Pythagoras

1
Pythagoras
2
What is Pythagoras Theorem?
a2 b2 c2

• The famous British mathematician and science
  writer Jacob Bronowski considered this the most
  important mathematical result of all times
  because it connects space with numbers in a
  dramatic way

3
Was Pythagoras the Only One Who Knew?

• We have evidence that the Babylonians knew this
  relationship some 1000 years earlier.
• Plimpton 322, a Babylonian mathematical tablet
  dated back to 1900 B.C., contains a table of
  Pythagorean triplets.

4
Following that
There have been many (well over 300) proofs of
the theorem Liu Hui ? Leonardo Da Vinci ?
President J.A. Garfield
Shearing Similarity Dissection
Three main types of proofs
5
Shearing
6
Shearing
(ABCD) (AB1C1D)
(ABC) (ABC1)
7
Shearing
(ABCD) 2(ABC1)
8
Shearing
AC2 BC2 AB2
This is proved by showing,
(ACQP)
(BRSC)
(AXYB)
9
Shearing
ACQP is a square. Therefore,
AC2 (ACQP)
10

(ABP) ½(base)(height)
½(AP)(AC) 2(ABP) (AP)(AC)
(ACQP)

2(ABP) (ACQP)
base
11
Shearing
ACQP is a square.
Therefore, AC2 (ACQP)
1. 2(ABP) (ACQP)
12
Y
U
X
V
(AXC) ½(base)(height)
½(AX)(AV) 2(AXC) (AX)(AV)
(AXUV)
2(AXC) (AXUV)
13
Shearing
ACQP is a square.
Therefore, AC2 (ACQP)
1. 2(ABP) (ACQP)
2. 2(AXC) (AXUV)
14
Shearing
15
AP AC
AB AX
?PAB 90O ?BAC
?CAX 90O ?BAC
?PAB ?CAX
ABP and AXC are congruent triangles.
(SAS)
(ABP) (AXC)
16
Shearing
Y
ACQP is a square.
Therefore, AC2 (ACQP)

U
B
X

V
1. 2(ABP) (ACQP)
2. 2(AXC) (AXUV)
A
C
3. (ABP) (AXC)
2(ABP) 2(AXC) (ACQP)
(AXUV)
Q
P
17

Shearing
Y

U

B
(ACQP) (AXUV)
X

V
A
C
Q
P
18
Shearing
BRSC is a square. Therefore, BC2
(BRSC)


19




(ABR) ½(base)(height)
½(BR)(BC) 2(ABR) (BR)(BC)
(BRSC)

2(ABR) (BRSC)
20
Shearing
BRSC is a square. Therefore, BC2
(BRSC)
1. 2(ABR) (BRSC)


21


(YBC) ½(base)(height) ½
(YB)(BV) 2(YBC) (YB)(BV)
(VUYB)




2(YBC) (VUYB)
22
Shearing
BRSC is a square. Therefore, BC2
(BRSC)
V
1. 2(ABR) (BRSC)
2. 2(YBC) (VUYB)


23
(No Transcript)
24
BR BC
BA BY
ABP and AXC are congruent triangles. (SAS)
?ABR 90O ?CBA
?YBC 90O ?CBA
(ABR) (YBC)
?ABR ?YBX
25
Shearing
BRSC is a square. Therefore, BC2
(BRSC)
1. 2(ABR) (BRSC)
2. 2(YBC) (VUYB)
V
3. (ABR) (YBC)
2(ABR) 2(YBC) (BRSC)
(VUYB)
26
Shearing
(BRSC) (VUYB)
V
27
Shearing
(ACQP) (AXUV)

(BRSC) (VUYB)

(AXUV) (VUYB) (AXYB)
(ACQP) (BRSC) (AXYB)

AC2 BC2 AB2
28
Similarity
29
Similarity
3
2
4
1
2
5
1
2
6
1

• 1 ?2 90

?1 ?3 90
Therefore, ?3 ?2

• 1 ?5 90

?4 ?2 90
Therefore, ?5 ?2
Therefore, ?4 ?1

• 6 ?2 90

Therefore, ?6 ?1
30
All the 3 triangles have the same corresponding
angles. Therefore, they are SIMILAR.
31
Similarity
By Similarity, b/c x/b
x b2/c
32
Similarity
By Similarity, a/c y/a
y a2/c
33
Similarity
c x y
b2/c a2/c
c2 b2 a2
34
Dissection
35
Dissection
Let us begin with a right-angled triangle with
sides a, b, and c (where c is the hypotenuse).
?C90?
Angle sum of a triangle 180
?A?B90
36
Dissection
Let us rotate the triangle 90, 180 and 270 to
get four congruent triangles.
37
Let us now rearrange the 4 triangles...
Area of big square c2
38
The triangles fit neatly to form the square
because
Each angle of the square is formed of ?A?B
a
b
From the 1st slide ?A?B 90
39
Finally, a little calculation
c2 area of triangles area of small square
c2 4(½ab) (a-b)2
c² 2ab (a² 2ab b²)
c² a² b²