemma 4.4.
−→ −−→
Lemma 4.11. Take points O ∈ (A, B) on a line l. Then l = OA ∪ OB.
−−→
−→
Proof. Take X 6∈ OB. We must show that X ∈ OA. If X = A we are done, so
assume that X 6= A. Then X, A and B are distinct points in l. Since O ∈ (X, B)
−→
and O ∈ (A, B), Lemma 4.8 implies that O 6∈ (X, A). Thus X ∈ OA.
−−→
−−→
OB is called the opposite ray of OB in l.
The collinear case of Proposition 4.1 was essentially done in Lemma 4.8.
Lemma 4.12. Let A, B and C be collinear points in the complement in the plane
of a line l. Then A ∼l B, A ∼l C or B ∼l C.
Proof. If two of the points A, B and C are equal, we are done since the relation ∼l
is reflexive. So assume A, B and C are distinct. Let m = AC. If l ∩ m = ∅, then
we are done, as all three statements are true. So assume l ∩ m 6= ∅. Then l 6= m,
since A 6∈ l. Thus l ∩ m = {X}, for some point X, by Lemma 2.1. Now Lemma 3.1
translates Lemma 4.8 to our result.
Lemma 4.12 and Proposition 4.1 together immediately imply
Corollary 4.13. Let A, B and C be points in the complement in the plane of a
line l. Then at least one of the statements A ∼l B, A ∼l C and B ∼l C is true.
We call a set U convex if A, B ∈ U implies (A, B) ⊂ U .
Now we prove what Venema calls his plane separation postulate [16, p. 408].
Proposition 4.14. The complement in the plane of a line l is a disjoint union of
two nonempty convex sets H1 and H2 . If P ∈ H1 and Q ∈ H2 , then (P, Q) ∩ l 6= ∅.
Proof. By axiom I3, there is a point A 6∈ l. Take a point E ∈ l. Then AE ∩ l = {E}
by Lemma 2.1, since A 6∈ l. By axiom B2’, there exists a point B such that
E ∈ (A, B). Then B 6∈ l, since B 6= E. By definition A l B, since E ∈ (A, B) ∩ l.
Define H1 = {X 6∈ l | X ∼l A} and H2 = {X 6∈ l | X ∼l B}. H1 and H2 are
disjoint nonempty sets, since ∼l is an equivalence relation (Lemma 3.2).
A MINIMAL VERSION OF HILBERT’S AXIOMS FOR PLANE GEOMETRY
9
If C 6∈ l, then by Corollary 4.13, either A ∼l C or B ∼l C, as A l B. Thus the
complement of l is the union H1 ∪ H2 . If P ∈ H1 and Q ∈ H2 , then P l Q since
∼l is an equivalence relation. Thus (P, Q) ∩ l 6= ∅. Take P, Q ∈ H1 , so P ∼l A and
Q ∼l A. Since ∼l is an equivalence relation, P ∼l Q.
Let X ∈ (P, Q). Then (P, X) ⊂ (P, Q) by Lemma 4.6. So (P, X)∩l ⊂ (P, Q)∩l =
∅. Thus X ∼l P , and so X ∼l A, since ∼l is an equivalence relation. Thus X ∈ H1 .
Hence (P, Q) ⊂ H1 . Thus H1 is convex. Similarly H2 is convex.
The disjoint convex sets H1 = {X 6∈ l | X ∼l A} and H2 = {X 6∈ l | X ∼l B}
above are called the half-planes bounded by l. Since ∼l is an equivalence relation,
H1 and H2 do not depend on the choice of points A ∈ H1 and B ∈ H2 .
Take a line l = OA, and let H one of the two half-planes bounded by l. We
define the angle ]AOB to be on the half-plane H if B ∈ H. By Lemma 4.3, this
−−→
does not depend on the point B, but only the ray OB.
−−→
Take non-collinear points O, A and B and a point D. The ray OD is between
−→
−−→
OA and OB if D ∈ int ]AOB. By Lemma 4.3, the betweenness relation of rays
−→ −−→
−−→
does not depend on the points A, B and D, but only the rays OA, OB and OD.
A triangle 4ABC is defined as an ordered triple hA, B, Ci of three non-collinear
points. In §7 we need definitions [7, p. 69 & 77] of triangle interiors and angle order.
Definition 4.15. Given non-collinear points P , Q and R, let p = QR, q = P R,
and r = P Q. We define the interior of the triangle as the set of points in the plane
int 4P QR := {X | X 6∈ p, X 6∈ q, X 6∈ r, X ∼p P, X ∼q Q, and X ∼r R}.
We say ]ABC < ]P QR if there exists G ∈ int ]P QR so that ]ABC ∼
= ]P QG.
Greenberg [7, Prop. 3.21] proves a trichotomy law for this angle ordering relation
<, and shows < is preserved under congruence. We will cite Greenberg’s result,
but more often we will use Lemma 3.6 instead.
5. quadrilaterals
Moise has a nice treatment on quadrilaterals [15, §4.4] which we expand on.
With Proposition 7.5 in mind, we first generalize the notion of a quadrilateral.
We define a tetralateral to be an ordered 4-tuple ABCD of distinct points in
the plane, no three of which are collinear. Take a tetralateral ABCD. We define
(for this section) the lines a = AB, b = BC, c = CD, d = DA, l = AC, and
m = BD. Since no three points of the tetralateral ABCD are collinear, we have
C, D 6∈ a, A, D 6∈ b, A, B 6∈ c, B, C 6∈ d, B, D 6∈ l, and A, C 6∈ m. There are six
intersections of the four open intervals (A, B), (B, C), (C, D) and (D, A). Then
Lemma 5.1. For a tetralateral ABCD, we have four empty intersections:
(A, B) ∩ (B, C) = (B, C) ∩ (C, D) = (C, D) ∩ (D, A) = (D, A) ∩ (A, B) = ∅.
Proof. (A, B)∩(B, C) ⊂ a∩b = {B}, by Lemma 2.1, since A 6∈ b. Since B 6∈ (A, B),
by axiom B1’, (A, B) ∩ (B, C) = ∅. The other intersections are similarly empty. We define a quadrilateral ABCD [15, p. 69] to be a tetralateral ABCD where
(A, B)∩(C, D) = ∅ and (B, C)∩(A, D) = ∅. Given a quadrilateral ABCD, all six
intersections of the four open intervals (A, B), (B, C), (C, D) and (D, A) are empty,
by Lemma 5.1. A quadrilateral ABCD is convex [15, p. 69] if A ∈ int ]BCD, B ∈
int ]CDA, C ∈ int ]DAB, and D ∈ int ]ABC. The term convex quadrilateral
10
WILLIAM RICHTER
indicates that the “inside” of the quadrilateral is a convex set, but we will not pursue
this. Moise explains the term convex quadrilateral is inconsistent: the union of four
non-collinear line segments (or four points) cannot be a convex set.
We begin by proving an exercise of Moise [15, Ex. 4, §4.4].
Lemma 5.2. Take a quadrilateral ABCD. Then either A ∼c B or C ∼a D.
Proof. Assume that C a D. Then there exists a point G ∈ a ∩ (C, D). Then by
Lemma 2.1, a ∩ c = {G}, since A 6∈ c. But G 6∈ (A, B), since (A, B) ∩ (C, D) = ∅.
Furthermore A, B ∈ a − {G}, since A, B 6∈ c. Thus A ∼c B, by Lemma 3.1.
We generalize Moise’s [15, Thm. 1, §4.4] slightly, with the same proof.
Lemma 5.3. Take a tetralateral ABCD with B l D and A m C. Then there
is a point G ∈ (A, C) ∩ (B, D), and we have a convex quadrilateral ABCD.
Proof. Since A m C, there exists a point G ∈ (A, C) ∩ m. So G ∈ l ∩ m. Then
l ∩ m = {G}, by Lemma 2.1, since A 6∈ m. Thus B, D ∈ m − {G}, since B, D 6∈ l.
Hence G ∈ (B, D), since B l D, by Lemma 3.1, and a short proof by contradiction.
Thus G ∈ (A, C) ∩ (B, D).
−−→
Then B 6∈ (D, G), by axiom B3’, so D ∈ BG − {B}. Now G ∈ int ]ABC, by
Lemma 3.4. Thus D ∈ int ]ABC, by Lemma 3.5. A similar argument shows that
A ∈ int ]BCD, B ∈ int ]CDA, and C ∈ int ]DAB.
We prove Moise’s result on the diagonals of a convex quadrilateral.
Lemma 5.4. Take a convex quadrilateral ABCD. Then B l D and A m C.
There is a point G ∈ (A, C) ∩ (B, D), and ABDC is not a quadrilateral.
Proof. By the Crossbar Theorem 3.8, B l D, since A ∈ int ]BCD. Similarly
A m C. Thus there exists a point G ∈ (A, C) ∩ (B, D), by Lemma 5.3.
ABDC is not a quadrilateral, because (C, A) ∩ (B, D) 6= ∅.
We need this tetralateral result for Proposition 7.5.
Lemma 5.5. Take a tetralateral ABCD with C ∼a D. Then either ABCD
is a convex quadrilateral, ABDC is a convex quadrilateral, D ∈ int 4ABC, or
C ∈ int 4DAB.
Proof. Either C ∈ int ]DAB or D ∈ int ]CAB, by Lemma 3.6, since C, D 6∈ a,
C ∼a D, and C 6∈ d. We study these two cases.
Case 1: Assume that C ∈ int ]DAB. Then C ∼d B. Either D ∈ int ]CBA or
C ∈ int ]DBA, by Lemma 3.6, since D 6∈ b. We study these two cases.
Case 1.1: Assume D ∈ int ]CBA. Then D ∈ int ]ABC and A ∼b D, by
Lemma 3.2. By the Crossbar Theorem 3.8, C m A, so C 6∈ int ]BDA. Thus
B ∈ int ]CDA, by Lemma 3.6, since B, C 6∈ d, B ∼d C, and C 6∈ m. Thus
A ∼c B, by Lemma 3.2. Hence A ∈ int ]BCD, since A 6∈ b, and A 6∈ c. Thus
ABCD is a convex quadrilateral.
Case 1.2: Assume C ∈ int ]DBA. Then C ∈ int 4DAB, since C ∈ int ]DAB.
Thus in case 1, either ABCD is a convex quadrilateral or C ∈ int 4DAB.
Case 2: Assume that D ∈ int ]CAB. By the proof of case 1, either ABDC
is a convex quadrilateral or D ∈ int 4ABC. We can obtain the proof by switching
C and D is the above proof, because our lemma is symmetric in C and D.
We specialize Lemma 5.5 to quadrilaterals.
A MINIMAL VERSION OF HILBERT’S AXIOMS FOR PLANE GEOMETRY
11
Lemma 5.6. Take a quadrilateral ABCD. Then either ABCD is a convex
quadrilateral, A ∈ int 4BCD, B ∈ int 4CDA, C ∈ int 4DAB or D ∈ int 4ABC.
Furthermore either B l D or A m C.
Proof. Either A ∼c B or C ∼a D, by Lemma 5.2. We will apply to Lemma 5.5 to
the quadrilaterals ABCD and CDAB.
CDAB is a quadrilateral. ABDC is not a convex quadrilateral, by Lemma 5.4
and a short proof by contradiction, since ABCD is a quadrilateral. Similarly,
CDBA is a not a convex quadrilateral, since CDAB is a quadrilateral.
If C ∼a D, then either ABCD is a convex quadrilateral, D ∈ int 4ABC, or
C ∈ int 4DAB, by Lemma 5.5. Similarly, if A ∼c B, then either CDAB is a
convex quadrilateral, B ∈ int 4CDA, or A ∈ int 4BCD. But CDAB is convex
iff ABCD is convex. This proves our first assertion.
Thus either ABCD is a convex quadrilateral, A ∈ int ]BCD, B ∈ int ]ADC,
C ∈ int ]BAD, or D ∈ int ]ABC. For each of these five possibilities, either
B l D or A m C, or both, by Lemma 5.4 and the Crossbar Theorem 3.8.
6. The congruence relation ∼
=
For points A 6= B, the segment AB is defined as the subset of the plane
AB = {A, B} ∪ (A, B).
∼ on segments. If
Note that AB = BA. We have the undefined binary relation =
AB ∼
= CD, we say that the segment AB is congruent to the segment CD. Hilbert’s
axioms for congruence of segments are
C1. Given a segment AB and a ray r from C, there is a unique point D on r
such that CD ∼
= AB.
∼
C2. = is an equivalence relation on the set of segments.
C3. If A ∗ B ∗ C, A0 ∗ B 0 ∗ C 0 , AB ∼
= A0 C 0 .
= B 0 C 0 , then AC ∼
= A0 B 0 and BC ∼
We write AB < CD if there exists a point E ∈ (C, D) such that AB ∼
= CE.
Greenberg [7, Prop. 3.13] proves the segment order relation < satisfies a trichotomy
law, a transitive property and is well defined up to congruence of segments.
We have the undefined binary relation ∼
= on angles, and Hilbert’s axioms are
−−
→
C4. Given an angle ]AOB, a ray O0 A0 , and a half-plane H of the line O0 A0 ,
−−−→
there is a unique ray O0 B 0 on the half-plane H so that ]AOB ∼
= ]A0 O0 B 0 .
∼
C5. = is an equivalence relation on the set of angles.
C6. Take triangles 4ABC and 4A0 B 0 C 0 . If AB ∼
= A0 B 0 , AC ∼
= A0 C 0 and
0 0 0
0 0 0
∼
∼
]BAC = ]B A C , then ]ABC = ]A B C .
We write 4ABC ∼
= 4A0 B 0 C 0 , and say the two triangles are congruent, if the
corresponding segments and angles are congruent. Now we prove that the SAS
criterion implies congruent triangles, following Hilbert’s proof.
Theorem 6.1 (SAS). Take triangles 4ABC and 4A0 B 0 C 0 . If AB ∼
= A0 B 0 , AC ∼
=
0 0
0 0 0
0 0 0
∼
∼
A C and ]BAC = ]B A C , then 4ABC = 4A B C .
Proof. Let l = AB, m = BC and n = AC. Since A, B and C are non-collinear
points, l ∩ m = {B} and m ∩ n = {C}, by Lemma 2.1. By axiom C6 we know
−−→
that ]ABC ∼
= ]A0 B 0 C 0 . By axiom C1, there is a unique point D ∈ BC so
that BD ∼
= B 0 C 0 . Then ]ABD = ]ABC, so by the reflexive property of angle
congruence (axiom C5), ]ABD ∼
= ]ABC. So ]ABD ∼
= ]A0 B 0 C 0 , by axiom C5.
12
WILLIAM RICHTER
So by axiom C6, ]BAD ∼
= ]B 0 A0 C 0 . By hypothesis and axiom C5 symmetry
∼
and transitivity, ]BAD = ]BAC. But B 6∈ (C, D), by definition of a ray, since
−−→
−→
−−→
D ∈ BC ⊂ m. Therefore C ∼l D, so the rays AC and AD are on the same
−→ −−→
half-plane of l. Since ]BAD ∼
= ]BAC, axiom C4 implies that AC = AD. Then
D ∈ n ∩ m. Thus C = D. We now have five of the six congruences we need, and
axiom C6 implies the last congruence ]ACB ∼
= ]A0 C 0 B 0 .
Greenberg uses Lemma 6.1 instead of axiom C6. However, his proof [7, Prop. 3.17]
of the ASA theorem is similar to Hilbert’s proof of the SAS theorem. His proof [7,
Prop. 3.19] of the angle addition result seems difficult, and does not follow Hilbert’s
treatment. Trying to follow Hilbert, we instead prove an angle subtraction result.
Lemma 6.2. Take triangles 4AOB and 4A0 O0 B 0 , and points G ∈ int ]AOB and
G0 ∈ int ]A0 O0 B 0 . Suppose we have the angle congruences ]AOB ∼
= ]A0 O0 B 0 and
0 0 0
0 0 0
∼
∼
]AOG = ]A O G . Then ]BOG = ]B O G .
Proof. We may replace the points A, B, G, A0 , B 0 , and G0 by other points on their
rays. By axiom C1, we may assume that OA ∼
= O0 A0 and OB ∼
= O0 B 0 . By the
0
Crossbar Theorem 3.8, we may assume that G ∈ (A, B) and G ∈ (A0 , B 0 ). By the
SAS Theorem 6.1, 4AOB ∼
= 4A0 O0 B 0 , so AB ∼
= A0 B 0 , ]OAB ∼
= ]O0 A0 B 0 and
0
0
0
0
0
0
]OBA ∼
= ]O B A . By ASA [7, Prop. 3.19], 4AOG ∼
= 4A O G , so AG ∼
= A0 G0 .
0
0
Then BG ∼
= B G by the segment subtraction result [7, Prop. 3.11]. By SAS,
4BOG ∼
= ]B 0 O0 G0 .
= 4B 0 O0 G0 . Therefore we have ]BOG ∼
We state and prove the angle analogue of a result of Greenberg [7, Prop. 3.11],
as his exercise proof of his angle ordering result [7, Prop. 3.21] seems to require it.
Lemma 6.3. Take triangles 4AOB and 4A0 O0 B 0 with ]AOB ∼
= ]A0 O0 B 0 and a
0
0 0 0
point G ∈ int ]AOB. There exists G ∈ int ]A O B such that ]AOG ∼
= ]A0 O0 G0 .
Proof. We sketch the proof, as it is similar to that of Lemma 6.2. We may assume
that OA ∼
= A0 B 0 , ]OAB ∼
= ]O0 A0 B 0 and G ∈ (A, B). By
= O0 B 0 , AB ∼
= O0 A0 , OB ∼
a result of Greenberg [7, Prop. 3.11], there exists a unique point G0 ∈ (A0 , B 0 ) such
that AG ∼
= A0 G0 . Then G0 ∈ int ]A0 O0 B 0 . By the SAS Theorem 6.1, 4OAG ∼
=
0 0 0
4O A G .
We deduce the angle addition result from Lemma 6.2 and Lemma 6.3.
Lemma 6.4. Take triangles 4AOB and 4A0 O0 B 0 , and points G ∈ int ]AOB and
G0 ∈ int ]A0 O0 B 0 . Suppose we have the angle congruences ]AOG ∼
= ]A0 O0 G0 and
0 0 0
0 0 0
∼
∼
]BOG = ]B O G . Then ]AOB = ]A O B .
Proof. Let l = O0 A0 . Then G0 ∼l B 0 . By axiom C4, there exists a unique ray
−−0→
O X so that X ∼l G0 and ]AOB ∼
= ]A0 O0 X. By Lemma 6.3, there exists a
0 0
point Y ∈ int ]A O X so that ]AOG ∼
= ]A0 O0 Y . Then Y ∼l X. So Y ∼l G0 by
−−0−→0 −−0→
Lemma 3.2. By axioms C4 and C5, O G = O Y . So we can assume that Y = G0 .
Hence G0 ∈ int ]A0 O0 X. By Lemma 6.2, ]BOG ∼
= ]XO0 G0 .
Let m = O0 G0 . By the crossbar theorem, the intervals (A0 , X) and (A0 , B 0 ) both
−−−→
intersect the ray O0 G0 . So A0 m X and A0 m B 0 . Thus X ∼m B 0 by Lemma 4.12.
−−→
−−−→
Thus O0 X and O0 B 0 are on the same half-plane of m, and ]XO0 G0 ∼
= ]B 0 O0 G0 , by
−−0→ −−0−→0
0 0
∼
axiom C5. Thus O X = O B by axiom C4. Hence ]AOB = ]A O B.
A MINIMAL VERSION OF HILBERT’S AXIOMS FOR PLANE GEOMETRY
13
Given an angle ]AOB and a point A0 with O ∈ (A, A0 ), the angle ]BOA0 is
called the supplement of ]AOB, and we say ]AOB and ]BOA0 are supplementary
angles. We say an angle is a right angle if it is congruent to its supplement. Then
Lemma 6.5. Supplements of congruent angles are congruent. An angle congruent
to a right angle is a right angle. All right angles are congruent. Take a line h = OH
and two right angles ]AOH and ]HOB, with A h B. Then O ∈ (A, B).
Proof. Greenberg proves the first statement [7, Prop. 3.14]. This implies the second
statement, by axiom C5. The third statement, due to Hilbert, is [7, Prop. 3.23].
Let v = OA. Then h ∩ v = {O}, by Lemma 2.1, since A 6∈ h. By axiom B2’,
there exists a point X ∈ v with O ∈ (A, X). Then X 6∈ h, since X 6= O, and
A h X. Thus X ∼h B, by Corollary 4.13, since B 6∈ h.
]HOX ∼
= ]AOH, since ]AOH is a right angle with supplement ]HOX. By
−−→ −−→
axiom C5, ]HOX ∼
= ]HOB, since all right angles are congruent. Thus OX = OB,
by axiom C4, since X ∼h B. Then B ∈ v. Hence O ∈ (A, B), by Lemma 3.1 and a
short proof by contradiction, since A, B ∈ v − {O}.
7. Euclid, Hilbert and the parallel postulate
Hartshorne [10] does an excellent job explaining how Hilbert’s work rigorizes
book I of Euclid’s Elements [6]. Here we fill some gaps in Hartshorne’s proofs.
Hartshorne gives the mild Hilbert rigorization [10, p. 97] needed for Euclid’s
Prop. I.5 (an isosceles triangle has congruent base angles). A proof requiring no
Hilbert rigorization [15, §6.2, Thm. 1] is due to Pappus [11, p. 254] in 300 C.E.
Lemma 7.1. Take a triangle with 4BAC with AB ∼
= AC. Then ]CBA ∼
= ]ACB.
Proof. 4BAC ∼
= 4CAB, by SAS, since ]BAC ∼
= ]BAC, by axiom C5.
Greenberg [7, Prop. 3.22] and Moise [15, §6.2, Thm. 3] give the same Hilbert
rigorization of Euclid’s Prop. I.8, the SSS Theorem about congruent triangles.
Theorem 7.2. Take triangles 4ABC and 4A0 B 0 C 0 with AB ∼
= A0 B 0 , AC ∼
= A0 C 0 ,
0 0
0 0 0
∼
∼
and BC = B C . Then 4ABC = 4A B C .
Hartshorne’s proof [10, Prop. 10.1] of the SSS Theorem 7.2 is perhaps too sketchy.
Hartshorne’s Hilbert rigorization of Euclid’s Prop. I.9 [10, p. 100], the existence
of angle bisectors, has a gap which we fill in the second paragraph of our proof
Lemma 7.3. Given an angle ]BAC, there exists a point F ∈ int ]BAC so that
]BAF ∼
= ]F AC.
−−→
Proof. By axiom B2’, there exists a point D with B ∈ (A, D), so D ∈ AB − {A}.
−→
−−→ −−→
By axiom C1, there exists a point E ∈ AC − {A} with AE ∼
= AD. Then AD = AB
−→ −→
and AE = AC, by Lemma 4.3. Thus ]BAC = ]DAE. Then ]EDA ∼
= ]AED,
by Lemma 7.1. Let d = AD, e = AE, and h = DE. Then d ∩ e = {A}, by
Lemma 2.1, since C ∈ e = AC, and C 6∈ d = AB. Thus E 6∈ d, since E 6= A. Hence
d ∩ h = {D}, by Lemma 2.1, so A 6∈ h, since A 6= D. By axioms C4 and C1, there
exists a point F so that ]F DE ∼
= ]EDA, F 6∈ h, F h A, and DF ∼
= DA. By the
∼
SAS Theorem 6.1, 4DF E = 4DAE, since DE ∼
= DE by axiom C2, so F E ∼
= AD
and ]DEF ∼
= ]AED. Let v = AF . We must show D, E 6∈ v and D v E.
Assume (for a contradiction) that D ∈ v. Then v∩h = {D}, by Lemma 2.1, since
A 6= D. Then D ∈ (A, F ), by Lemma 3.1 and a short proof by contradiction, since
14
WILLIAM RICHTER
A, F ∈ v − {D}. Thus ]EDA is a right angle, as it is congruent to its supplement
]F DE. By Lemma 6.5, ]AED and ]DEF are right angles and E ∈ (A, F ). Thus
−→
D = E, by axiom C1, since AE ∼
= AD and D, E ∈ AF , since A 6∈ (D, F ) and
A 6∈ (E, F ), by axiom B3’. This is a contradiction, since D 6= E. Thus D 6∈ v. By
a similar proof by contradiction, E 6∈ v. Thus D, E 6∈ v.
By the SSS Theorem 7.2, 4F AD ∼
= 4F AE. Thus ]DAF ∼
= ]F AE. Since
D, E 6∈ v, either D ∼v E or D v E. Assume (for a contradiction) that D ∼v E.
Then D = E by axioms C4 and C1, since ]DAF ∼
= ]EAF . This is a contradiction,
so D v E. Thus there exists a point G ∈ (D, E) ∩ v. Then G ∈ int ]DAE, by
Lemma 3.4, and h ∩ v = {G}, by Lemma 2.1, since A 6∈ h. Thus G ∈ (A, F ), by
Lemma 3.1 and a short proof by contradiction, since A, F ∈ v − {G}, and F h A.
−→
By axiom B3’, F 6∈ (A, G), so F ∈ AG − {A}. Thus F ∈ int ]DAE, by Lemma 3.5.
Thus, F ∈ int ]BAC and ]BAF ∼
= ]F AC.
Take distinct points B and C. Hartshorne constructs [10, Prop. 10.2] an isosceles
triangle 4BAC, and then proves the existence of midpoints by using the Crossbar
Theorem 3.8, Lemma 7.3 and SAS to find the midpoint of (A, B). This, together
with axiom B2’, proves Hilbert’s axiom II.2, which is stronger than our axiom B2:
II.2. Given two distinct points A and C, there exists points B and D such that
B ∈ (A, C) and C ∈ (A, D).
Greenberg, who takes II.2 as an axiom himself, gives a much simpler but less
appealing proof [7, p. 86, Ex. 6] of II.2 using only axiom B2’.
Hartshorne gives a nice Hilbert rigorization of Euclid’s Prop. I.16 [10, Prop. 10.3]
Theorem 7.4. Take a triangle 4ABC with a point D such that C ∈ (B, D). Then
]CAB < ]ACD.
This result (see also [16, Thm. 6.3.2]) is called the Exterior Angle Theorem,
because ]ACD, the supplement of ]BCA, is called an exterior angle of 4ABC.
Hartshorne’s Hilbert rigorization of Euclid’s Prop. I.7 [10, p. 35 & Ex. 9.4] is
insufficient, so we reprove Prop. I.7, adding details to Hartshorne’s proof. Afterward
we will give a much simpler proof of the result.
Proposition 7.5. Take distinct points A and B in the plane, and let a = AB.
Take distinct points C and D in the plane with C, D 6∈ a and C ∼a D, and suppose
that AC ∼
6 BD.
= AD. Then BC ∼
=
Proof. 4CAD is isosceles, so ]CDA ∼
= ]ACD, by Lemma 7.1. Define the lines
b = BC, c = CD, d = AD, l = AC and m = BD. By Lemma 2.1, l ∩ a = {A},
since C 6∈ a. Hence B 6∈ l. Similarly B 6∈ d, A 6∈ m, and A 6∈ b.
Assume (for a contradiction) that C ∈ d. Then A 6∈ (D, C), since C ∼a D,
−−→
so C ∈ AD. Then C = D, by axiom C1, since AC ∼
= AD. But C 6= D, so we
have a contradiction. Hence C 6∈ d. We similarly prove that D 6∈ l. Since C 6∈ d,
Lemma 2.1 implies c ∩ d = {D}. Hence A 6∈ c, since A 6= D, since D 6∈ a.
If D ∈ b or C ∈ m, then a similar argument shows that BC ∼
6 BD, and we
=
are done. Hence we may assume that D 6∈ b and C 6∈ m. Then b ∩ c = {C}, by
Lemma 2.1, since D 6∈ b. Thus B 6∈ c, since B 6= C, since C 6∈ a.
Thus A, B, C and D are distinct points, and C, D 6∈ a, A, D 6∈ b, A, B 6∈ c,
B, C 6∈ d, B, D 6∈ l, and A, C 6∈ m. Thus ABCD is a tetralateral with C ∼a D.
By Lemma 5.5, either ABCD is a convex quadrilateral, C ∈ int 4DAB,
ABDC is a convex quadrilateral, or D ∈ int 4ABC. We study the four cases.
A MINIMAL VERSION OF HILBERT’S AXIOMS FOR PLANE GEOMETRY
15
Case 1: Assume ABCD is a convex quadrilateral. Then B ∈ int ]CDA and
A ∈ int ]BCD. By Definition 4.15, ]CDB < ]CDA and ]ACD < ]BCD, and
]CDA ∼
6 ]BCD, since < is
= ]ACD. Thus ]CDB < ]BCD, and so ]CDB ∼
=
transitive, preserves congruence and has a trichotomy property [7, Prop. 3.21]. By
a short proof by contradiction using Lemma 7.1, so BD ∼
6 BC.
=
Case 2: Assume C ∈ int 4DAB. By axiom B2’ there exists a points E ∈ d so
that D ∈ (A, E). By axiom I1, d = DE. Since C 6∈ d and D 6∈ b, the angles ]CDE
and ]BCD exist. B ∈ int ]CDE, by Lemma 3.7, since C ∈ int ]ADB. By the
−→
Crossbar Theorem 3.8, there exists a point F ∈ (B, D) ∩ AC, since C ∈ int ]BAD.
Thus A 6∈ (F, C), and F ∈ int ]BCD, by Lemma 3.4. A, C, F ∈ l are distinct
points, because ABDC is a tetralateral. F 6∈ (A, C), since C ∼m A, since
C ∈ int ]ADB. Thus C ∈ (A, F ) by axiom B3’. Supplements of congruent angles
are congruent [7, Prop. 3.14], so ]CDE ∼
= ]F CD, since ]CDA ∼
= ]ACD.
Thus ]CDB < ]CDE ∼
= ]F CD < ]BCD, since B ∈ int ]CDE and F ∈
int ]BCD. The argument of case 1 shows that BC 6∼
= BD.
We have cases 3 and 4, where ABDC is a convex quadrilateral or D ∈
int 4ABC. In both cases we prove that BC 6∼
= BD by the arguments of cases
1 and 2. This works because our proposition is symmetric in C and D.
Remark 7.6. Hartshorne explains Euclid’s proof of Prop. I.7 has a gap which
must be filled by showing that A ∈ int ]DCB and B ∈ int ]CDA in case 1,
but he ignores the other 3 cases. Fitzpatrick’s literal translation of the Elements
(http://farside.ph.utexas.edu/euclid.html) gives only Hartshorne’s one case. The
version of the Elements we use [6] gives our cases 1 and 2. In what is considered
to be the definitive version of the Elements, Heath explains that [11, p. 259–260]
a literal translation of Euclid’s statement of Prop. I.7 is “apparently obscure” and
“vague”1, that this second case of [6] is due to the 5th Century mathematician
Proclus, and that Euclid had “the general practice of giving only one case, and
that the most difficult, and leaving the others to be worked out by the reader.” For
Prop. I.7 this seems unfortunate, especially because of extra Hilbert theory needed.
Now we give a simple proof of Euclid’s Prop. I.7 which we think is well known.
Proof of Proposition 7.5. Assume (for a contradiction) that BC ∼
= BD. We have
triangles 4ACB and 4ADB because C, D 6∈ a. Then 4ACB ∼
= 4ADB, by the
SSS Theorem 7.2, since AB ∼
= AB, by axiom C2. Hence ]BAC ∼
= ]BAD. By
−→ −−→
axiom C4, AC = AD, since C ∼b D. Then C = D, by axiom C1, since AC ∼
= AD.
Since C and D are distinct, we have a contradiction. Hence BC 6∼
BD.
=
We discuss Hartshorne’s treatment of circles [10, §11]. Given a point O and a
segment AB, let Γ be the circle with center O and radius AB defined as
Γ = {C | OC ∼
= AB}.
A point X is inside Γ if OX < OA and outside Γ if OA < OX.
As Hartshorne explains, Euclid’s Prop. I.7 says that two circles can not have
two intersection points in the same half-plane of the line between their centers.
1Fitzpatrick translation is “On the same straight-line, two other straight-lines equal, respectively, to two (given) straight-lines (which meet) cannot be constructed (meeting) at a different
point on the same side (of the straight-line), but having the same ends as the given straight-lines.”
16
WILLIAM RICHTER
We work Hartshorne’s circle convexity exercise [10, Ex. 11.1(a)], which Greenberg
proves in an exercise [7, Ex. 4.29]. Our proof is similar but slightly simpler.
Lemma 7.7. Let Γ be a circle points with center O and radius OR. Assume the
points A and B are inside Γ. Then the open interval (A, B) lies inside Γ.
Proof. We have OA < OR and OB < OR. Take X ∈ (A, B). We must show that
OX < OR. As in [10, Prop. 11.2], we use Theorem 7.4 and Euclid’s Proposition
I.19 (which relates side-lengths of a triangle to their opposite angles), and we also
need I.18 [6, p. 16–22]. We have two cases depending on whether OA ∼
= OB (the
easier case) or not.
Assume that OA ∼
6 OB. We may assume that OA < OB. Applying I.18
=
to the triangle 4OAB yields ]OBA < ]OAB. Applying I.16 to 4OAX yields
]OAB < ]OXB. Then ]OBA < ]OXB by the transitive property of angle
ordering [7, Prop. 3.21]. Applying I.19 to the triangle 4OXB yields OX < OB.
Then OX < OR by the transitive property of segment ordering [7, Prop. 3.13].
Assume OA ∼
= OB. Then 4AOB is isosceles, so ]OBA = ]OAB, by Lemma 7.1.
The proof is now similar to the case above, but we do not need I.18.
We say lines l and m in a plane are parallel, and write l k m, if l ∩ m = ∅. The
Exterior Angle Theorem 7.4 allows us to construct parallel lines.
Take lines l and m, distinct points A, C ∈ l − m and distinct points B, E ∈ m − l.
The line x = AB is called a transversal which cuts l and m. Suppose also that
E x C. Then ]CAB and ]EBA are called alternate interior angles [16, p. 107].
Venema gives a nice proof of the Alternate Interior Angles Theorem [16, Thm. 6.8.1]
Lemma 7.8. If ]CAB ∼
= ]EBA, then l k m.
Proof. Assume (for a contradiction) that l ∦ m, so there exists a point G ∈ l ∩ m.
Then G 6= A and G 6= B. By Lemma 2.1, x ∩ l = {A} and x ∩ m = {B}, since
B 6∈ l and A 6∈ m. Thus C 6∈ x, G 6∈ x, and E 6∈ x.
−→ −→
Assume (for a contradiction) that G ∼x C. Then A 6∈ (G, C). Thus AG = AC,
−→
by Lemma 4.3, since G ∈ AC − {A}. Hence ]GAB = ]CAB, so ]GAB ∼
= ]EBA.
By Lemma 3.2 and a short proof by contradiction, G x E, since C ∼x G and
C x E. Thus B ∈ (E, G), by Lemma 3.1 and a short proof by contradiction, since
x∩m = {B} and E, G ∈ m−{B}. Hence ]EBA is the supplement of ]ABG and so
an exterior angle of triangle 4GAB. Thus ]GAB < ]EBA, by the Exterior Angle
Theorem 7.4. Thus ]GAB ∼
6 ]EBA, by Greenberg’s [7, Prop. 3.21] trichotomy
=
law for <. This is a contradiction. Hence G x C.
Then G ∼x E, by Corollary 4.13, since G x C and C x E. By axiom C5,
]EBA ∼
= ]CAB. We obtain a contradiction by the argument above, switching the
pairs l and m, A and B, and E and C. This switch works because of the symmetry
in our lemma. Hence l k m.
As Hartshorne points out [10, Thm. 10.24], the result is Euclid’s Prop. I.27.
Hartshorne however fails to point out that the proof requires Hilbert betweenness
fixing, and that Euclid’s Prop. I.27 is vaguely stated, as Euclid lacks the betweenness
axioms needed to define alternate interior angles. As Hartshorne points out [10,
Ex. 10.10], Lemma 7.8 implies, without using the parallel postulate,
Proposition 7.9. A quadrilateral ABCD with AB ∼
= CD and BC ∼
= AD is a
parallelogram.
A MINIMAL VERSION OF HILBERT’S AXIOMS FOR PLANE GEOMETRY
17
Proof. By Lemma 5.6, either B x D or A y C. We may assume that B x D.
So when we cut the lines l and r by the transversal x, ]CAB and ]ACD are
alternate interior angles. Also cut the lines t and b by the transversal x. Then
B x D again implies that ]CAD and ]ACB are alternate interior angles.
By SSS, we have congruent triangles 4CAB ∼
= 4ACD so ]CAB ∼
= ]ACD and
∼
]CAD = ]ACB. Thus AB k CD and AD k BC by Lemma 7.8.
A typical reader of Hartshorne’s could not be expected to work this exercise [10,
Ex. 10.10], as Hartshorne does not mention the betweenness issues we dealt with,
and merely says “Join the midpoints of AB and CD, then use (I.27).”
We prove part of Euclid’s Prop. I.21, as Hartshorne [10, Thm. 10.4] fails to
explain that Euclid’s proof requires Hilbert betweenness fixing.
Lemma 7.10. Take a triangle 4ABC with a point D ∈ int 4ABC. Then ]BAC <
]BDC.
Proof. Let m = AC. Then D ∈ int ]ABC and D ∼m B, since D ∈ int 4ABC.
−−→
There exists a point E ∈ (A, C) ∩ BD, by the Crossbar Theorem 3.8. Apply the
Exterior Angle Theorem 7.4 to 4BAE. Then ]BAE < ]BEC, since E ∈ (A, C).
−→ −→
By Lemma 4.3, AC = AE, so ]BAC = ]BAE. Thus ]BAC < ]BEC.
−−→
B, D and E are distinct, and B 6∈ (E, D), since E ∈ BD. But E 6∈ (B, D), since
B ∼m D and (A, C) ⊂ m. Thus D ∈ (E, B), by axiom B3’. Apply the Exterior
Angle Theorem 7.4 to 4DEC. Then ]DEC < ]BDC.
−−→
−−→
EB = ED, by Lemma 4.3, since D ∈ (E, B). Thus ]BEC = ]DEC. Hence
]BAC < ]BDC, by Greenberg’s [7, Prop. 3.21] transitivity law for <.
For the rest of the paper we adopt Venema’s version of Euclid’s parallel postulate.
P. Given a line l in the plane and a point P 6∈ l, there exists a unique line m
in the plane so that P ∈ m and m k l.
We prove a version of the triangle sum theorem (Lemma 10.1).
Lemma 7.11. Take a triangle 4ABC, with lines l = AC, x = AB and y = BC.
There exists a line m 3 B and points E, F ∈ m − {B} with m k l, B ∈ (E, F ),
E x C, F y A, C ∈ int ]ABF , ]EBA ∼
= ]CAB and ]CBF ∼
= ]BCA.
−−→
Proof. By axiom C4, there exists a unique ray BE with E x C and ]EBA ∼
=
]CAB. Let m = BE. Then ]EBA and ]CAB are alternate interior angles, so
l k m by Lemma 7.8. By axiom B2’, there exists F ∈ m with B ∈ (E, F ). Then
E x F and E y F , since B ∈ (E, F ). Then C ∼x F , by Corollary 4.13, since
E x C. C ∼m A, since (C, A) ∩ m ⊂ l ∩ m = ∅, since l k m. Thus C ∈ int ]ABF .
−−→
−−→
The Crossbar Theorem 3.8 implies BC ∩ (F, A) 6= ∅. Then F y A since BC ⊂ y.
−−→
By axiom C4, there exists a unique ray BF 0 with F 0 y A and ]CBF 0 ∼
= ]BCA.
Let m0 = BF 0 . Then ]CBF 0 and ]BCA are alternate interior angles, so l k m0 by
Lemma 7.8. Since F y A and F 0 y A, we have F 0 ∼y F , by Corollary 4.13.
Axiom P implies m0 = m, because B ∈ m, l k m and l k m0 . Then B 6∈ (F, F 0 ),
−−→
−−→
by Lemma 3.1, since F, F 0 ∈ m − {B} and F 0 ∼y F . Thus BF 0 = BF 0 , and so
]CBF = ]CBF 0 . Thus ]CBF ∼
= ]BCA, by Lemma 3.2.
We used axiom P only at the end, to essentially prove the converse of the Alternate Interior Angles theorem [16, Thm. 6.8.1]. The SMSG proof of the triangle
18
WILLIAM RICHTER
sum theorem [1, Thm. 9-13] is fairly rigorous, showing that C ∈ int ]ABF , but
not showing ]EBA ∼
= ]CAB and ]CBF ∼
= ]BCA are alternate interior angle
pairs. The SMSG proof [1, Thm. 9-5] of Lemma 7.8 is also reasonably rigorous,
considering one of the two cases, and ignoring betweenness issues.
8. 3-dimensional geometry
Hartshorne constructs the 5 Platonic solids and the 13 Archimedean solids [10,
Ch. 8], but does not “take the time to set up axiomatic foundations for this solid
geometry,” and recommends Euclid’s book XI [6]. We explain Euclid’s work with
Hilbert’s 3-dimensional axioms [16, App. B1]. We give the SMSG [1, Thm. 8.2]
proof of Euclid’s Proposition XI.4, which improves on Euclid’s proof by not using
the converse of the Pythagorean theorem. Following the SMSG insight, we similarly
improve Euclid’s proofs of his Propositions XI.6 and XI.8.
We explain Hilbert’s 3-dimensional axioms [16, App. B1]. We are given a set of
points called Space. We are given subsets of Space called lines and planes. Points
in the set Space are called coplanar if there is some plane in Space containing them.
All but two of our axioms for the set Space together with its subsets are
J1. Given distinct points A and B in Space, there is a unique line l such that
A ∈ l and B ∈ l.
J2. Given a line l in Space, there exists two distinct points A and B in the
plane such that A ∈ l and B ∈ l.
I4. If the points A, B and C are not contained in some line, there exists a
unique plane α containing A, B and C.
I5. Take a plane α and distinct points A, B ∈ α, and let l be the unique line
containing A and B. Then l ⊂ α.
I6. If two distinct planes intersect, their intersection is a line.
I7. There exists four distinct non-coplanar points that do not lie on a line.
Axiom I5 requires axiom J1. A plane α in Space is a set of points, as it is a subset
of Space, and α contains lines, the lines in Space that are subsets of α. By axioms
J1, J2 and I5, α satisfies axioms I1 and I2. We can now use the notation AB
unambiguously to mean the line guaranteed by axiom J1 or axiom I1. We assume
every plane α has a betweenness relation ∗. We assume the axiom (M for model)
M1. Every plane α in Space satisfies axioms I3, B1–4 and P.
The results of §3 and §4 now hold in the plane α, and define segments, rays, angles
and half-planes in α. We assume our other model axiom
M2. Every plane α in Space satisfies axioms C1–6.
Thus every plane α in Space satisfies axioms I1–3, B1–4, C1–6 and P.
Our axioms differ from Hilbert’s, because we did not want to now change our
earlier axioms, and claim, “an easy modification of our proofs show our earlier
results hold in any plane in Space.” The wording of most of our earlier axioms
would remain the same, but the meanings would change, as now the points would
be in Space, rather than in the plane. Our axiom B4 would be now changed to
• Take a plane α, a line l ⊂ α, and non-collinear points A, B, C ∈ α − l.
If there exists D ∈ l so that A ∗ D ∗ C, there exists an X ∈ l such that
A ∗ X ∗ B or B ∗ X ∗ C.
A MINIMAL VERSION OF HILBERT’S AXIOMS FOR PLANE GEOMETRY
19
To avoid this trouble, we do not now modify our earlier axioms. Another option
we did not take is correctly stating Hilbert’s axioms from the outset. Our approach
seems suited for learning plane geometry first and then 3-dimensional geometry.
Suppose distinct lines l and m intersect. By axioms I4 and I5, they determine a
plane. We say l and m are perpendicular, and write l ⊥ m, if they meet at a right
angle in this plane. Take a plane α, a line l and a point O with α ∩ l = {O}. We
say l is perpendicular to α, and write l ⊥ α, if l is perpendicular to every line m
such that O ∈ m ⊂ α. We prove Euclid’s [6, Prop. XI.4], following [1, Thm. 8.2].
Lemma 8.1 (Euclid). Take a plane α, lines p, q, n and a point O such that p ⊂ α,
q ⊂ α, p ∩ q = {O} and n ∩ α = {O}. If n ⊥ p and n ⊥ q, then n ⊥ α.
Proof. Take a line x ⊂ α through O, with x 6= p and x 6= q. Take X ∈ x − {O} so
that x = OX. Then X 6∈ p and X 6∈ q. Choose points P, P 0 ∈ p so that O ∈ (P, P 0 ).
Since P q P 0 , then X q P or X q P 0 , by Lemma 3.2. We may assume that
X q P , switching P and P 0 if necessary. Thus there is a point Q ∈ (P, X) ∩ q.
Take points N, N 0 ∈ n so that O ∈ (N, N 0 ) and N O ∼
= N 0 O. By axiom C2,
OP ∼
= OP and OQ ∼
= OQ. We have right angles ]N OP and ]N OQ, since n ⊥ p
and n ⊥ q. Thus ]N OP ∼
= ]N 0 OP and ]N OQ ∼
= ]N 0 OQ. Therefore 4N OP ∼
=
0
0
N OP and 4N OQ ∼
= N 0 P and
= N OQ by the SAS Theorem 6.1. Hence N P ∼
NQ ∼
= N 0 Q. Thus 4N P Q ∼
= 4N 0 P Q, by SSS [7, Prop. 3.22], so ]N P Q ∼
=
−−→
−−→
0
0
∼
]N P Q. Hence ]N P X = ]N P X, since P Q = P X, since P 6∈ (Q, X). Hence
4N P X ∼
= 4N 0 P X by SAS, so N X ∼
= N X 0 . Thus 4N OX ∼
= 4N 0 OX by SSS, so
0
∼
]N OX = ]N OX. So ]N OX is a right angle, and n ⊥ x. Therefore n ⊥ α. We prove [6, Prop. XI.8] without the converse of the Pythagorean theorem.
Lemma 8.2 (Euclid). Take planes α and β and lines l, m ⊂ β such that l k m and
l ⊥ α. Then m ⊥ α.
Proof. Since l ⊥ α, α and β are distinct planes, so they intersect in a line z = α ∩ β,
by axiom I5. Take A ∈ l ∩ α. Then z ∩ l = {A}, by axiom I1, since l ⊥ z. By axiom
P, m ∩ z 6= ∅, or else lines z and l are distinct lines through A parallel to m. Then
m ⊥ z, by the converse to the Alternate Interior Angles Theorem [16, Thm. 6.8.1],
since l k m. Take B ∈ m∩z. Then z = AB. Choose C ∈ m−{B}, and then C 0 ∈ m
with B ∈ (C, C 0 ) and BC ∼
= BC 0 . Then 4ABC and 4ABC 0 are congruent right
triangles by the SAS Theorem 6.1, so CA ∼
= C 0 A. Choose X ∈ α such that ]XAB
is a right angle, and let n = AX. Then n ⊥ z, and n ⊥ l, since l ⊥ α. Thus n ⊥ β,
by Lemma 8.1, since z ∩ l = {A}. Thus ]CAX and ]C 0 AX are right angles. Thus
4CAX ∼
= 4C 0 AX by SAS, and so CX ∼
= C 0 X. Hence 4CBX ∼
= 4C 0 BX by SSS,
0
so the supplementary angles ]CBX and ]C BX are congruent. Hence ]CBX is
a right angle. Let p = BX, so m ⊥ p. Then p ∩ z = {B}, by Lemma 2.1, since
X 6∈ z. Then m ⊥ α, by Lemma 8.1.
Using Lemma 8.2, Euclid proves [6, Prop. XI.11] that given a plane α and a
point A 6∈ α, there exists a line l 3 A such that l ⊥ α. Using his Proposition XI.11,
Euclid proves [6, Prop. XI.12] that given a point B ∈ α, there exists a line m 3 B
such that m ⊥ α. Euclid’s simple proof by contradiction proves [6, Prop. XI.13]
that given B ∈ α, there exists at most one line m 3 B such that m ⊥ α. We now
remove the converse of the Pythagorean theorem from the proof of a third result
of Euclid [6, Prop. XI.6], using Lemma 8.2 and ideas of its proof.
20
WILLIAM RICHTER
Lemma 8.3 (Euclid). Take a plane α and distinct lines l and m with l ⊥ α and
m ⊥ α. Then l k m.
Proof. There exists a points A, B ∈ α such that l ∩ α = {A} and m ∩ α = {B}. Let
z = AB. By axiom I4 there is a unique plane β containing the lines l and z. There
exists a line n ⊂ β such that B ∈ n and n ⊥ z. Then l k n by the Alternate Interior
Angles Theorem [16, Thm. 6.8.1]. Then n ⊥ α by Lemma 8.2. Then m = n, by
Euclid’s Proposition XI.13, since B ∈ n. Therefore l k m.
9. Birkhoff’s alternative to Hilbert’s work
Venema uses Birkhoff’s [4] ruler and protractor postulates [16, Axioms 5.4.1 &
5.6.2], which measure segments and angles with the real line R. Venema’s book is
primarily a text on Hilbert’s axioms, where Birkhoff’s two real line axioms relieve us
of the labor [10, §4] of using Hilbert’s axioms to construct the real line. We explain
how Birkhoff’s ingenious ideas give a rigorous treatment of geometry different from
Hilbert’s, following MacLane [14] improvements on Birkhoff’s work.
We use MacLane’s [14] set R/360 of equivalence classes [x] of points x ∈ R, under
the equivalence relation x ∼ x + k360 for any integer k.
Our first four Birkhoff-MacLane type postulates are
BM1. There is a unique line containing any two distinct points A and B.
BM2. For any line l, there is a 1-1 correspondence from R to l. If x, y ∈ R map to
points P, Q ∈ l, then the distance from P to Q is d(P Q) := |x − y|. Given
segments P Q and AB, we say that P Q ∼
= AB iff d(P Q) = d(AB) ∈ R.
−−→
BM3. There is a 1-1 correspondence from R/360 to the set of rays OP starting at
−→ −−→
O. If α, β ∈ R/360 map to OA, OB, then ]AOB := α − β ∈ R/360. Two
angles are said to be congruent if they are equal in R/360.
−→
−−→
BM4. ]AOB = [180] iff OA and OB are opposite rays on a line.
Birkhoff defines betweenness using the ruler postulate: B is between A and C iff
A, B and C are distinct collinear points and d(AC) = d(AB) + d(BC). Given two
points A and B, we have the segment AB and the open interval (A, B) as before.
Birkhoff’s angles satisfy ]AOB = −]BOA, and are called directed angles.
Two triangles 4ABC and 4A0 B 0 C 0 are similar if there is a positive real number k and number = ±1 such that ]ABC = ]A0 B 0 C 0 , ]BCA = ]B 0 C 0 A0 ,
]CAB = ]C 0 A0 B 0 , kd(AB) = d(A0 B 0 ), kd(BC) = d(B 0 C 0 ), kd(CA) − d(C 0 A0 ).
The triangles are congruent if this holds with k = 1. MacLane’s SAS postulate is
BM5. The triangles 4ABC and 4A0 B 0 C 0 are similar if ]ABC = ]A0 B 0 C 0 ,
d(AB) = kd(A0 B 0 ) and d(BC) = kd(B 0 C 0 ), for = ±1 and k ∈ R positive.
For any α ∈ R/360 we have α = [x] for some x ∈ [−180, 180], and x will be
unique unless α = [180], in which case x must be either 180 or -180. So for any
α ∈ R/360, we have α = [x] where exactly one of the four statements is true:
x = 0, x = 180, x ∈ (0, 180), and x ∈ (−180, 0). By axiom BM3, ]AOB = [0]
−→ −−→
iff OA = OB. Thus by axiom BM4, if A, O and B are non-collinear points, then
]AOB = [x] where exactly one of the statements x ∈ (0, 180) and x ∈ (−180, 0) is
true. MacLane calls ]AOB a proper angle if x ∈ (0, 180), and an improper angle if
x ∈ (−180, 0), and orders the proper angles by the order of (0, 180). We will abuse
notation by replacing [x] by its preferred representative x ∈ (−180, 180].
A MINIMAL VERSION OF HILBERT’S AXIOMS FOR PLANE GEOMETRY
21
Birkhoff’s protractor postulate is stronger than our axiom BM3, and MacLane
defines a separate axiom which amounts to the crossbar theorem and its converse.
−−→
BM6. Let ]AOB be proper. If 0 < ]AOC < ]AOB, then OC ∩ (A, B) 6= ∅.
Conversely, if D is between A and B, then 0 < ]AOD < ]AOB.
Birkhoff’s directed angles allow a short proof of the angle addition theorem [4,
−→ −−→
−−→
p. 332]. Given rays OA, OB and OC, axiom BM3 implies ]AOB + ]BOC =
]AOC, regardless of betweenness of rays. Notice that the proof Lemma 6.4 is
much more difficult. Birkhoff [4, Thm. 5] also has no difficulty summing the angles
of a triangle. He uses axiom BM5 and a proof of existence of a centroid (divide
4ABC into four congruent triangles similar to 4ABC with scale factor 12 ) to show
(2)
]ABC + ]BCA + ]CAB = 180.
But now we pay for using directed angles, having to work quite hard to prove
Birkhoff’s result [4, 3, Thm. 5], whose proof MacLane improves on [14, Thm. 6]
Lemma 9.1. Given a triangle 4ABC, either all three angles are proper, or else
all three angles are improper.
Proof. Assume that 4ABC has both a proper and an improper angle. We will
obtain a contradiction. We may assume that ]ABC is proper and that ]CAB is
improper. By (2) we know that ]ABC + ]CAB 6= 0, for otherwise ]BCA = 180,
but this violates axiom BM4, as A, B and C are non-collinear. We may assume
that ]ABC + ]CAB ∈ (0, 180), so that 0 < −]CAB < ]ABC. By axiom BM6,
there exists a point X ∈ (A, C) so that ]ABX = −]CAB. Applying (2) to the
triangle 4ABX, we have ]ABX +]BXA+]XAB = 180. But ]XAB = ]CAB,
so ]BXA = 180. As A, B and X are also non-collinear, this violates axiom BM4.
Thus 4ABC can not have both a proper and an improper angle.
Birkhoff shows that any model of his axioms is essentially the Cartesian (x, y)
plane R2 . Hartshorne [10, §21] also proves that any model of his collection of
Hilbert’s axioms is R2 , and uses Hilbert theory to construct a version of R geometrically.
Birkhoff’s protractor postulate has an additional requirement that a function to
be continuous. Unfortunately continuous functions taking values in R/360 are only
treated in a Point-set Topology course, long after Calculus. Birkhoff only uses his
continuity requirement in his proof of Lemma 9.1, which contained a mistake, and
his erratum [3] is also very difficult to read. MacLane used his axiom BM6 to show
that a function is continuous, and used this to give a clean version of Birkhoff’s
proof. Our proof of Lemma 9.1 involves no mention of continuous functions.
10. Venema’s axioms and US Geometry textbooks
We now adopt Venema’s axioms [16, App. C]: axiom BM2, his Ruler Postulate;
his SAS axiom (our SAS Theorem 6.1); his Protractor Postulate, this version of
axiom BM3.
−→
−−→
V. For three distinct points A, O and B with OA and OB not opposite rays,
there is a real number µ(]AOB), called the measure of ]AOB, such that
the following conditions are satisfied.
(a) 0◦ ≤ µ(]AOB) < 180◦ .
−→ −−→
(b) µ(]AOB) = 0 iff OA = OB.
22
WILLIAM RICHTER
(c) For each real number r ∈ [0, 180), and for each half-plane bounded by
−−→
OA there exists a unique ray OE such that E ∈ H and µ(]AOB) = r◦ .
(d) If D ∈ int ]AOB then µ(]AOB) = µ(]AOD) + µ(]DOB).
Venema calls (d) the Angle Addition Postulate. Venema does not define degree, as
in r◦ , and degrees are only used in this paper to synchronize with [2]. It would be
better to always write r instead of r◦ . If one wants to use degrees, one needs to
remember that one degree is the real number π/180, so that r◦ := rπ/180.
It’s an exercise of Venema that his axioms imply our axioms I1–3, B1–4, C1–6,
where two angles are called congruent if their measures are equal, and betweenness
is following Birkhoff’s definition. So all our earlier results still hold.
Venema partly leaves this well-known result as an exercise [16, Thm. 7.2.3].
Lemma 10.1. The sum of the measures of a triangle 4ABC is 180◦ .
Proof. Let l = AC, x = AB and y = BC, and let m be the unique (by axiom P) line
such that B ∈ m and m k l. By Lemma 7.11, there exists points E, F ∈ m − {B}
with B ∈ (E, F ), E x C, F y A and C ∈ int ]ABF . Let ]1 = ]EBA,
]2 = ]ABC and ]3 = ]CBF . By Lemma 7.11, ]1 ∼
= ]CAB and ]3 ∼
= ]BCA.
Thus µ]1 = µ]CAB, and µ]2+µ]3 = µ]F BA, by the Angle Addition Postulate.
Since B ∈ (E, F ), ]F BA and ]1 are supplementary angles, and so form a linear
pair. By Venema’s Linear Pair Theorem [16, Thm. 5.7.18], µ]1 + µ]F BA = 180◦ .
Hence µ]1 + µ]2 + µ]3 = 180◦ . Thus µ]CAB + µ]ABC + µ]BCA = 180◦ . The UCSMP high school Geometry text [2] makes an effort at mathematical
rigor, but fails to prove the triangle sum theorem, and two parallelogram results.
Possibly Birkhoff played a role in the lack of rigor in US high school Geometry
texts, as his book for high school teachers [5] minimizes (as MacLane points out)
his two concepts of directed angles and the the continuity requirement of his Protractor Postulate, and does not prove the triangle sum theorem. Although perhaps
Birkhoff’s book is not to blame, given the relatively high Hilbert rigor of SMSG [1],
and Moise’s out of print high school geometry text with Downs, which based on [1],
gives a nice proof of the Crossbar Theorem, although not the triangle sum theorem.
We contend there is little point in an axiomatic Geometry course unless all results
are proved rigorously using only the axioms, logic and set theory.
All widely used US Geometry texts seem to follow Birkhoff by assuming that the
real line measures distances and angles. For UCSMP these are postulates Number
Line Assumption, Distance Postulate and Angle Measure Postulate. UCSMP does
not however have a postulate like Venema’s [16, p. 408] Plane Separation Postulate,
our Proposition 4.14.
UCSMP [2, p. 116] gives an imprecise definition of the interior of a triangle
• Every non-zero angle separates the plane into two nonempty sets other than
the angle itself. If the angle is not a straight angle, exactly one of those
sets is convex. This convex set is called the interior of the angle. The
nonconvex set is called the exterior of the angle. For a straight angle, both
sets are convex and either set may be considered the interior.
This definition has no precise meaning. The last sentence about a straight angle
amounts to Venema’s Plane Separation Postulate, and there is no such UCSMP
postulate. UCSMP defines convex [2, p. 62], but there is no justification for the last
sentence of the above definition, and no proof that a non-straight angle separates
the plane into two sets, one convex and the other nonconvex.
A MINIMAL VERSION OF HILBERT’S AXIOMS FOR PLANE GEOMETRY
23
UCSMP [2, p. 126 ] gives a correct definition of adjacent angles, modulo their
imprecise definition of the interior of a triangle.
• Two nonstraight angles are adjacent if a common side is in the interior of
the angle formed by the noncommon sides.
UCSMP [2, p. 127] correctly states their Angle Addition Assumption postulate,
modulo the new adjacent angle imprecision.
• If ]AV C and ]CV B are adjacent, then µ]AV C + µ]CV B = µ]AV B.
UCSMP [2, p. 271] gives an imprecise definition of alternate interior angles.
• We say that ]4 and ]6 are alternate interior angles because they are interior
angles on opposite sides of the transversal and they have different vertices.
Opposite side here has no precise meaning since UCSMP has no plane separation
postulate. Thus their Corresponding Angles Postulate [2, p. 146] is also imprecise.
Otherwise UCSMP has an adequate proof [2, p. 271] that if two parallel lines are
cut by a transversal, then the alternate interior angles are congruent.
Let’s turn to the UCSMP [2, p. 285 ] proof of Lemma 10.1. It’s same as the proof
we gave, except for the missing Plane Separation Postulate, Crossbar Theorem,
and rigorous definitions of angle interiors and alternate interior angles. Therefore
UCSMP does not have a rigorous proof of the triangle sum theorem Lemma 10.1.
UCSMP also falsely claims two parallelogram results are trivial: Proposition 7.9
(a quadrilateral with congruent opposite sides is a parallelogram), and [2, p. 429]
Proposition 10.2. A quadrilateral ABCD with ]ABC ∼
=
= ]CDA and ]DAB ∼
]BCD is a parallelogram.
Proof. By Lemma 5.6, either ABCD is a convex quadrilateral, A ∈ int 4BCD,
B ∈ int 4ACD, C ∈ int 4ABD or D ∈ int 4ABC.
Assume A ∈ int 4BCD. By Lemma 7.10, ]BCD < ]DAB. Thus ]BCD ∼
6
=
]DAB, by Greenberg’s [7, Prop. 3.21] trichotomy law for <. This is a contradiction.
Hence A 6∈ int 4BCD. The other three cases are similar. B ∈ int 4ACD implies
]CDA < ]ABC, C ∈ int 4ABD implies ]DAB < ]BCD, and D ∈ int 4ABC
implies ]ABC < ]CDA. We then have contradictions in all three cases, so we
conclude that B 6∈ int 4ACD, C 6∈ int 4ABD, and D 6∈ int 4ABC.
Hence ABCD is a convex quadrilateral. Divide the quadrilateral into triangles
4ABC and 4CDA. By Lemma 10.1, the sum of all six angles of both triangles
is 360◦ . By axiom V(d), ]BCA + µ]ACD = µ]BCD, since A ∈ int ]BCD,
and µ]DAC + ]CAB = µ]DAB, since C ∈ int ]DAB. Thus the sum of the
four angles of ABCD is 360◦ . But by hypothesis, µ]BCD = µ]DAB, and
µ]CDA = µ]ABC. Dividing by 2 yields µ]ABC + µ]BCD = 180◦ . Thus
]ABC and ]BCD are supplementary angles. Furthermore they are alternate
interior angles, since ABCD is a convex quadrilateral. Therefore by Lemma 7.8,
ABCD is a parallelogram.
The UCSMP proof attempt amounts to the last paragraph above, using their
quadrilateral-sum theorem [2, p. 289] that the sum of angles in a convex quadrilateral is 360◦ . It seems clear that UCSMP made the unwarranted assumption that
the quadrilateral was convex in both Propositions 7.9 and 10.2.
24
WILLIAM RICHTER
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