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CHAPTER 4
Parallel Lines in Euclidean Geometry
The most important of Euclid’s postulates to the development of geometry is
Euclid’s Fifth Postulate. We’ll talk about this for awhile. Before we do that, I’d
like you to look at the bottom of page 187 in Euclid’s Elements, and find the word
isosceles. Next to it is a Greek word: ισoσκλης. Here’s another one of those ancient
Greek words that we still use.
1. Euclid’s Fifth Postulate
Euclid’s Fifth Postulate states that if α + β < 180◦ in Figure 1, then there is a
point C to the right (on the same side of t as α and β) that lies on both l and m.
t
m
B
β β
0
l
α0 α
C
A
Figure 1. The lines l and m are cut by a transversal t. The angles α,
β, α0 , and β 0 are interior angles. A, B, and C are points of intersection.
This postulate is sometimes called the Parallel Postulate, even though it really
is a non-parallel postulate. Keep in mind that our definition of parallel is the same
as Euclid’s. Two lines in a plane are parallel, if they do not intersect.
Again, Euclid’s Fifth Postulate is actually an axiom of non-parallelism. As we
will see, we don’t need an extra axiom or postulate to conclude that α + β = 180◦
72
2. HILBERT’S AXIOMS
73
implies that l and m are parallel. For some reason, Euclid felt the need to include
this axiom of non-parallelism, and this is, I believe, the single most important act in
geometry. Great effort was expended trying to prove that the Fifth Postulate could
be proved from more basic assumptions. All of these efforts failed for good reason. A
proof is impossible.
1.1. Exercises.
–1– On pages 204-219 of Euclid’s Elements, there are notes on the Fifth Postulate. Skim these pages, and write down the names of all the mathematicians
mentioned. Indicate with a check mark the ones you’ve heard of before.
2. Hilbert’s Axioms
Interpreted as an axiom system, Euclid’s Elements uses postulates/axioms that
were perhaps too obvious to state explicitly, and so he didn’t mention them. He also
overlooked axioms that he used without realizing that he did. Hilbert essentially
took Euclid’s postulates, started calling them axioms, and threw in Euclid’s phantom
axioms. He had to add a lot of axioms (I counted twenty in [Hilbert]).
Among Hilbert’s axioms is his Axiom IV (Euclid’s Axiom) [Hilbert, p 25]
Let a be any line and A a point not on it. Then there is at most one
line in the plane, determined by a and A, that passes through A and
does not intersect a.
This is pretty much the same as Euclid’s Fifth Postulate, which states that for all
cases except when α + β = 180◦ the lines are not parallel. In other words, there can
be at most one parallel line. Both Euclid and Hilbert, who follows Euclid’s logic for
the most part, use an axiom to specify when lines are not parallel. They don’t need
an axiom to ensure the existence of parallels, since this can be proved from the other
axioms (but they do need an axiom to make sure there aren’t too many). Let’s look
at the existence of parallels now.
3. IF α + β = 180◦ , THEN l AND m ARE PARALLEL
74
3. If α + β = 180◦ , then l and m are parallel
One of the new axioms in Hilbert’s system is Axiom I,2 on page 3, which states
that at most one line may pass through two points. Axioms I,1 and I,2 together
ensure that two distinct points determine a line uniquely. OK. Very basic.
Hilbert’s Axiom III,5 on page 12 leads quickly to Theorem 12 on page 14, which
you should recognize as the SAS (side-angle-side) criterion. SAS says that if you have
two triangles, and two pairs of corresponding sides are congruent, and in addition,
the included pair of angles are congruent, then all parts of the triangle are congruent.
t
m
l
C0
B
β0 β
α0 α
A
C
Figure 2. We assume here that α + β = 180◦ .
We want to prove that in the configuration of Figure 1, if α + β = 180◦ , then l
and m are parallel. This means that l and m will never intersect.
We will assume the opposite. Suppose that α +β = 180◦ and l and m do intersect.
Therefore, there is a point of intersection C. Let’s assume that C lies to the right.
Consider a new point C 0 that lies on m to the left so that the distance from B to C 0
is the same as the distance from A to C.
We have that AC = BC 0 by definition. Segments AB = BA, because they’re the
same segment. Since
(150)
α + β = 180◦
We also have that
(151)
β + β 0 = 180◦
4. NEUTRAL GEOMETRY
75
Subtracting equations gives us
(152)
α − β 0 = 0◦ ,
and
(153)
α = β 0.
By SAS, the triangle 4ABC is congruent to 4BAC 0. Therefore, β = α0 . It follows
that α + α0 = 180◦ . The line AC 0 must be the same as l. Now, l and m have two
points in common. If B ∈
/ l, which is what we’re generally assuming, then we’ve got
a problem. Our original assumption that l and m intersect is wrong, and therefore, l
and m must in fact be parallel.
Note that the number 180◦ comes from how we measure angles in degrees. No
matter what units of measure we use for angles, the relevant thing is that the number
that corresponds to two right angles (or one straight angle) goes here. We will see
that the statement “the angle sum of a triangle is the same as the measure of a
straight angle” is equivalent to the Fifth Postulate and Euclid’s Axiom.
3.1. Exercises.
–1– Did we use the Fifth Postulate or Euclid’s Axiom in this proof?
–2– Did we use the fact that the angle sum of a triangle is 180◦ ?
–3– Did we use the fact that the measure of a straight angle is 180◦ ?
4. Neutral geometry
You may have heard of Fermat’s Last Theorem (I say “fur MAH” or “fair
MAH”). It was one of the big unsolved problems in mathematics until fairly recently,
and these big problems drive a lot of the development in mathematics. Although
mathematics was different then, we might say that the big problem between Euclid’s
time and the time of Gauss, Bolyai, and Lobachevski (early 1800’s) was the ProveEuclid’s-Fifth-Postulate Problem. Gauss, Bolyai, and Lobachevski finally realized
that you can’t prove Euclid’s Fifth Postulate, and that’s when things started to get
interesting.
4. NEUTRAL GEOMETRY
76
I want to continue exploration of this Prove-Euclid’s-Fifth-Postulate Problem for
awhile. For convenience, let’s talk about things within Hilbert’s axiom system, even
though this came much later. Hilbert’s axiom system has an axiom called Euclid’s
Axiom, as we’ve seen. Euclid’s Axiom (or Hilbert’s Parallel Axiom) is equivalent
to Euclid’s Fifth Postulate, so if you’re talking about one, you’re also talking about
the other. If you take Hilbert’s axiom system and throw out Euclid’s Axiom, we have
an axiom system for something called neutral geometry. A traditional modern
geometry course might spend two-thirds of a semester on neutral geometry. We
won’t, because I don’t think it’s that important. It is a little bit important, however.
The Prove-Euclid’s-Fifth-Postulate Problem, therefore, can be stated this way:
Prove Euclid’s Axiom as a theorem in neutral geometry.
4.1. Exercises.
–1– State Euclid’s Axiom.
–2– Is it actually possible to prove Euclid’s Axiom as a theorem in neutral geometry?
–3– We will eventually find that the statement “the angle sum of a triangle is
180◦ ” is equivalent to Euclid’s Axiom. Based on what I’ve said so far, is it
possible to prove this statement as a theorem in neutral geometry?
–4– What is neutral geometry?
If Hilbert’s axiom system is an axiom system for Euclidean geometry, and we could
prove Euclid’s Axiom as a theorem in neutral geometry, then neutral geometry and
Euclidean geometry would be the same. We know that Euclidean geometry satisfies
all the axioms of neutral geometry, so if they’re not the same, then there must be
at least one other geometry that also satisfies the axioms of neutral geometry. What
Gauss, Bolyai, and Lobachevsky did was find another geometry. That geometry is
usually called hyperbolic geometry, and today, we’re still talking about it as the
other geometry. That’s not good. There are other geometries. On the other hand,
hyperbolic geometry is an important geometry. We’ll talk about that later.
In the previous section, we showed the following is a theorem in neutral geometry.
If in Figure 3 ∠2 + ∠4 = 180◦ , then l is parallel to m.
4. NEUTRAL GEOMETRY
77
Figure 3. The lines l and m are cut by a transversal t.
Since all the axioms of neutral geometry are also axioms in Hilbert’s Euclidean geometry, we must be able to prove this statement in Euclidean geometry, as well.
Therefore, this last statement is a theorem of Euclidean geometry.
The contrapositive of a statement is equivlalent logically, so the following is also
a theorem in neutral geometry (and therefore also in Euclidean geometry).
If in Figure 3 l is not parallel to m, then ∠2 + ∠4 6= 180◦ .
Neither of these statements is equivalent to Euclid’s Fifth Postulate, which
can be phrased as follows.
If in Figure 3 ∠2 + ∠4 6= 180◦ , then l is not parallel to m.
We have not established this as a theorem in neutral geometry, and we will see that
this is impossible. This statement, therefore, is only true in Euclidean geometry, and
not in neutral geometry. Note that the contrapositive of this statement is also only
true in Euclidean geometry.
If in Figure 3 l is parallel to m, then ∠2 + ∠4 = 180◦ .
4.2. Exercises.
–1– In Figure 3, suppose ∠2 = 150◦ and ∠4 = 27◦ . In Euclidean geometry, can
you determine whether l is parallel to m? How about in neutral geometry?
5. THE ALTERNATE INTERIOR ANGLES THEOREM
78
–2– In Figure 3, suppose ∠2 = 100◦ and ∠4 = 80◦ . In Euclidean geometry, can
you determine whether l is parallel to m? How about in neutral geometry?
–3– In Figure 3, suppose ∠1 = 100◦ and l is parallel to m. In Euclidean geometry,
can you determine the measure of ∠3? How about in neutral geometry?
5. The Alternate Interior Angles Theorem
Given two lines l and m cut by a transversal t, we showed earlier that if the
interior angles on one side of the transversal sum exactly to 180◦ , then l and m are
parallel. We did not use Euclid’s Axiom (or any of the other equivalent parallel
axioms). Therefore, this would be a theorem of neutral geometry.
5.1. Exercises.
–1– Look at Figure 3. Suppose ∠1 ∼
= ∠4. You may assume that a straight angle
◦
measures 180 . What is ∠1 + ∠2?
–2– What is ∠2 + ∠4?
–3– In neutral geometry, what can you conclude about l and m? How do you
know?
You have proved a version of the interior angles theorem that is true in neutral
geometry.
The Alternate Interior Angles Theorem. (neutral geometry) Given two
lines l and m cut by a transversal t and alternate interior angles are equal, then l and
m are parallel.
In neutral geometry, whenever alternate interior angles are congruent (equal), we
know that l and m are parallel. Since we are not assuming Euclid’s Axiom, it’s
conceivable that we could tilt m a little bit, and m would still be parallel to l, but
the alternate interior angles would no longer be congruent after tilting m. In neutral
geometry, therefore, we cannot conclude that parallel lines imply congruent alternate
interior angles. This sort of dead end is a recurring outcome in the history of geometry
leading up to the discovery of non-Euclidean geometry.
6. ANGLE SUMS OF TRIANGLES IN NEUTRAL GEOMETRY
79
5.2. Exercises.
–1– In neutral geometry, if l and m are cut by a transversal, and alternate interior
angles are congruent, must l and m be parallel?
–2– In neutral geometry, if l and m are cut by a transversal, and l and m are
parallel, must alternate interior angles be congruent?
6. Angle sums of triangles in neutral geometry
We are accustomed to assuming that the three angles of a triangle always sum
to 180◦ (or π radians). We will see, however, that the angle sum of a triangle
is closely related to the various parallel axioms, and once we give these away, the
number 180◦ goes away also.
Using our neutral geometry Alternate Interior Angles Theorem, we can try to
prove that the angle sum of a triangle is 180◦ .
Figure 4. The lines l and m are parallel to the base of the triangle AB.
In Figure 4 is a triangle ∆ABC. We can construct lines l and m, as in the
picture, so that ∠1 = ∠10 and ∠2 = ∠20 . According to the Alternate Interior Angles
Theorem, we can conclude that l is parallel to line AB, and we can also conclude
that m is parallel to line AB. In Euclidean geometry, we have Euclid’s Axiom
(or the Fifth Postulate), which says that through C there can be at most one line
parallel to AB. In Euclidean geometry, therefore, l and m must be the same line,
and therefore, ∠10 , ∠3, and ∠20 together form a straight angle, which measures 180◦ .
It would follow that ∠1 + ∠2 + ∠3 = 180◦ .
7. WHAT EUCLID DID
80
In neutral geometry, however, there may be multiple parallel lines through C,
and so l and m may be different lines. If this is the case, then ∠10 + ∠3 + ∠20 6= 180◦
7. What Euclid did
If you read Euclid’s Elements carefully (I haven’t), you might get the feeling that
Euclid suspected the existence of something like hyperbolic geometry. For example,
people have noticed that he seems to try to avoid use of the Fifth Postulate. Therefore,
he has many neutral geometry results among his propositions. Let’s follow Euclid’s
investigation of angle sums.
When two lines cross, they form four angles, and the opposite pairs are called
vertical angles. Consider Proposition 15 on page 277 of Euclid’s Elements. Here
Euclid shows that vertical angles are congruent (e.g., ∠AED ∼
= ∠CEB). This result
follows from the assumption that straight angles measure 180◦ .
Euclid’s Proposition 16 states that, given a triangle, an exterior angle is always
bigger than each of the interior angles opposite of it. I have drawn Figure 5 to look
similar to the one in the The Elements, but the letters are switched around.
Figure 5. Start with the triangle ∆ABC.
In Figure 5, we’re starting with the triangle 4ABC. The side AB is extended
to a point D. The angle ∠CBD is an exterior angle, and the angles ∠BAC and
∠ACB are the angles opposite. Note that
(154)
∠ABC + ∠CBD = 180◦
8. THE SUM OF TWO ANGLES
81
Let E be the midpoint of side BC. I’ve put single slash marks through the segments
CE and BE to indicate they’re congruent. We can then draw a segment through A
and E to a point F so that E is the midpoint of AF . Then we connect the points B
and F with a line segment. Note that ∠4 and ∠40 are vertical angles, so they must
be congruent.
From the SAS criterion (which is one of Hilbert’s axioms other than Euclid’s
axiom, so is in neutral geometry), triangles 4AEC and 4F EB are congruent. In
particular, ∠3 = ∠30 . Since ∠CBD = ∠30 + ∠6, we clearly have that
(155)
∠CBD > ∠ACB,
and the exterior angle is larger than one of the opposite angles. I’ll let you show that
this exterior angle is also larger than the other opposite angle.
7.1. Exercises.
–1– Draw a new picture without E and F . You still should have the exterior
angle ∠CBD. Now extend side CB downwards to a point G. This will form
∠ABG as an exterior angle.
–2– Your new picture is kind of like a mirror image of Figure 5. Applying what
we just figured out, which opposite angle is smaller than this exterior angle,
∠ABG?
–3– OK. How are angles ∠ABG and ∠CBD related?
–4– How are we doing with Proposition 16?
8. The sum of two angles
Finally, let’s nail down Euclid’s Proposition 17. This one says that any two
angles of a triangle sum to less than 180◦ .
Let’s go back to the triangle 4ABC with the extension to D. We know that
(156)
∠ACB < ∠CBD.
We also know that
(157)
∠ABC + ∠CBD = 180◦ .
9. THE SUM OF THREE ANGLES
82
It follows that
(158)
∠ABC + ∠ACB < 180◦ .
There isn’t anything special about which two angles we’re talking about, so clearly,
any pair of angles from this triangle must sum to less than 180◦ .
As a look ahead, note that in Figure 5, AC and BF are parallel and congruent.
Therefore, 4ABC and 4ABF have the same base and the same height. The areas
are the same, but the angle at A is approximately cut in half. Imagine repeating
this construction on 4ABF . The angle at A gets smaller still, but the area stays the
same. Keep going. What’s special about the triangles as you go out in this sequence?
9. The sum of three angles
This is where we are. In neutral geometry, we can show that any two angles of a
triangle sum to less than 180◦ .
Theorem 6. In neutral geometry, the sum of any two angles of a triangle sum
to less than 180◦ .
Recall that for a set of numbers S, if x is greater than or equal to every element of
S, then we will say that x is an upper bound for S. If y is the smallest possible upper
bound for S, then y is said to be a least upper bound (a.k.a. an infimum). For
example, the number 7 is an upper bound for the set [ 0, 1 ) = { x ∈ R | 0 ≤ x < 1 },
since it’s bigger than everything in [ 0, 1 ). The least upper bound is 1. Note that
this set has no largest element. Therefore, the least upper bound lies outside of the
set. It only makes sense to talk about a largest element, if the least upper bound lies
inside the set (e.g., 1 is the least upper bound for [ 0, 1 ]).
As an example, suppose that you have a triangle 4ABC, and you know that one
of the angles measures 2◦ . Let’s say that ∠A = 2◦ . The other two angles must sum
to less than 180◦ by Theorem 6. Therefore,
(159)
∠B + ∠C < 180◦ ,
and
(160)
∠A + ∠B + ∠C < 2◦ + 180◦ = 182◦ .
Similarly, if ∠A = ◦, then ∠A + ∠B + ∠C < 180◦ + ◦ .
10. SOME TRIANGLES WILL HAVE THE SAME ANGLE SUM
83
Apparently, if a triangle has one really tiny angle, then its angle sum can’t be
much bigger than 180◦ . Our plan of attack will be to show that every triangle has the
same angle sum as a triangle with one really tiny angle. We will use this observation
to show that the angle sum of a triangle in neutral geometry can’t be bigger than
180◦ at all.
Keep in mind that in neutral geometry, the angle sum of one triangle may be
different from the angle sum of another. We can’t show that every triangle has angle
sum equal to 180◦ after all. Eventually, we’ll see that if the area of the two triangles
are the same, then the angle sums will be the same. We won’t assume this fact,
however.
10. Some triangles will have the same angle sum
We will start with some generic triangle 4ABC, and we want to show something
about its angle sum. We know a little something about triangles with one tiny angle,
so we need to make some sort of connection. We can find the midpoint of side
BC, which is labeled E in Figure 6. We can then extend AE to a point F so that
AE ∼
= EF . The angles ∠4 and ∠40 are vertical angles, so they are congruent, and we
have satisfied the SAS criterion for triangles 4AEC and 4F EB. This is just like
what we did with Figure 5.
Figure 6. 4ABC has the same angle sum as 4AF B.
We’re after the angle sum of 4ABC, which is
X
(161)
4ABC = ∠1a + ∠1b + ∠2 + ∠3.
11. ANGLE SUMS OF TRIANGLES IN NEUTRAL GEOMETRY CANNOT EXCEED 180◦
84
We will show that this is the same as the angle sum for 4ABF , which is
X
(162)
4ABF = ∠1b + ∠2 + ∠30 + ∠5.
P
Since 4AEC and 4F EB are congruent triangles, ∠1a = ∠5, so
4ABC =
P
4F EB.
Furthermore, let’s suppose that ∠1 = ∠1a + ∠1b is the smallest angle of 4ABC.
Then the smallest angle of 4ABF is either ∠1b or ∠5. If ∠1b = ∠5, then they are
both exactly half the measure of angle ∠1. If ∠1b 6= ∠5, then one of these angles
must be less than half the measure of ∠1.
In any case, we know that the smallest angle of 4F EB is less than or equal to
half the smallest angle of 4ABC.
11. Angle sums of triangles in neutral geometry cannot exceed 180◦
In the last section, we proved the following statement.
Theorem 7. Given any triangle T in neutral geometry, there is a triangle T 0
with the same angle sum whose smallest angle measures less than or equal to half the
measure of the smallest angle of T .
Suppose you have a triangle T0 in neutral geometry with angle sum 190◦ and
smallest angle 40◦ . Then by Theorem 7 there is another triangle T1 that has the
same angle sum, 190◦ , and whose smallest angle measures less than or equal to 20◦ .
Similarly, there is a triangle T2 with angle sum 190◦ and smallest angle smaller than
10◦ . There’s something wrong, however. The two larger angles of T2 sum to less than
180◦ . Therefore, the angle sum of T2 must be less than 190◦ . This contradicts what
we just found. Therefore, T0 can’t have an angle sum of 190◦ !
As you can see, anytime we assume that the angle sum of a triangle in neutral geometry is greater than 180◦ , we can use Theorems 6 and 7 to arrive at a contradiction.
Let’s have you do the proof in general.
Theorem 8. The angle sum of a triangle in neutral geometry is less than or equal
to 180◦ .
Note that the number 180 is arbitrary. The number would be π, if we used that
for the measure of a straight angle. Without numbers, we might say that the angle
sum is less than or equal to the measure of a straight angle.
12. SUMMARY
85
Here is the proof of the theorem.
Let T0 be a triangle in neutral geometry, and assume that its angle sum is greater
than 180◦ . In particular, there exists some number > 0 such that the angle sum of
T0 is 180◦ + ◦. The triangle T0 has a smallest angle, and let’s say that it measures
α◦ .
By Theorem 6, we know that α < 90, because it plus one of the other angles sums
to less than 180◦ , and since α is the measure of the smallest angle, that other angle
must be bigger than α.
By Theorem 7, there exists triangle T1 with angle sum 180◦ +◦ and smallest angle
measuring less than α2 ◦.
There is also a triangle T2 with angle sum 180◦ + ◦ and smallest angle measuring
α◦
.
4
We can continue this indefinitely, and in general, triangle Tn has angle sum 180◦ +
◦ and the smallest angle will go to zero.
In particular, there exists some n such that the smallest angle measures less than
. The triangle Tn , therefore, will have smallest angle smaller than ◦ , while the
other two angles sum to less than 180◦ . The total angle sum, therefore, must be less
than 180◦ + ◦. That’s our contradiction.
◦
Note that we can only prove that the angle sum is less than or equal to 180◦ .
12. Summary
Before Gauss, Bolyai, and Lobachevski, everyone was trying to show that neutral
geometry and Euclidean geometry were the same. That is, they wanted to prove
Euclid’s Fifth Postulate as a theorem rather than assume it as an axiom. If you could
prove in neutral geometry that the angle sum of a triangle was exactly the same as
the measure of a straight angle, you could prove Euclid’s Fifth Postulate. The best
you can do, however, is to show that the angle sum is 180◦ or smaller.
Gauss, Bolyai, and Lobachevski, for the most part independently and about
the same time, had the obvious thought, “Hey, maybe you can’t prove Euclid’s Fifth
Postulate.” But how could you prove that doing this was impossible? One way is to
come up with some odd geometry that satisfies all of the axioms of neutral geometry,
but in this geometry, Euclid’s Fifth Postulate is false. If you could do that, then any
12. SUMMARY
86
neutral geometry proof would work in this geometry, and you certainly couldn’t prove
Euclid’s Fifth Postulate.
Two questions arise immediately. Does such a geometry exist? and if so, How
real is that geometry?
If we take all of Hilbert’s axioms, including Euclid’s Axiom, we (supposedly) have
an axiom system for Euclidean geometry. If we replace Euclid’s Axiom by the following axiom, we get an axiom system for a geometry called hyperbolic geometry.
I’ve also seen references like Lobachevski’s and/or Bolyai’s geometry.
The Hyperbolic Axiom. Given a line l and a point P not on l, there are at
least two lines through P that are parallel to l.
Next, we’ll explore a model of hyperbolic geometry. In Bonola’s Non-Euclidean
Geometry, a description of such a model starts on page 164. I don’t see this mentioned
in the text, but this model is generally associated with the mathematician Eugenio
Beltrami. I plan on using a different model named after Henri Poincaré.
12.1. Exercises.
–1– Assuming that hyperbolic geometry exists (it does), and keeping in mind the
things you’ve proven, is it possible that the angle sums of all triangles in
hyperbolic geometry is 180◦ ?
–2– Can there be any triangles in hyperbolic geometry that have angle sums
greater than 180◦ ?