Title: WARM UP EXERCSE
1WARM 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.
2Early Beginnings
- In ancient times the special relationship between
a right triangle and the squares on the three
sides was known.
2
3Early Beginnings
3
4Early 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(No Transcript)
63-4 Pythagorean Theorem 2
Pythagorean dissection proof.
a
b
a
a
c
c
b
c
b
c
c
a
b
73-4 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
83-4 Pythagorean Theorem 4
Garfields dissection proof.
a
b
c
c
a
b
½ (a b) (a b) 2 ½ a b ½ c2
9http//www.usna.edu/MathDept/mdm/pyth.html
10Extensions Semicircles
Prove it for homework.
11Extensions Golden Rectangles
Prove it for homework.
12THE 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.
13WARMU 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.
14Pythagoras Revisited
From the previous slide
And of course then, a 2 b 2 cx c(c x)
cx c 2 cx c2
14
15Euclids Proof
http//www.cut-the-knot.org/pythagoras/morey.shtml
15
16First Assignment
Find another proof of the Pythagorean Theorem.
17Cevas 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.
18Cevas 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.
19Cevas Theorem
We will be using the fact that triangles that
have the same altitudes have areas in proportion
to their bases.
h
20Cevas Theorem
Three Cevians concur iff the following is true
We will prove the if part first.
21Cevas 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.
22Cevas 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
23Cevas Theorem
We just saw that
Using the same argument
Thus
24Cevas Theorem
We now move to the only if part.
25Cevas 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
26Cevas Theorem
This is an important theorem and can be used to
prove many theorems where you are to show
concurrency.
27Cevas Theorem
There is also a proof of this theorem using
similar triangles. If you want it I will send you
a copy.
28Menelaus 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!
29Menelaus 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
30Menelaus 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.
31Menelaus 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
32Menelaus Theorem
We now move to the only if part.
Given
Show that L, M, and N are collinear.
33Menelaus 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
34Wrap-up
We looked at several proofs and some history of
the Pythagorean Theorem.
We proved Cevas Theorem.
We proved Menelaus Theorem.
35Next Class
We will cover the lesson Transformations 1.
36Second Assignment Learn the proofs for Cevas
and Menelaus theorems.