Samuel Otten
Grand Valley State University
N O N -E U C L I D E A N G E O M E T R Y
Euclid’s Axioms
I. For every point P and every point Q not equal to P there exists a unique line l that passes
through P and Q.
There is a unique line that passes through any two distinct points.
II. For any two segments AB and CD, there exists a unique point E such that A*B*E and the
segment CD is congruent to segment BE.
Given any two segments, one segment can be extended by the length of the other.
Any line segment can be extended indefinitely.
III. For every point O and every point A not equal to O there exists a circle with center O and
radius OA.
Any two distinct points determine a circle.
IV. All right angles are congruent to each other.
Regardless of how far away right angles are, they are the “same size” and offer a basis
of measurement.
V. For every line l and for every point P that does not lie on l there exists a unique line m
through P that is parallel to l.
If two lines are intersected by a transversal in such a way that the sum of the measures of
two interior angles on one side of the transversal is less than 180º, then the two lines meet on
that side of the transversal.
EQUIVALENTS TO THE PARALLEL POSTULATE

If two lines are parallel, then the AIA formed by a transversal are congruent.

If a line intersects one of two parallel lines, then it must also intersect the other.

If a line is perpendicular to one of two parallel lines, then it is perpendicular to the other.

The measures of the angles of all triangles sum to 180º.

Similar triangles exist.

Any triangle can be circumscribed by a circle.

Rectangles exist.
1
ASSUMPTIONS ADDED TO EUCLID’S AXIOMS
In examining the proofs presented by Euclid it was discovered that he was assuming things that
were not stated explicitly as assumptions. Therefore, mathematicians (David Hilbert among
them) presented additional postulates or axioms that are to be assumed along with Euclid’s.
BETWEENNESS
(1) If A*B*C, then A, B, and C are three distinct points lying on the same line, and C*B*A.
(2) Given any two distinct points B and D, there exist points A, C, and E lying on BD such that
A*B*D, B*C*D, and B*D*E.
(3) If A, B, and C are three distinct points lying on the same line, then one and only one of the
points is between the other two.
(4) For every line l and any three points A, B, and C not lying on l: If A and B are on the same
side of l and B and C are on the same side of l, then A and C are on the same side of l. If A and B
are on opposite sides of l and B and C are on opposite sides, then A and C are on the same side.
CONGRUENCE
(1) If A and B are distinct points and if A’ is any point, then for each ray r emanating from A’
there is a unique point B’ on r such that AB  AB  .
(2) Congruence of segments is reflexive and transitive.
(3) The “addition” of pairs of congruent segments results in congruent segments.
(4) Given any BAC and given any ray AB  emanating from a point A’, then there is a unique
ray AC  on a given side of line AB  such that BAC   BAC .
(5) Congruence of angles is reflexive and transitive.
(6) If two sides and the included angle of one triangle are congruent respectively to two sides and
the included angle of another triangle, then the two triangles are congruent. (SAS)
CONTINUITY
(Circular) If a circle  has one point inside and one point outside another circle   , then the two
circles intersect in two points.
(Elementary) If one endpoint of a segment is inside a circle and the other outside, then the
segment intersects the circle.
ARISTOTLE’S AXIOM
Given any side of an acute angle and any segment AB, there exists a point Y on the given side
of the angle such that if X is the foot of the perpendicular from Y to the other side of the angle,
XY > AB.
2
RESULTS IN NEUTRAL GEOMETRY
“Neutral Geometry” refers to the collection of mathematical results derived from geometric
axioms with the exception of Euclid’s fifth postulate. Theorems in neutral geometry may be
used as a basis for proving conjectures in both Euclidean and hyperbolic geometries, as we will
explore later. Elliptic geometry is built on a different set of axioms and therefore lies separate
from neutral geometry.
Pasch’s Theorem. If A, B, C are distinct noncollinear points and l is any line intersecting AB in
a point between A and B, then l also intersects either AC or BC (and both only in the case
that C lies on l).
Crossbar Theorem. If AD is between AC and AB , then AD intersects segment BC.
Isosceles Triangle Theorem. If in ABC we have AB  AC , then B  C .
Trichotomy of Segments. For segments AB and CD: AB  CD, AB  CD, or AB  CD.
Congruence of Supplemental Angles. Supplements of congruent angles are congruent.
Vertical Angle Theorem. When two lines intersect, the vertical angles formed are congruent.
Alternate Interior Angle Theorem (AIA). If two lines cut by a transversal have a pair of
congruent alternate interior angles, then the two lines are parallel.
Corollary 1. Two lines perpendicular to the same line are parallel.
Corollary 2. If l is any line and P is any point not on l, there exists at least one line m
through P parallel to l.
Angle-Side Correspondence of Triangles. In a triangle ABC , the greater angle lies opposite the
greater side and the greater side lies opposite the greater angle.
Saccheri-Legendre Theorem. The sum of the degree measures of the three angles in any triangle
is less than or equal to 180º.
Corollary 1. The sum of the degree measures of two angles in a triangle is less than or
equal to the degree measure of their remote exterior angle.
Corollary 2. The sum of the degree measures of the angles in any convex quadrilateral is
at most 360º.
Rectangle–180º Triangle Theorem. If a triangle exists whose angle sum is 180º, then a rectangle
exists. If a rectangle exists, then every triangle has angle sum equal to 180º.
Though the parallel postulate is never used as the basis for a proof in neutral geometry, there is
still work that occurs surrounding it. For instance, it can be proven in neutral geometry that
certain statements are logically equivalent to Euclid V (see the bottom of page 1). The following
page presents a proof of the logical equivalence between the parallel postulate and the statement
“Rectangles exist.” The reason for our particular focus on this equivalence will become apparent
when the Universal Hyperbolic Theorem is introduced.
3
RESULTS IN NEUTRAL GEOMETRY
“Rectangles exist.”  “Given any line l and any point P not on l, there is a unique line m that
passes through P and is parallel to l.”
Proof. First, let us assume the parallel postulate to be true. We begin with two parallel lines l
and m, and two points P and Q on l. We construct the perpendicular line segments from P to m
and from Q to m.
P
Q
R
S
l
But the parallel postulate implies that a line
perpendicular to one of two parallel lines is also
perpendicular to the other, so R and S must be
right angles, thus forming rectangle PQSR.
m
Now, let us assume that rectangles exist. It follows from the AIA theorem that the lines
determined by the opposite sides of a rectangle are parallel. We will show that no other lines
through a particular point are parallel, making the pair of lines forming the rectangle uniquely
parallel as claimed in the parallel postulate.
We have a rectangle. Let l and m be the
P
Q
l
lines shown at the right. We can introduce a
diagonal within the rectangle and, by the
Rectangle–180º Triangle Theorem and AAS
congruence of triangles, prove that opposite sides
of a rectangle are congruent.
Therefore,
m
S
R
PR  QS . This further implies that l and m are
always a distance of PR apart (simply vary Q and S).
Let n be any line through P distinct from l. Since all triangles have interior angles that
sum to 180º it must be the case that one side of n from P forms an acute angle with PR .*
Therefore we can use Aristotle’s axiom, and the point X can be chosen on the acute side of n
(with Q being the foot of the perpendicular from X to l and S being the foot of the perpendicular
from X to m) so that XQ > PR. This means that XQ > QS. We also know that X, S, and Q are
collinear because of the uniqueness of a perpendicular to a pair of parallel lines through a given
point not on the lines (X in our
P
Q
Q1
Q2
case). Since S is on m and
l
X*S*Q, it must be the case by
rules of continuity that n
x1
intersected m at some point
between P and X. Hence, l is the
x2
only line through P that is
S
m
parallel to m. This completes the
S
S
2
1
R
proof of the logical equivalence
X
n
of our statements.
□
*see Lemma 1 in Appendix A.
4
RESULTS IN EUCLIDEAN GEOMETRY
Law of Sines. For any triangle with vertices A, B, C and sides a, b, c respectively opposite the
vertices, the following is true:
a sin A

b sin B
and congruently
a
b
c


.
sin A sin B sin C
Proof. Consider the arbitrary triangle shown below with the altitude constructed from point C
intersecting side c at point D (note: the existence of a perpendicular through a given point is not
debated between geometries, only the existence of a unique parallel).
B
By the definition of an altitude we know that both CDA and
DC
DC
.
and sin A 
CDB are right. This means that sin B 
a
b
By solving each equation for DC and setting them equal to each
a sin A
.□
other we see that a  sin B  b  sin A which implies 
b sin B
D
a
C
A
b
Law of Cosines. For any ABC with sides a, b, c respectively opposite the vertices:
c 2  a 2  b 2  2ab  cos C.
Proof. We begin with an arbitrary triangle where h is the pictured altitude and x represents the
segment CD. We know that b sin C  h and b  cos C  x.
A
Looking at ADB we see that h and a  x are the legs, and
since this is a right triangle it must be true that
2
c 2  h 2  a  x  (note: at this point we have firmly entered
Euclidean Geometry).
Substituting for h and x leads us to the following:
c  b sin C   a  b cos C 
2
2
c
h
b
2
c 2  b 2 sin 2 C  a 2  2ab cos C  b 2 cos 2 C

x

c 2  b 2 sin 2 C  cos 2 C  a 2  2ab cos C
B
D
a
C
□
c 2  a 2  b 2  2ab cos C
This result leads to the converse of the Pythagorean Theorem, because if c 2  a 2  b 2 then it
must be true that C  90 so that cos C  cos90  0.
5
ATTEMPTS TO PROVE THE PARALLEL POSTULATE
The following are “proofs” by Legendre and Farkas Bolyai concerning the parallel
postulate. The errors in reasoning are illuminated subsequently.
Adrien Marie Legendre (1752-1833)
Given P not on line l. Drop perpendicular PQ from P to l at Q. Let m be the line through
P perpendicular to PQ . Then m is parallel to l, since l and m have the common perpendicular
PQ . Let n be any line through P distinct from m and PQ . We must show that n meets l. Let
PR be a ray of n between PQ and a ray of m emanating from P. There is a point R’ on the
opposite side of PQ from R such that QPR   QPR . Then Q lies in the interior of RP R  .
n
P
m
R'
R
B
l
A
Q
Since line l passes through the point Q interior to RP R  , l must intersect one of the
sides of this angle. If l meets side PR then certainly l meets n. Suppose l meets side PR at a
point A. Let B be the unique point on side PR such that PA  PB . Then PQA  PQB by
SAS congruence; hence PQB is a right angle so that B lies on l and n, a point of intersection.
ERRORS: Legendre makes several errors in the proof, though they all seem reasonable
because we instinctively assume things that are only true in Euclidean Geometry. The first error
in Legendre’s proof occurs while he is still describing the geometric situation. He claims that m
and l are parallel because they are both perpendicular to the same line. He is applying the
Alternate Interior Angles Theorem, which does not hold in Elliptic Geometry.
Another error occurs when Legendre makes the following assumption: If a line passes
through a point on the interior of an angle, then the line must intersect one of the rays forming
the angle. He uses this to say that line l must intersect PR or PR . However, this conditional
statement is not true in hyperbolic geometry and therefore not true in Neutral Geometry.
Legendre makes one final error when he says that B must lie on l because both are out
from PQ at a right angle. There is nothing in Neutral Geometry that guarantees the uniqueness
of a line perpendicular to a given line through a certain point on the given line; therefore, QB
and l might both be perpendicular to PQ but distinct so that the point B lies off of l.
6
ATTEMPTS TO PROVE THE PARALLEL POSTULATE
Farkas Bolyai (1775-1856)
Given P not on line l, PQ perpendicular to l at Q, and line m perpendicular to PQ at P.
Let n be any line through P distinct from m and PQ . We must show that n meets l. Let A be
any point between P and Q. Let B be the unique point such that A*Q*B and AQ  QB . Let R
be the foot of the perpendicular from A to n. Let C be the unique point such that A*R*C and
AR  RC .
Then A, B, and C are not collinear
(else R = P); hence there is a unique circle
C
P
passing through them.
Since l is the
perpendicular bisector of chord AB and n is
m
R
the perpendicular bisector of AC, l and n
meet at the center of the circle. This can be
A
n
Q
done for any n, and therefore m is uniquely
l
parallel to l through the point P.
B
ERRORS: The error that takes place in Bolyai’s attempted proof is assuming that a circle can be
drawn to include all three vertices of any triangle. This is only true if the Parallel Postulate is
already assumed to be true (see below). The otherwise clever proof falls flat without the
existence of the circle.
To find the center of the circumscribing circle for an arbitrary triangle, you would construct the
perpendicular bisectors of two of the triangle’s sides and their point of intersection would be equidistant from all
vertices and thus be the center of the circumscribing circle. However, there is no guarantee in Neutral Geometry
that this point of intersection exists. Saying that the two perpendicular bisectors intersect is using the assumption
that two lines cut by a transversal will meet on the side of the transversal with interior angles that sum to less than
180 degrees. However, only the Parallel Postulate will guarantee this intersection, and therefore saying that every
triangle can be inscribed inside a circle is equivalent to Euclid’s Fifth because we need the Fifth Postulate to
determine the existence of the center of the circle.
C?
The transv ersal that intersects m and n may
hav e angles on one side that sum to less than
180 degrees, but we cannot assume that they
intersect on that side.
m
n
7
DEVELOPMENT OF NON-EUCLIDEAN GEOMETRIES
As doubts began to rise about whether Euclid’s Fifth Axiom could be proven at all,
investigations were conducted surrounding the negation of the axiom. It is important to note that
the Parallel Postulate can be negated in two different ways. The postulate reads as follows:
Given a line l and a point P not on l, exactly one line exists through P parallel to l. To say that
more than one line exists would negate the statement, and to say that no lines exist would also
negate the statement. (The “no lines” negation is used to form Elliptical geometry, which will
not be thoroughly addressed in this paper.)
János Bolyai (1802-1860), Carl Friedrich Gauss (1777-1855), and Nikolai Lobachevsky
(1792-1856) contributed to the discovery of hyperbolic geometry which assumes everything
from neutral geometry and also assumes that there exist a line l and a point P not on l such that at
least two distinct lines parallel to l pass through P. This statement, a negation of Euclid V, only
claims that some line and some P have multiple parallels, but this can quickly be generalized into
the Universal Hyperbolic Theorem which states that for all lines l and all points P not on l,
infinitely many lines parallel to l pass through P. (note: In hyperbolic geometry we assume the
negation of the parallel postulate, which means we may also assume the negation of “rectangles
exist” since we proved their logical equivalence. Therefore, rectangles do not exist in hyperbolic
geometry.)
Proof. Let l be a line and P a point not on l. We will show that there are infinitely many lines
parallel to l that pass through the point P.
Let Q be the foot of the perpendicular from P to l, and let m be the line through P
perpendicular to PQ. Now, let R be any point on l distinct from Q. The line t can be
constructed perpendicular to l and containing the point R. Furthermore, let S be the foot of the
perpendicular from P to t.
t
P
m
S
R
Q
l
We know that m and PS are both parallel to l by Corollary 1 of the AIA theorem. It is also the
case that m and PS are distinct lines. Assume that S  m. This would mean that a rectangle
exists (PQRS). However, a rectangle cannot exist in hyperbolic geometry. Therefore, S  m
and PS must be distinct from m. This proves that at least two lines parallel to l contain the point
P. By varying R it follows that infinitely many parallel lines exist, thus proving the theorem.
8
□
RESULTS IN HYPERBOLIC GEOMETRY
Hyperbolic Triangle Theorem. All triangles have angle sum less than 180°.
Corollary 1. The defect of a triangle is 180 minus the angle sum, and is always positive.
Corollary 2. All convex quadrilaterals have angle sum less than 360°.
AAA Triangle Congruence Theorem. If two triangles are similar, then they are congruent.
Parallel Distance Theorem. If l and l  are any distinct parallel lines, then any set of points on l
equidistant from l  has at most two points in it.
Common Perpendicular Theorem. If lines l and l  have a common perpendicular segment
MM  , then they are parallel and MM  is unique.
Corollary. If A and B are any points on l such that M is the midpoint of AB, then A and
B are equidistant from l  . Further, if AM  MB, then A and B are not
equidistant from l  .
Proof. The theorem is a direct result of the Alternate Interior Angle Theorem and the fact
that rectangles do not exist in hyperbolic geometry (any other common perpendicular
could be used to form a rectangle between l and l  ).
The first statement in the corollary follows simply, as the assumption creates congruent triangles
on l which in turn can be used to identify congruent triangles on l  .
For the second statement in the corollary, assume to the contrary that M is not the midpoint of A
and B, but A and B are equidistant from l  ; that is to say AA  BB.
A'
M'
B'
A
M
B
l'
l
We lay off a distance of AM  on the opposite side of M  and we label the new point C . Let
us also construct the segment that leaves l  from C  at a right angle and intersects l at the point
C. Now we consider the triangles formed by joining A  to M and C  to M. By SAS,
A' M ' M  C' M ' M . This means
A'
C'
l'
M'
It also means that
A' M  C' M .
corresponding angles within these
triangles are congruent and their
corresponding complements are conl
gruent; therefore, AA' M  CC' M
C
A
M
and AMA'  CMC'.
It follows that AA' M  CC' M (ASA), so AA  CC  and these two both equal BB . But the
Parallel Distance Theorem states that a set of equidistant segments between two parallel lines can
have at most two segments in it, while we have identified three equidistant segments. This
contradiction guarantees that if M is not the midpoint of A and B on l, then A and B are not
equidistant from l  and we have proven the corollary.
□
9
RESULTS IN HYPERBOLIC GEOMETRY
PARALLELS: Two lines are asymptotal parallels if they meet at infinity in one direction and
diverge in the other direction. Two lines are ultraparallel if they have a common perpendicular
and diverge in both directions from the perpendicular.
HYPERBOLIC TRIGONOMETRY:
e t  e t
t3 t5
sinh t 
 t    ...
2
3! 5!
cosh t 
e t  e t
t2 t4
 1    ...
2
2! 4!
cosh 2 t  sinh 2 t  1 *
DISTANCE: It is necessary to rethink the calculation of distance for the models of hyperbolic
geometry.† If you consider the figure at the right, Euclid II requires that a point Q exist on line m
such that RS  PQ . However, it is quite clear that Q
A
cannot exist with “ordinary” or Euclidean measurement of
R
distance. Therefore, the following distance function has been
developed:
l
AR  BS
d R, S   k  ln
,
BR  AS
P
where the distances within the function are Euclidean and k is
a universal hyperbolic constant, usually 1.
This distance function satisfies the intuitive properties
of distance.
 d R, S  is defined for all points R and S.
 d R, S  is never negative.
 d R, S   0 when R and S are the same point.
 d R, S   d S , R for all points R and S.
 d R, S  is continuous and its range is all real numbers.
It now becomes possible to find a Q on m such that RS  PQ .
*see Appendix A for development.
†
m
see diagrams of the models in Appendix B.
10
S
B
REFERENCES
Greenberg, M. J. (1993). Euclidean and Non-Euclidean Geometries: Development and
History (Third Edition). New York, NY: W.H. Freeman and Company.
Hilbert, D., & Cohn-Vossen, S. (1990). Geometry and the Imagination. New York, NY:
Chelsea Publishing Co.
Martin, G.E. (1975). The Foundations of Geometry and the Non-Euclidean Plane. New
York, NY: Intext Educational Publishers.
Meschkowski, H. (1964). Noneuclidean Geometry. New York, NY: Academic Press.
Stillwell, J. (1996). History of Mathematics: Sources of Hyperbolic Geometry (Volume
10). Providence, RI: American Mathematical Society.
11
APPENDIX A
P
Lemma 1.
Let PQ be a line and m be the line perpendicular to
m
PQ at the point P. If rectangles exist and n is any line through P
distinct from PQ and m, then exactly one of N1 PQ and
Q
N 2 PQ forms an acute angle.
Since m and n share the point P and are distinct lines, they must have no other points in
common (Euclid I). Therefore, looking around the point P, line n must deviate from line m on
each side. There are four possible ways this could occur without violating Euclid I:
n
m
P
P
CASE 1
m
n
CASE 2
n
m
P
P
m
n
CASE 4
CASE 3
Let us consider for each case the triangle formed with the point Q and a line segment along m, as
well as the triangle formed with a congruent angle at Q but instead a line segment along n.
N1
N2
P
M1
M2
Q
CASE 1
P
M1
N1
In this situation PN1Q must be smaller in measure
than PM 1Q , and PN 2 Q must be smaller in measure
than PM 2 Q. With this being the case, it is impossible
for N1QN 2 and M 1QM 2 to have an equal sum of
interior angles. But, since we assumed rectangles exist,
the Rectangle-180º Triangle theorem guarantees that all
triangles have interior angles that sum to 180º. This
contradiction implies that CASE 1 is impossible.
In this situation PN1Q is larger in measure than PM 1Q ,
and PN 2 Q is also larger than PM 2 Q. This means that
N1QN 2 and M 1QM 2 cannot have an equal sum of
interior angles and therefore cannot both be 180º triangles.
Hence, CASE 2 is impossible.
M2
N2
Q
CASE 2
M1
P
N2
M2
N1
In this situation PN1Q is larger than PM 1Q , but
PN 2 Q is smaller than PM 2 Q. These angle inequalities could balance out so that both triangles contained
180º. Therefore, CASE 3 is possible under the assumption
that rectangles exist. A similar argument shows that CASE
4 is also possible.
Q
CASE 3
12
APPENDIX A
Moving back to the larger situation, we now know that one of the following is true for
any line n through P distinct from m (assuming rectangles exist):
N2
P
N1
n
m
P
m
n
N1
N2
Q
Q
We can conclude that either N1 PQ lies within the right angle or N 2 PQ lies within the right
□
angle. Hence, one of N1 PQ and N 2 PQ must be acute.
We will arrive algebraically at the hyperbolic trigonometric identity cosh 2 t  sinh 2 t  1
using the definitions of cosh and sinh.
 e t  e t
cosh t  sinh t  
 2
2
2
2
  e t  e t
  
  2
2
e

t
 e t
4
e

t
 e t e t  e t  e t  e t e t  e t
4



  e



2
t
 e t
4


2
 

e
2t
 2  e 2t  e 2t  2  e 2t
4
 
e
2t
 e 2t  e 2t  e 2t  4
4
 
004
4
1
13



APPENDIX B
One-sheet and two-sheet hyperboloids. Surfaces with constant negative curvature that can
be conceptualized as a surface which hyperbolic geometry takes place on.
A triangle in the Poincaré disc, illustrating
the hyperbolic characteristic that triangles
contain less than 180.
Diagrams describing the development of
the Klein and Poincaré disc models of the
hyperbolic plane.
Euclid
Lobachevsky
Beltrami
14
Klein
Poincaré
15