WARM UP EXERCSE
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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. A C B b a c h c - x x shortside a h x hypotenuse c b a long side b c x h hypotenuse c b a shortside a h x long side b c x h
description
WARM UP EXERCSE. C. 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. b. a. h. c - x. x. A. c. B. Early Beginnings. - PowerPoint PPT Presentation
Transcript of WARM UP EXERCSE
WARM UP EXERCSEthe 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.
A
C
B
b a
c
hc - x x
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
Early Beginnings
In ancient times the special relationship between a right triangle and the squares on the three sides was known.
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Early Beginnings
OR
3
Indeed, the Assyrians had knowledge of the general form before 2000 b.c.
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Early Beginnings
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)
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§3-4 Pythagorean Theorem 2Pythagorean dissection proof.
a
a
a
a
b
b
bb
cc
c
c
c
=
a
a
a
a
b
b
b
b
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§3-4 Pythagorean Theorem 3Bhaskara’s dissection proof.
a
a
a
bb
b
b
a
c
c
c
c
c2 = 4 · ½ · a · b + (b – a)2
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§3-4 Pythagorean Theorem 4Garfield’s dissection proof.
a
b
b
ac
c
½ (a + b) · (a + b) = 2 · ½ · a · b + ½ · c2
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ExtensionsSemicircles
Prove it for homework.
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ExtensionsGolden Rectangles
Prove it for homework.
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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.
WARMU UP EXERCSEthe 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.
A
C
B
b a
c
hc - x x
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
Pythagoras Revisited
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From the previous slide:
A
C
B
b a
c
h
c - x x
2short side a h xa cx
hypotenuse c b a
2long side b c x hb c(c x)
hypotenuse c b a
And of course then,
a 2 + b 2 = cx + c(c – x) = cx + c 2 – cx = c2
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Eu
clid’s P
roof
http://www.cut-the-knot.org/pythagoras/morey.shtml
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First Assignment
Find another proof of the Pythagorean Theorem.
Ceva’s Theorem
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.
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.
Ceva’s Theorem
Theorem – Triangles with the same altitudes have areas in proportion to their bases.
Proven on the next slide.
To prove this theorem we will need the two following theorems.
Theorema c a e a c e
IF and thenb 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.
A N
C
B1
h ANk ANC 21k NBarea C h NBNBC
area
N
A C N
B
A
2
N
h
Ceva’s Theorem
Three Cevians concur iff the following is true:
We will prove the if part first.
A BN
C
LM D
AN BL CM1
NB LC MA
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
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Given: AL, BM, CN concur
Theorem.
Theorem
(1) & (2) Transitive
What is given? What will we prove?
Why?
Why?
Ratio property and subtraction of areas.
Look at what we have just shown.
A BN
C
LM D
Prove:AN BL CM
1NB LC MA
(1) AN k ANC
NB k NBC
(2) AN k ADN
NB k DBN
(3) AN k ANC k ADN
NB k NBC k DBN
(4)k ANC k ADN
k NBC k DBN
AN k CD
k CD
A
NB B
A BN
C
LM D
D
Why
Why
Continued
Ceva’s Theorem We just saw that:
Using the same argument: A BN
C
LM DD
(4)AN K CDA
NB K CDB
A BN
C
LM D
(5)BL K ABD
LC K CDA
(6)CM K CDB
MA K ABD
ThusAN BL CM k ACD k ABD k CDB
1NB LC MA k CDB k CDA k ABD
Ceva’s Theorem
We now move to the only if part.
Ceva’s Theorem
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First half Ceva’s Theorem
Given.
(1) & (2) Transitive
What is given? What will we prove?
Why?
Why?
Simplification.
Which is true only if N’ = N
Prove: AN, BM, CL concurGiven: AN BL CM1
NB LC MA
(1) AN' BL CM1
N'B LC MA
(2)AN BL CM
1NB LC MA
(3)
(4) AN' AN
N'B NB
Why
Why
Assume AL and BM intersect at D and the other line through D is CN’.
A BN
C
LM D
H
AN' BL CM AN BL CM
N'B LC MA NB LC MA
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
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, 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.
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
A B N
C
LM
AN BL CM1
NB LC MA
Note we will use the convention that NB = - BN
Menelaus’ Theorem
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.
A B N
C
LM
D
We will prove the if part first.
AMN BDN andBDL CML by AA
Menelaus’ Theorem
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Given: M, L, N, Collinear
Previous slide.
Previous slide.
(1) · (2)
What is given? What will we prove?
Why?
Why?
Simplify
Prove: AN BL CM1
NB LC MA
(1) AN AM
NB DB
(2) BL DB
LC CM
(3) AN BL AM DB
NB LC DB CM
(4) AN BL CM1
NB LC AM
Why
Why
A B N
C
LM
D
(5) AN BL CM1
NB LC MA Why MA = - AM
QED
Menelaus’ Theorem
We now move to the only if part.
AN BL CM1
NB LC MAGiven
Show that L, M, and N are collinear.
Menelaus’ Theorem
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First half Menelaus’ Theorem
Given.
(1) & (2) Transitive
What is given? What will we prove?
Why?
Why?
Simplification.
Which is true only if N’ = N
Prove: L, M, N collinear.Given: AN BL CM1
NB LC MA
(1)
(2)
(3)
(4)AN' AN
N'B NB
Why
Why
Assume L, M, N’ ≠ N collinear.
AN' BL CM AN BL CM
N'B LB MA NB LB MA
AN' BL CM1
N'B LC MA
AN BL CM1
NB LC MA
A B N’
C
LM
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.
