An ellipse is the locus (set) of points the sum of whose distances from two fixed
points (called foci) is constant.
Construction steps:
1. Construct a circle with center A.
2. Choose point P on the circle.
3. Construct radius PA.
4. Choose point B inside the circle.
5. Construct line segment PB.
6. Construct the perpendicular bisector of PB
through midpoint M.
7. Label the intersection of PA and the perpendicular
bisector, point E.
8. Display the lengths of AE and BE, and their sum.
9. Trace point E
10. Animate
point P. around circle A.
Prove that AE +BE is constant
as P moves
11. Point E will trace an ellipse with foci A and B.
EM ⊥ BP by construction.
∠EBM and ∠EPM are congruent right angles.
MP ≅ MB by the definition of bisector.
EM ≅ EM by the reflexive property.
△PME ≅ △BME by SAS.
AP is the radius of a circle and is constant.
AP = AE + EP.
BE ≅ EP.
BE + AE ≅ AP, and is constant.
A Brief Introduction to
Non-Euclidean Geometry
Euclid’s fifth postulate (the Parallel Postulate) states:
Given a line and a point not on the line, exactly one line
parallel to the given line may be drawn through the point.
What if we disagreed with Euclid’s Parallel Postulate?
Would there be any theorems in Euclidean Geometry that
we could no longer trust?
Another way of asking this question is “What theorems in
Euclidean Geometry depend upon the parallel postulate
for proof?”
For example, consider the statement:
Alternate interior angles of parallel lines are congruent.
m
P
l
Q
If we allow a second line n parallel to line l through P,
Then the two angles at point P would be congruent to each
other, which they cannot be if n and m are different lines.
m
n
P
l
Q
It turns out that congruent alternate interior angles depend
on Euclid’s Parallel Postulate so, if we deny Euclid’s Parallel
Postulate, we can no longer trust the alternate interior angles
theorem.
Now consider our proof of the theorem:
The sum of the measures of the angles of a triangle is 180°.
So, since the proof that the sum of the measures of
Proof:
Construct
parallel
AB. rests on Qcongruent
the
angles
of a PQ
triangle
isto180
C
alternate interior angles
which rests Congruent
on Euclid’s
alternate
P
interior
angles
Parallel Postulate, we can no longer trust
the
sum of
the alternate
angles of a triangle theorem if we deny
Congruent
B
interior angles
Euclid’s Parallel Postulate.
A
B
A + ACB
PCA
+ QCB
C
= 180°
Consider the following proof of the theorem:
The sum of the measures of the angles of a quadrilateral is 360°.
Since the proof that the sum of the measures of
The measures
the
anglesof the
of a quadrilateral is 360 rests on the
angle of ABD is 180
Theof
measures
of the
sum of the measures of the angles
a triangle
is
angle of CBD is 180
180 which rests on congruent alternate interior
angles, which rests on Euclid’s Parallel Postulate,
we no longer accept that the sum of the angles
of a quadrilateral is 360 if we deny
Euclid’s Parallel Postulate.
Therefore, the sum of the measures of the angles of quadrilateral ABCD is
180 + 180 = 360
Is this quadrilateral a rectangle?
In order to prove that this quadrilateral is a rectangle, we need
to prove that the fourth angle is a right angle, and to do this
we need to know the sum of the angles is 360.
Therefore, we cannot even assume rectangles
exist if we deny Euclid’s Parallel Postulate.
For centuries, mathematicians attempted to prove Euclid’s
fifth postulate (the parallel postulate) using the first four. The
development of Non-Euclidean Geometry was an outgrowth
of these attempts, most often using an indirect proof (proof by
contradiction).
Euclid’s
Parallel
One
Negation
of Postulate
Euclid’s Parallel Postulate
more than
Given a line and a point not on the line, exactly
oneone
lineline
parallel to the given line may be drawn through the point.
So what contradictions did mathematicians come up with in
assuming the above negation of the Parallel Postulate?
Lets’ examine one attempted proof of the parallel postulate
by the Hungarian mathematician Janos Bolyai (1802-1860).
Bolyai’s “proof” makes use of the following well known
relationship that we discussed in class a few days ago:
Given a triangle, the circumcenter (center of the circle passing through the vertices
of the triangle) is the intersection of the perpendicular bisectors of the sides.
This also assures that a circle can be constructed through any three non-collinear points.
C


A




B
Begin with line PQ and construct
lines m and l perpendicular to PQ
at points P and Q, respectively.
Then l and m are parallel because
they are perpendicular to the
same line.
Let n be any other line through P.
We must prove that n is not
parallel to l (i.e. we must prove
that n intersects l ).
P
m
n
Q
l
Choose a point A between P and Q.
Let B be the unique point on line PQ
such that Q is between A and B and
AQ = QB.
Construct a perpendicular from A
to line n meeting n at a point R.
Let C be the unique point on ray AR such
that R is between A and C and AR = RC.
C
P
m

R
A
n
Q
l
B
Since points A, B, and C are the
vertices of a triangle, there is a
unique circle passing through
points A, B, and C.
The center of this circle is the
intersection of the perpendicular
bisectors of sides AB and AC.
C
P
m

R
n
A
Q

l
Is there a flaw in Bolyai’s
“Proof”?
B
But the perpendicular bisector of side AB is l ,
and the perpendicular bisector of side AC is n,
so l and n must intersect at the center of this circle.
Therefore, m is the only line through P that is parallel to l.
77 Statements
whose
proofs depend
on Euclid’s
Postulate.
77 Statements
Equivalent
to Euclid’s
Parallel Parallel
Postulate
C

A


B
By using these,
Bolyai was implicitly
assuming the
parallel postulate
in attempting to prove
the parallel postulate!
For centuries, mathematicians attempted to prove Euclid’s
fifth postulate (the parallel postulate) using the first four. The
development of Non-Euclidean Geometry was an outgrowth
of these attempts, most often using an indirect proof (proof by
contradiction).
Euclid’s
Parallel
One
Negation
of Postulate
Euclid’s Parallel Postulate
more than
Given a line and a point not on the line, exactly
oneone
lineline
parallel to the given line may be drawn through the point.
What other contradictions did mathematicians come up with in
assuming the above negation of the Parallel Postulate?
Theorem: Every
right
angle
is obtuse.
Parallel
lines
do not
form congruent alternate interior s.
Theorem: Rectangles do not exist.
Theorem: Every
right
angle
is obtuse.
Parallel
lines
do not
form congruent alternate interior s.
Theorem: Rectangles do not exist.
Theorem: Parallel lines are not always the same distance apart.
Theorem: Every
right
angle
form
iscongruent
obtuse.
alternate
interiorinterior
angles.s.
Parallel
lines
do
not
form congruent
alternate
Theorem: Rectangles do not exist.
Theorem: Parallel lines are not always the same distance apart.
Theorem: In the right triangle shown,
a2
+
b2

c 2.
c
a
b
Theorem: Every
right
angle
form
iscongruent
obtuse.
alternate
interiorinterior
angles.s.
Parallel
lines
do
not
form congruent
alternate
Theorem: Rectangles do not exist.
Theorem: Parallel lines are not always the same distance apart.
Theorem: In the right triangle shown,
a2
+
b2

c 2.
c
a
b
Theorem: The ratio of the circumference of a
circle to its diameter is not constant
and, therefore, not .
d
C

d
Theorem: Every
right
angle
form
iscongruent
obtuse.
alternate
interiorinterior
angles.s.
Parallel
lines
do
not
form congruent
alternate
Theorem: Rectangles do not exist.
Theorem: Parallel lines are not always the same distance apart.
Theorem: In the right triangle shown,
a2
+
b2

c 2.
c
a
b
Theorem: The ratio of the circumference of a
circle to its diameter is not constant.
and, therefore, not .
d
C

d
Theorem: In a triangle, the 3 angle measures never add up to 180
Theorem: Every
right
angle
form
iscongruent
obtuse.
alternate
interiorinterior
angles.s.
Parallel
lines
do
not
form congruent
alternate
Theorem: Rectangles do not exist.
Theorem: Parallel lines are not always the same distance apart.
Theorem: In the right triangle shown,
a2
+
b2

c 2.
c
a
b
C
But are they really contradictions?
d
Theorem: The ratio of the circumference of a
circle to its diameter is not constant.
and, therefore, not .
d

Theorem: In a triangle, the 3 angle measures never add up to 180
Theorem: All similar triangles are congruent.
77 Statements
whose
proofs depend
on Euclid’s
Postulate.
77 Statements
Equivalent
to Euclid’s
Parallel Parallel
Postulate
Theorem: Every
right
angle
form
iscongruent
obtuse.
alternate
interiorinterior
angles.s.
Parallel
lines
do
not
form congruent
alternate
Theorem: Rectangles do not exist.
Theorem: Parallel lines are not always the same distance apart.
Theorem: In the right triangle shown,
a2
+
b2

c 2.
c
a
b
Theof
ratio
of the circumference
a
C
SoTheorem:
NONE
these
is really ofcontradictions!!
d

circle to its diameter is not constant.
and, therefore, not .
d
Theorem: In a triangle, the 3 angle measures never add up to 180
Theorem: All similar triangles are congruent.
Theorem: Every
right
angle
form
iscongruent
obtuse.
alternate
interiorinterior
angles.s.
Parallel
lines
do
not
form congruent
alternate
Theorem: Rectangles do not exist.
Theorem: Parallel lines are not always the same distance apart.
Theorem: In the right triangle shown,
a2
+
b2

c 2.
c
a
b
Theorem: The ratio of the circumference of a
circle to its diameter is not constant.
and, therefore, not .
d
C

d
Theorem: In a triangle, the 3 angle measures never add up to 180
Theorem: All similar triangles are congruent.
Letter from the mathematician Farkas Bolyai (1775-1856) to his son,
mathematician Janos Bolyai (1802-1860)
You must not attempt this approach to parallels. I know this way to its very
end. I have traversed this bottomless night, which extinguished all light and joy of
my life. I entreat you, leave the science of parallels alone….I thought I would
sacrifice myself for the sake of truth. I was ready to become a martyr who would
remove the flaw from geometry and return it purified to mankind. I accomplished
monstrous, enormous labors; my creations are far better than those of others and
yet I have not achieved complete satisfaction….I turned back when I saw that no
man can reach the bottom of the night. I turned back unconsoled, pitying myself
and all mankind.
I admit that I expect little from the deviation of your lines. It seems to me that I
have been in these regions; that I have traveled past all reefs of this infernal Dead
Sea and have always come back with broken mast and torn sail. The ruin of my
disposition and my fall date back to this time. I thoughtlessly risked my life and
happiness.
er
It is now my definite plan to publish a work on parallels as
soon as I can complete and arrange the material and an
opportunity presents itself; at the moment I still do not clearly
see my way through, but the path which I have followed gives positive
evidence that the goal will be reached, if it is at all possible; I have not
quite reached it, but I have discovered such wonderful things that I was
amazed, and it would be an everlasting piece of bad fortune if they
were lost. When you, my dear father, see them, you will understand; at
present I can say nothing except this: that out of nothing I have created
a strange new universe.
Euclidean Geometry
Euclid’s five postulates and all the theorems
that derive from them.
Euclid's 5 postulates
1. A line segment can be drawn joining any two points.


Euclid's 5 postulates
1. A line segment can be drawn joining any two points.
2. Any line segment can be extended indefinitely in a line.


Euclid's 5 postulates
1. A line segment can be drawn joining any two points.
2. Any line segment can be extended indefinitely in a line.
3. Given any line segment, a circle can be drawn having the segment as radius
and one endpoint as center.


Euclid's 5 postulates
1. A line segment can be drawn joining any two points.
2. Any line segment can be extended indefinitely in a line.
3. Given any line segment, a circle can be drawn having the segment as radius
and one endpoint as center.
4. All right angles are congruent.
Euclid's 5 postulates
1. A line segment can be drawn joining any two points.
2. Any line segment can be extended indefinitely in a line.
3. Given any line segment, a circle can be drawn having the segment as radius
and one endpoint as center.
4. All right angles are congruent.
5. Given a line and a point not on the line, exactly one line parallel to the given line
may be drawn through the point. (Euclid’s Parallel Postulate)


In the early years of the twentieth century, the German
mathematician David Hilbert attempted to fill in the gaps in
Euclid’s Geometry.
David Hilbert
(1862-1943)
What gaps in Euclid’s Geometry do we mean ?
Euclid's 5 postulates
1. A line segment can be drawn joining any two points.
2. Any line segment can be extended indefinitely in a line.
3. Given any line segment, a circle can be drawn having the segment as radius and one
endpoint as center.
4. All right angles are congruent.
5. Given a line and a point not on the line, exactly one line parallel to the given line may be
drawn through the point. (Euclid’s Parallel Postulate)
Here is how we proved the isosceles triangle theorem
• Construct the angle bisector of C.
•
The two triangles are congruent by SAS.
• Therefore, A is congruent to B.

P
But, which one of Euclid’s postulates says that the diagram can’t be like …
We can still prove the triangles
congruent by SAS, but now,
B  CAP (not CAB).
P
In the early years of the twentieth century, the German
mathematician David Hilbert attempted to fill in the gaps in
Euclid’s Geometry. He proposed a series of postulates divided
into four categories: Incidence, Betweenness, Congruence, and
Continuity axioms. These axioms were designed to replace
Euclid’s first four postulates and still enable us to prove all the
theorems of Euclidean Geometry., except those based on
Euclid’s fifth postulate, (the parallel postulate).
David Hilbert
(1862-1943)
The incidence axioms
Incidence Axiom 1: For every point P and every point Q not
equal to P, there exists a unique line l incident with P and Q.
Equivalent to Euclid’s first two postulates
Incidence Axiom 2: For every line l there exist at least two
distinct points incident with l.
Incidence Axiom 3: There exist three distinct points with
the property that no line is incident with all three of them.
Any system with lines and points suitably interpreted that satisfies
all three incidence axioms is called an “Incidence Geometry.”
I-1: For every point P and every point Q not equal to P, there exists a unique line l incident with P and Q.
I-2: For every line l there exist at least two distinct points incident with l.
I-3: There exist three distinct points with the property that no line is incident with all three of them.
The simplest candidate for model of an incidence geometry:
Interpret points as three people: Jay, Kaye, and Emma.
Each puts in a cell phone call (line) to the other.
Cell phones do not have a “conference call” feature.
Is this a model of incidence geometry?
K

J
M
I-1: For every point P and every point Q not equal to P, there exists a unique line l incident with P and Q.
I-2: For every line l there exist at least two distinct points incident with l.
I-3: There exist three distinct points with the property that no line is incident with all three of them.
In this example, the 4 vertices of the tetrahedron are the
points and the 6 edges are the lines. Is this a model of
incidence geometry?
C
B
D
A
I-1: For every point P and every point Q not equal to P, there exists a unique line l incident with P and Q.
I-2: For every line l there exist at least two distinct points incident with l.
I-3: There exist three distinct points with the property that no line is incident with all three of them.
In this example, the 8 vertices of the cube are the points and
the 14 visible edges are the lines. Is this a model of incidence
geometry?
I-1: For every point P and every point Q not equal to P, there exists a unique line l incident with P and Q.
I-2: For every line l there exist at least two distinct points incident with l.
I-3: There exist three distinct points with the property that no line is incident with all three of them.
Interpret points to be the letters in the set {B, A, L, D}.
Interpret lines as all possible subsets of size three. Is this
a model of an incidence geometry?
Points B, A, L, D
Lines {B,A,L} , {B,A,D} , {B,L,D} , {A,L,D}
I-1: For every point P and every point Q not equal to P, there exists a unique line l incident with P and Q.
I-2: For every line l there exist at least two distinct points incident with l.
I-3: There exist three distinct points with the property that no line is incident with all three of them.
In this example, the 5 vertices of the figure are the points and
the 10 connectors are the lines. Is this a model of an
incidence geometry?
Why didn’t I draw the
diagram like this?
Hilbert’s incidence, betweenness, congruence, and
continuity axioms can replace Euclid’s first four
postulates. Hilbert’s axioms prevent ...
David Hilbert
(1862-1943)
But what about Euclid’s fifth postulate, the parallel postulate?
Euclid’s Parallel Postulate
Given a line and a point not on the line, exactly one line
parallel to the given line may be drawn through the point.
Euclid’s Parallel Postulate
Given a line and a point not on the line, exactly one line parallel to the given
line may be drawn through the point.
In this example, the 4 vertices of the tetrahedron are the
points and the 6 edges are the lines. Does Euclid’s parallel
postulate hold in this model?
C
B
D
A
Euclid’s Parallel Postulate
Given a line and a point not on the line, exactly one line parallel to the given
line may be drawn through the point.
In this example, the 5 vertices of the figure are the points and
the 10 connectors are the lines. Does Euclid’s parallel
postulate hold in this model?
Euclid’s Parallel Postulate
Given a line and a point not on the line, exactly one line parallel to the given
line may be drawn through the point.
Does Euclid’s parallel postulate hold in this model?
K

J
M
Proclus (410-485) – seven centuries after Euclid (and 14 centuries before Bolyai)
This [Euclid’s parallel postulate] ought to be struck out of the postulates
altogether; for it is a statement involving many difficulties…. It is alien to the
special character of postulates.
Here is Proclus “proof” of the Parallel Postulate
(given a modern twist)
Given two parallel lines l and m. Suppose line n intersects line m at point P,
and suppose n is also parallel to l.
Let Q be the foot of the perpendicular from P to l.
Since m and n intersect at point P, a point Y can be chosen on n such that
ray PY lies between ray PQ and a ray PR on line m.
Take X to be the foot of the perpendicular from Y to m.
P

X

R

m

Y
Q
n
l
The further Y is chosen from point P, the longer segment XY will become.
As Y recedes along n endlessly from P, the length of segment XY will
eventually become greater than segment PQ.
At that stage, Y must be on the other side of l, so that n must have crossed l.
Contradiction, since n was assumed parallel to l.
Where is the hidden assumption (the flaw in the proof)?
P
X


Y
Q
X


Y
R

m
n
l
The further Y is chosen from point P, the longer segment XY will become.
As Y recedes along n endlessly from P, the length of segment XY will
eventually become greater than segment PQ.
At that stage, Y must be on the other side of l, so that n must have crossed l.
Contradiction, since n was assumed parallel to l.
Where is the hidden assumption (the flaw in the proof)?
P
X
X




Y
Y
R

m
n
l
Q
l
In attempting an indirect proof in this way, mathematicians for
centuries replaced Euclid’s Parallel Postulate
(they hoped temporarily) with:
Given a line and point not on the line, more than one parallel
can be drawn to the line through the point.
No contradiction to the other four Euclidean postulates (or to any
theorem based on them) has ever been reached.nor
norcan
canititbe.
be.
A system in which Hilbert’s incidence, betweenness, congruence,
and continuity axioms hold (they are equivalent to Euclid’s first
four postulates), AND
…………………………………………..
the new postulate
is called Hyperbolic Geometry
And you can prove a lot of strange theorems in Hyperbolic Geometry