What is Pythagoras’ Theorem?
c
a
a2 + b2 = c2
b
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
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.
Following that…
There have been many (well over 300) proofs of the
theorem
Liu Hui  Leonardo Da Vinci  President J.A. Garfield
Three main types of proofs:
Shearing
Similarity
Dissection
Shearing
Shearing
C
B
C1
B1
(ABCD) = (AB1C1D)
D
A
C
C1
(ABC) = (ABC1)
B
A
Shearing
D
C
C1
(ABCD) = 2(ABC1)
B
A
Shearing
Y
R
S
B
X
C
A
Q
P
AC2 + BC2 = AB2
This is proved by showing,
(ACQP) + (BRSC) = (AXYB)
Shearing
Y
B
X
ACQP is a square.
Therefore, AC2 = (ACQP)
C
A
Q
P
B
Q
C
C
B
A
rotate
Q
P
P
A
(ABP) = ½(base)(height)
= ½(AP)(AC)
2(ABP) = (AP)(AC)
height
= (ACQP)
2(ABP) = (ACQP)
base
Shearing
Y
ACQP is a square.
B
X
Therefore, AC2 = (ACQP)
1. 2(ABP) = (ACQP)
C
A
Q
P
Y
U
C
B
V
X
V
height
A
C
U
A
(AXC) = ½(base)(height)
= ½(AX)(AV)
2(AXC) = (AX)(AV)
= (AXUV)
2(AXC) = (AXUV)
base
X
Shearing
Y
U
B
ACQP is a square.
X
Therefore, AC2 = (ACQP)
V
1. 2(ABP) = (ACQP)
C
A
Q
P
2. 2(AXC) = (AXUV)
Shearing
Y
Y
U
B
X
B
X
V
C
A
C
A
Q
P
Q
P
Y
B
B
X
P
C
Q
X
A
P
AP = AC
A
C
A
AB = AX
PAB = 90O + BAC
CAX = 90O + BAC
PAB = CAX
ABP and AXC are congruent triangles. (SAS)
(ABP) = (AXC)
Shearing
Y
ACQP is a square.
U
Therefore, AC2 = (ACQP)
B
X
V
1. 2(ABP) = (ACQP)
2. 2(AXC) = (AXUV)
C
A
3. (ABP) = (AXC)
2(ABP) = 2(AXC)
Q
P
(ACQP) = (AXUV)
Shearing
Y
U
B
X
V
(ACQP) = (AXUV)
C
A
Q
P
Shearing
Y
U
R
BRSC is a square.
B
X
V
S
C
A
Therefore, BC2 = (BRSC)
S
B
C
R
A
S
B
C
A
(ABR) = ½(base)(height)
= ½(BR)(BC)
A
S
R
C
2(ABR) = (BR)(BC)
height
= (BRSC)
2(ABR) = (BRSC)
B
base
R
Shearing
Y
U
R
B
X
V
BRSC is a square.
Therefore, BC2 = (BRSC)
1. 2(ABR) = (BRSC)
S
C
A
Y
Y
U
B
U
X
B
V
C
V
A
(YBC) = ½(base)(height)
= ½ (YB)(BV)
C
V
U
C
2(YBC) = (YB)(BV)
height
= (VUYB)
2(YBC) = (VUYB)
Y
base
B
Shearing
Y
U
R
BRSC is a square.
B
X
V
Therefore, BC2 = (BRSC)
V
1. 2(ABR) = (BRSC)
S
C
A
2.
2(YBC) = (VUYB)
Y
U
R
B
X
Y
V
U
S
C
A
R
S
B
C
X
A
Y
R
U
R
B
X
V
S
C
BR = BC
A
B
C
A
Y
BA = BY
ABR = 90O + CBA
YBC = 90O + CBA
ABR = YBX
ABP and AXC are congruent
triangles. (SAS)
(ABR) = (YBC)
B
Shearing
Y
BRSC is a square.
U
R
B
X
V
S
C
Therefore, BC2 = (BRSC)
1. 2(ABR) = (BRSC)
2. 2(YBC) = (VUYB)
A
3. (ABR) = (YBC)
2(ABR) = 2(YBC)
(BRSC) = (VUYB)
Shearing
Y
U
R
B
X
V
V
S
C
A
(BRSC) = (VUYB)
Shearing
Y
(ACQP) = (AXUV)
U
R
B
X
(AXUV) + (VUYB) = (AXYB)
V
S
C
Q
(BRSC) = (VUYB)
A
(ACQP) + (BRSC) = (AXYB)
AC2 + BC2 = AB2
P
Similarity
Similarity
32
141
52
1 + 2 = 90°
1 + 3 = 90°
Therefore, 3 = 2
4 + 2 = 90°
Therefore, 4 = 1
1
16
1 + 5 = 90°
Therefore, 5 = 2
6 + 2 = 90°
Therefore, 6 = 1
2
b
c
a
x
y
2
b
c
x
a
1
2
2
b
a
1
1
y
All the 3 triangles have the same corresponding angles.
Therefore, they are SIMILAR.
Similarity
2
c
b
2
x
a
1
b
1
By Similarity,
b/c = x/b
x = b2/c
Similarity
2
b
c
a
2
a
1
1
y
By Similarity,
a/c = y/a
y = a2/c
Similarity
c
b
a
x
c = x+y
= b2/c + a2/c
c2 = b2 + a2
y
Dissection
Dissection
Let us begin with a right-angled triangle with
sides a, b, and c (where c is the hypotenuse).
A
c
b
C
C=90
a
B
Angle sum of a triangle =180°
A+B=90°
Dissection
Let us rotate the triangle 90°,
180° and 270° to get four
congruent triangles.
Let us now rearrange the 4 triangles...
c
a b
c
a-b
Area of big square = c2
Recall…
The triangles fit neatly to form the square
because:
Each angle of the square is
formed of A+B…
B
c
A
c
a
1st
From the slide
A+B = 90°
b
Finally, a little calculation:
c2 = area of triangles + area of small square
c
c
a
b
c2 = 4(½ab) + (a-b)2
a-b
c² = 2ab + (a² – 2ab + b²)
c² = a² + b²