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WARM UP EXERCSE

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Title: WARM UP EXERCSE


1
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 triangle. List all
the proportions you can among these three
triangles.
2
Early Beginnings
  • In ancient times the special relationship between
    a right triangle and the squares on the three
    sides was known.

2
3
Early Beginnings
  • OR

3
4
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
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6
3-4 Pythagorean Theorem 2
Pythagorean dissection proof.
a
b
a
a
c
c

b
c
b
c
c
a
b
7
3-4 Pythagorean Theorem 3
Bhaskaras dissection proof.
c
a
a
b
b
c
c
b
a
a
b
c
c2 4 ½ a b (b a)2
8
3-4 Pythagorean Theorem 4
Garfields dissection proof.
a
b
c
c
a
b
½ (a b) (a b) 2 ½ a b ½ c2
9
http//www.usna.edu/MathDept/mdm/pyth.html
10
Extensions Semicircles
Prove it for homework.
11
Extensions Golden Rectangles
Prove it for homework.
12
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.
13
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 triangle. List all
the proportions you can among these three
triangles.
14
Pythagoras Revisited
From the previous slide
And of course then, a 2 b 2 cx c(c x)
cx c 2 cx c2
14
15
Euclids Proof
http//www.cut-the-knot.org/pythagoras/morey.shtml
15
16
First Assignment
Find another proof of the Pythagorean Theorem.
17
Cevas 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 Hud, 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.
18
Cevas 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.
Proof as homework.
19
Cevas Theorem
We will be using the fact that triangles that
have the same altitudes have areas in proportion
to their bases.
h
20
Cevas Theorem
Three Cevians concur iff the following is true
We will prove the if part first.
21
Cevas 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.
We will begin with the ration AN/NB.
22
Cevas Theorem
What will we prove?
Given AL, BM, CN concur
What is given?
Theorem.
Why?
Theorem
Why?
(1) (2) Transitive
Why
Ratio property and subtraction of areas.
Why
Look at what we have just shown.
Continued
22
23
Cevas Theorem
We just saw that
Using the same argument
Thus
24
Cevas Theorem
We now move to the only if part.
25
Cevas Theorem
What is given?
What will we prove?
Prove AN, BM, CL concur
Assume AL and BM intersect at D and the other
line through D is CN.
First half Cevas Theorem
Why?
(1)
Given.
Why?
(2)
(1) (2) Transitive
Why
(3)
Simplification.
Why
Which is true only if N N
25
26
Cevas Theorem
This is an important theorem and can be used to
prove many theorems where you are to show
concurrency.
27
Cevas Theorem
There is also a proof of this theorem using
similar triangles. If you want it I will send you
a copy.
28
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 Cevas Theorem. Whereas
Cevas Theorem is used for concurrency, Menelaus
Theorem is used for colinearity of three points.
AND its proof is easier!
29
Menelaus Theorem
Given points A, B, C that form triangle ABC, and
points L, M, N that lie on lines BC, AC, AB,
then L, M, N are collinear if and only if
Note we will use the convention that NB - BN
30
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.
31
Menelaus Theorem
What will we prove?
Given M, L, N, Collinear
What is given?
Previous slide.
Why?
Previous slide.
Why?
(1) (2)
Why
Simplify
Why
Why
MA - AM
QED
31
32
Menelaus Theorem
We now move to the only if part.
Given
Show that L, M, and N are collinear.
33
Menelaus Theorem
What is given?
What will we prove?
Prove L, M, N collinear.
Assume L, M, N ? N collinear.
First half Menelaus Theorem
Why?
(1)
Given.
Why?
(2)
(1) (2) Transitive
(3)
Why
Simplification.
Why
Which is true only if N N
33
34
Wrap-up
We looked at several proofs and some history of
the Pythagorean Theorem.
We proved Cevas Theorem.
We proved Menelaus Theorem.
35
Next Class
We will cover the lesson Transformations 1.
36
Second Assignment Learn the proofs for Cevas
and Menelaus theorems.
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