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Hyperbolic Geometry
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Introduction and Set Up
As we already know from previous presentations, Euclidean geometry is based on the following five
postulates (p. 481):
1. To draw a line through two points
2. To extend a given line
3. To draw a circle with given center through a given point
4. All right angles are equal
5. If a line crossing two other lines makes the interior angles on the same side less
than two right angles, then these two lines will meet on that side when extended
far enough
Again, as we already saw in section 3 (p. 39), assuming all the first four axioms, we actually proved
that the fifth postulate above is equivalent to the Playfair’s axiom.
5’. For each point P and each line l, there exists at most one line through P parallel to
l
Mathematicians thought for two thousand years that the fifth postulate was actually a theorem and
so they tried to prove it assuming the other four. In the nineteenth century a progress was made by
working out the consequences of the denial of the parallel postulate instead of trying to prove it by
contradiction. What was discovered assuming
Given a line and a point not on it, there is more than one line going through the given
point that is parallel to the given line.
was a new coherent theory named non-Euclidean geometry . In particular Bolyai and Lobachevski
developed the Hyperbolic geometry. People think that also Euclid was not really convinced that
his fifth postulate was actually a postulate, therefore he proved his first 28 propositions avoiding to
use it. This means that all these propositions also apply in hyperbolic geometry (and more generally
in non-Euclidean geometry).
It is useful to list all tools that was already presented and that find here application.
• (I.5): In isosceles triangles the angles at the base equal one another.
• (I.27): If a line meets two other lines makes the alternate angles equal, then the two straight
lines are parallel.
• (I.28): If a line falling on two other lines makes the exterior angle equal to the interior and
opposite angle on the same side, then the straight lines are parallel to one other.
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• (34.1): In a Hilbert plane, suppose that two equal perpendiculars AC, BD stand at the ends
of the interval AB, and we join CD (this is the so called Saccheri quadrilateral). Then the
angles at C and D are equal, and furthermore, the line joining the midpoints of AB and CD,
the midline, is perpendicular to both.
• (34.6): Given a triangle ABC, there is a Saccheri quadrilateral for which the sum of its two top
angles is equal to the sum of the three angles of the triangle.
• (34.7): In any Hilbert plane:
There exists a rectangle ⇐⇒ There exists a triangle with angle sum = 2RA
• (34.11): Given rays Aa, Bb, Cc, if Aa ||| Bb andBb ||| Cc ⇒ Aa ||| Cc
• (ASAL): Given four rays Aa, Bb, A0 a0 , B 0 b0 with ^(BAa) = ^(B 0 A0 a0 ), AB = A0 B 0 , ^(ABb) =
^(A0 B 0 b0 ),
then Aa ||| Bb ⇐⇒ A0 a0 ||| B 0 b0
• (ASL): Given Aa ||| Bb and A0 a0 ||| B 0 b0 , assume ^(BAa) = ^(B 0 A0 a0 ) and AB = A0 B 0 . Then
^(ABb) = ^(A0 B 0 b0 ). The figure consisting of the segment AB and the two limiting parallel
rays Aa and Bb a limit triangle.
• (SAS): side-angle-side criterion for congruence of triangles.
The concept of limiting parallels will be fundamental so let’s recall the definition.
Definition 1.1. A ray Aa is limiting parallel to a ray Bb (Aa ||| Bb) if either they are coterminal
(i.e. if they lie on the same line and ”go in the same direction”), or if they lie on distinct lines not
equal to the line AB, they do not meet, and every ray in the interior of the angle BAa meets the ray
Bb.
A
a
B
b
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The Hyperbolic Geometry
The goal of this presentation is to give an introduction to hyperbolic geometry and to present the
proofs of the main theorems of this book’s section, i.e. the existence of a common perpendicular (40.5)
and the existence of the enclosing line (40.6).
We can start the axiomatic development of the hyperbolic geometry due to Lobachevski and Bolyai
using the Hilbert’s axioms of incidence, betweenness and congruence avoiding continuity since in this
book we want to avoid the use of R (This implies that we also won’t use neither Archimedes’ axiom (A)
nor the circle-circle intersection (E)). In addition to the Hilbert’s axiom we will assume the hyperbolic
axiom (L).
(L): For each line l and each point A not on l, there are two rays Aa and Aa0 from A,
not lying on the same line, and not meeting l, such that any ray An in the interior of
the angle aAa0 meets l.
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A
n
a’
a
l
Definition 2.1. A Hilbert plane satisfying (L) is called a hyperbolic plane (or a hyperbolic geometry).
We immediately note that a geometry with the (L) axiom cannot be Euclidean ((P) fails).
Remark 2.1. Using the definition of limiting parallel above, we see that if we choose a point B on the
line l and we construct the rays on l Bb and Bb0 , then Aa ||| Bb and Aa0 ||| Bb0 . Thus (L) implies
that for any point A and ray Bb, there exists a limiting parallel Aa to Bb.
Definition 2.2. For any segment AB, let b be a line perpendicular to AB at B; choose one ray
Bb on the line b, and let Aa be the limiting parallel ray to Bb, (which exists by (L)). Then we call
α = ^(BAa) the angle of parallelism of the segment AB, and denote it by α(AB).
A
α
a
b
B
Remark 2.2. Reflecting the ray Aa with respect to the line AB it will be again a limiting parallel to
the other ray on b. Since the two limiting parallel from A to b do not lie on the same line (by (L)),
α(AB) is always acute. In fact taking a straight line through A we get a right angle, but limiting
parallels on different lines exist and so we get this fact.
Proposition 2.1 (40.1). The angle of parallelism varies inversely with the segment:
1. AB ∼
= A0 B 0 ⇐⇒ α(AB) = α(A0 B 0 )
2. AB < A0 B 0 ⇐⇒ α(AB) > α(A0 B 0 )
Proof.
1. follows easily from exercise 34.10
2. Assume: AB < A0 B 0 . Mark C on the ray AB such that AC = A0 B 0 and draw the perpendicular
c to AC at C and let Aa0 be the limiting parallel from A to Cc. Then α0 = ^(CAa0 ) = α(AC) =
α(A0 B 0 ).
Let Bb be the ray perpendicular to AB at B on the same side of AC as a0 and c. If Bb does not
meet a0 , then the ray Bb would be in the interior of the angles CAa0 and ACc, meeting neither
the ray a0 nor c, and so it would be also limiting parallel to Aa0 and Cc by (34.12.1). But since
Bb ||| Cc and the angles at B and C are both right angles, we have a contradiction since the
parallelism angle has to be acute. So Bb has to meet Aa0 , and implies that the limiting parallel
from A to Bb makes an angle α greater than α0 , i.e., α(AB) > α(A0 B 0 ). By contraposition we
can show by the same argument the reverse implication.
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A
α
a’
b
B
c
C
Our next goal is to establish some results about limiting parallel rays, limit triangles and parallel lines
that are not limiting parallel.
Proposition 2.2 (Exterior angle theorem (40.2)). If AB is a segment, with limiting parallel rays
emanating from A and B, then the exterior angle β at B is greater than the interior angle α at A.
A
α
a
β
B
b
Proof. Since the ray passing through A making an angle β with AB is parallel to l (I.27) we already
know that α ≤ β.
Assume α = β. Let a0 and b0 be the opposite rays to a and b. Then the supplementary angle at A and
B would also be equal as well. We can now apply (ASAL) to AB obtaining that a0 is also limiting
parallel to b0 . But this is a contradiction to (L) since two limiting parallel from A to b cannot lie on
the same line. Therefore α < β.
A
α
a’
a
β
b’
b
B
Corollary 2.1 (40.3). In a hyperbolic plane the sum of the angles of any triangle is less than two
right angles.
Proof. By Proposition (34.4) we know that for any triangle there is a Saccheri quadrilateral whose
top two angles are equal to the angle sum of the triangle. So we only have to prove that the top two
angles of any Saccheri quadrilateral are acute.
Let ABCD be the Saccheri quadrilateral with base AB = l. Let a and b be limiting parallel from C
and D to l, with end ω by (L). Then by Proposition 2.1 the angles of parallelism are equal. By the
exterior angle theorem applied on the limit triangle CDω we have that β > γ. On the other hand, by
(34.1), the top angles α + β > α + γ = δ, and so δ must be acute.
C
D
α
γ
β
δ
α
a
b
A
B
Proposition 2.3 (AAL - 40.4). Given two limit triangles ABlm and A0 B 0 l0 m0 , suppose that the angles
A and B are equal respectively to the angles at A0 and B 0 . Then also the sides AB and A0 B 0 are equal.
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Proof. Assume w.l.o.g. AB > A0 B 0 . Choose a point C on AB such that CB = A0 B 0 , and draw a ray
n at C, on the same side of AB as l and m, making an angle equal to the angle A0 , which is also equal
to the angle at A. Now comparing C, B, n, m to the limit triangle A0 , B 0 , l0 , m0 , it follows from
(ASAL) that n is limiting parallel to m.
Then by transitivity (34.11) it follows also that l is limiting parallel to n. But this contradicts the
exterior angle theorem because the angle at C, which is exterior to the limit triangle ACln, is equal
to the angle at A. So we can conclude that AB = A0 B 0 .
A
C
l
n
B
m
The next theorem states the existence of a perpendicular for parallels which are not limiting parallel.
Theorem 2.1 (40.5). In a hyperbolic plane, if l and m are two parallel lines that are not limiting
parallels, then there is a unique line in the plane that is perpendicular to both of them.
Proof. Let l, m be two parallel lines, which are not limiting parallel and AB, CD be two perpendicular
lines from A, C on l to m. If AB = CD, then DBCA is a Saccheri quadrilateral and so (by 34.1)the
line joining the midpoints of AC and BD is perpendicular to l and m. If AB 6= CD w.l.o.g. assume
CD > AB. Take E on CD such that AB = ED and let n be a ray through E making the same angle
with ED as l makes with AB.
n
C
D
A
F
B
G
H
K
q
l
p
m
We claim that n meets l in F . Let p be a limiting parallel from B to l. l, m are not limiting parallel,
so p does not meet m. Let q be the ray through D making the same angle with m as p does at B.
By (I.28) p, q are parallel but not limiting parallel (by exterior angle theorem). Applying (ASAL) to
ABlp and EDnq, we find that q is limiting parallel to n. Therefore, n is not limiting parallel to p
and so n must meet l at F . Take now H on l and K on m such that AH = EF and BK = DG and
comparing the quadrilaterals EF DG and AHBK by (SAS) follows that F G = HK and HK ⊥ m.
So GKF H is a Saccheri quadrilateral and the line joining midpoints of F H and GK is perpendicular
to l and m. The uniqueness of the perpendicular line follows from the following. Assume that AB
and CD are both perpendicular to l and m. Then by (34.7) and (40.3) ABCD is a rectangle which
is a contradiction.
This theorem is actually crucial to prove the following proposition.
Proposition 2.4 (40.6). Given an angle in the hyperbolic plane, there is a unique line (called the
enclosing line of the angle) that is limiting parallel to both arms of the angle.
Proof. Let O be the vertex of the considered angle, choose points A, B on the arms of the angle at
equal distance from O. Denote by α the end of the ray OA (equivalence class of limiting parallel).
Draw the line Bα which is the ray through B limiting parallel to OA. So let α be the end of OA, β
be the end of OB and draw Bα and Aβ. Let a be the ray bisecting the angle αAB and b the angle
αBβ. By symmetry the angle bisected at A and B are equal. We distinguish now three cases.
1. ”a, b meet at a point C”. Then AC = BC and by (ASL) applied to ACβ and BCβ the angles
at C of these two triangles are equal, which is impossible.
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2. ”a, b are limiting parallel with end γ”. Bγ is in the interior of the angle ABβ, so it meets Aβ in
a point C. By (AAL=40.4) applied to ACγ and BCβ follows AC = BC. Using (I.5) the angles
BAC and ABC are equal, but it is also not possible since the angle BAC is also equal to the
angle ABα, which is properly contained in the angle ABC.
3. ”a, b are parallel, but not limiting parallel”. By (40.5) there exists a common perpendicular line
l meeting a at C and b at D. We claim that l is the enclosing line, i.e. l has ends α and β. By
symmetry it is enough to check that l has end β. If not, draw the lines Cβ and Dβ distinct
from l. We compare the limit triangles ACβ and BDβ. The angles at A and B are equal by
construction. The sides AC and BD are equal by symmetry. So by (ASL) the angles at C
and D are equal. It follows that Cβ and Dβ make equal angles with l at C and D. But this
is a contradiction to the exterior angle theorem (40.2). So we conclude that l has ends α and
β. Uniqueness follows directly by (L) since we cannot have two distinct line that are limiting
parallel at both ends.
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