Transcript WARM UP EXERCSE

WARM UP EXERCSE
the altitude to the right angle of a right
triangle forms two new right triangles
which are similar to the original right
A
triangle. List all the proportions you
can among these three triangles.
short side
a h x
  
hypotenuse c b a
long side
b cx h
 

hypotenuse c
b
a
short side a
h
x
 

long side
b cx h
C
b
c-x
c
h
a
x
B
Early Beginnings
In ancient times the
special relationship
between a right
triangle and the
squares on the three
sides was known.
2
Early Beginnings
OR
3
Early Beginnings
Indeed, the Assyrians had knowledge of the
general form before 2000 b.c.
The Babylonians had knowledge of all of the
Pythagorean triples and had a formula to
generate them.
(3,4,5)
( 5, 12, 13)
( 7, 24, 25)
( 8, 15, 17)
( 9, 40, 41)
(11, 60, 61)
(12, 35, 37)
(13, 84, 85)
(16, 63, 65)
(20, 21, 29)
(28, 45, 53)
(33, 56, 65)
(36, 77, 85)
(39, 80, 89)
(48, 55, 73)
(65, 72, 97)
4
5
§3-4 Pythagorean Theorem 2
Pythagorean dissection proof.
b
a
a
b
a
a
b
b
=
b
a
c
c
a
b
a
c
c
b
c
b
a
6
§3-4 Pythagorean Theorem 3
Bhaskara’s dissection proof.
c
a
c
b
a
b
b
a
b
c
a
c
c2 = 4 · ½ · a · b + (b – a)2
7
§3-4 Pythagorean Theorem 4
Garfield’s dissection proof.
a
b
c
a
c
b
½ (a + b) · (a + b) = 2 · ½ · a · b + ½ · c2
8
http://www.usna.edu/MathDept/mdm/pyth.html
9
Extensions
Semicircles
Prove it for homework.
10
Extensions
Golden Rectangles
Prove it for homework.
11
THE GENERAL EXTENSION TO PYTHAGORAS'
THEOREM: If any 3 similar shapes are drawn on the
sides of a right triangle, then the area of the shape on the
hypotenuse equals the sum of the areas on the other two
sides.
12
WARMU UP EXERCSE
the altitude to the right angle of a right
triangle forms two new right triangles
which are similar to the original right
A
triangle. List all the proportions you
can among these three triangles.
short side
a h x
  
hypotenuse c b a
long side
b cx h
 

hypotenuse c
b
a
short side a
h
x
 

long side
b cx h
C
b
c-x
c
h
a
x
B
Pythagoras Revisited
b
From the previous slide:
a
c-x
B
A
short side
a h x
    a 2  cx
hypotenuse c b a
long side
b cx h
2
 
  b  c(c  x)
hypotenuse c
b
a
And of course then,
a 2 + b 2 = cx + c(c – x) = cx + c 2 – cx = c2
14
Euclid’s Proof
http://www.cut-theknot.org/pythagoras/more
15
y.shtml
First Assignment
Find another proof of
the Pythagorean
Theorem.
16
Ceva’s Theorem
The theorem is often attributed to Giovanni Ceva, who
published it in his 1678 work De lineis rectis. But it was
proven much earlier by Yusuf Al-Mu'taman ibn Hűd,
an eleventh-century king of Zaragoza.
Definition – A Cevian is a line from a vertex of a triangle
through the opposite side.
Altitudes, medians and angle bisectors are examples of
Cevians.
Ceva’s Theorem
To prove this theorem we will need the two following
theorems.
Theorem – Triangles with the same altitudes have areas
in proportion to their bases.
Proven on the next slide.
Theorem
a c
a e
a ce
IF  and  then 
b d
b f
b df
Proof as homework.
Ceva’s Theorem
We will be using the fact that triangles that have the
same altitudes have areas in proportion to their bases.
C
h
A
N
B
1
h
AN
area ANC k ANC 2
AN



area NBC k NBC 1
NB
h NB
2
Ceva’s Theorem
Three Cevians concur iff the following is true:
C
M
A
D
L
AN BL CM
1
NB LC MA
N
We will prove the if part first.
B
Ceva’s Theorem
We will be using the fact that triangles that have the
same altitudes have areas in proportion to their bases
and the following three ratios to prove our theorem.
AN BL CM
,
,
NB LC MA
We will begin with the ration AN/NB.
Ceva’s Theorem
AN BL CM
What
we prove?
1
Prove:will
NB LC MA
Theorem.
Why?
Given:
BM, CN concur
What isAL,
given?
(1)
(2)
(3)
AN k ANC

NB k NBC
AN k ADN

NB k DBN
AN k ANC k ADN


NB k NBC k DBN
Theorem
Why?
(1)
& (2) Transitive
Why
Ratio property and
Why
subtraction of areas.
AN k ANC  k ADN k CDA


NB k NBC  k DBN
k CDB
(4)
Look at what we have just
shown.
C
Continued
M
A
D
D
N
C
L
B
M
D
A
N
L
B
22
Ceva’s Theorem
We just saw that:
(4)
AN K CDA

NB
K CDB
Using the same argument:
(5)
BL K ABD

LC K CDA
(6)
CM K CDB

MA
K ABD
C
A
L
N
B
C
Thus
AN BL CM
k ACD k ABD k CDB

1
NB LC MA
k CDB k CDA k ABD
DD
M
M
A
D
N
L
B
Ceva’s Theorem
We now move to the only if part.
Ceva’s Theorem
AN BL CM
1
NB LC MA
Given:
What is given?
Prove:
What will
AN,we
BM,
prove?
CL concur
Assume AL and BM intersect at D and the other line through D
is CN’.
First half Ceva’s Theorem
Why?
(2)
AN' BL CM
1
N'B LC MA
AN BL CM
 1
NB LC MA
(3)
AN' BL CM AN BL CM

N'B LC MA NB LC MA
(4)
AN' AN

N'B NB
(1)
Given.
Why?
(1) & (2) Transitive
Why
C
Simplification.
Why
Which is true only if N’ = N
M
A
L
D
H N25
B
Ceva’s Theorem
This is an important theorem and can be
used to prove many theorems where you are
to show concurrency.
Ceva’s Theorem
There is also a proof of this theorem using
similar triangles. If you want it I will send
you a copy.
Menelaus’ Theorem
Menelaus' theorem, named for Menelaus (70-130ad)
of Alexandria. Very little is known about Menelaus's
life, it is supposed that he lived in Rome, where he
probably moved after having spent his youth in
Alexandria. He was called Menelaus of Alexandria
by both Pappus of Alexandria and Proclus.
This is the duel of Ceva’s Theorem. Whereas
Ceva’s Theorem is used for concurrency,
Menelaus’ Theorem is used for colinearity of
three points.
AND its proof is easier!
Menelaus’ Theorem
Given points A, B, C that form triangle ABC, and
points L, M, N that lie on lines BC, AC, AB, then
AN BL CM
L, M, N are collinear if and only if NB LC MA   1
C
Note we will use the
convention that NB = - BN
M
L
A
B
N
Menelaus’ Theorem
We will prove the if part first.
Begin by constructing a line parallel to AC through
B. It will intersect MN in a new point D producing
two sets of similar triangles.
C
AMN  BDN and
BDL  CML by AA
M
L
A
D
B
N
Menelaus’ Theorem
AN BL CM
NB LC MA
What will we prove?
Prove:
 1
Given:
L, N, Collinear
What isM,
given?
(1) AN  AM
(2)
(3)
Previous slide.
Why?
NB DB
BL
DB

LC CM
AN BL AM DB

NB LC DB CM
(4)
AN BL CM
1
NB LC AM
(5)
AN BL CM
1
NB LC MA
Previous slide.
Why?
(1)
· (2)
Why
Simplify
Why
C
M
QED
A
Why MA = - AM
L D
B
31
N
Menelaus’ Theorem
We now move to the only if part.
Given
AN BL CM
1
NB LC MA
Show that L, M, and N are collinear.
Menelaus’ Theorem
AN BL CM
1
NB LC MA
Given:
What is given?
Prove:
What will
L, M,
weNprove?
collinear.
Assume L, M, N’ ≠ N collinear.
(1)
AN' BL CM
1
N'B LC MA
(2) AN BL CM   1
NB LC MA
(3)
AN' BL CM AN BL CM

N'B LB MA NB LB MA
(4)
AN' AN

N'B NB
First
Why?
half Menelaus’ Theorem
Why?
Given.
(1) & (2) Transitive
Why
Why
Simplification.
Which is true
only if N’ = N
C
M
A
L
B
33
N’
Wrap-up
We looked at several proofs and some
history of the Pythagorean Theorem.
We proved Ceva’s Theorem.
We proved Menelaus’ Theorem.
Next Class
We will cover the lesson Transformations 1.
Second Assignment
Learn the proofs for
Ceva’s and Menelaus’
theorems.