proof of Pythagorean theorem

proof of Pythagorean theorem

Let

a

and

b

be two lengths and let

c

denote the length of the

hypotenuse

of a

right triangle

whose legs have lengths

a

and

b

.

{xy}

,

(

0

,

0

)

;

(

0

,

40

)

*

*

@

-

;

(

30

,

0

)

*

*

@

-

;

(

0

,

0

)

*

*

@

-

,

(

-

2

,

15

)

*

a

,

(

20

,

-

2

)

*

b

,

(

21

,

16

)

*

c

Behold the following two ways of dissecting a square of length

a

+

b

:

{xy}

,

(

0

,

30

)

;

(

0

,

0

)

*

*

@

-

;

(

70

,

0

)

*

*

@

-

;

(

70

,

70

)

*

*

@

-

;

(

0

,

70

)

*

*

@

-

;

(

0

,

30

)

*

*

@

-

;

(

40

,

0

)

*

*

@

-

;

(

70

,

40

)

*

*

@

-

;

(

30

,

70

)

*

*

@

-

;

(

0

,

30

)

*

*

@

-

,

(

-

2

,

15

)

*

a

,

(

-

2

,

50

)

*

b

,

(

72

,

20

)

*

b

,

(

72

,

55

)

*

a

,

(

20

,

-

2

)

*

b

,

(

55

,

-

2

)

*

a

,

(

15

,

72

)

*

a

,

(

50

,

72

)

*

b

,

(

21

,

16

)

*

c

,

(

54

,

21

)

*

c

,

(

49

,

54

)

*

c

,

(

16

,

49

)

*

c

,

(

-

2

,

72

)

*

A

,

(

-

2

,

-

2

)

*

B

,

(

72

,

-

2

)

*

C

,

(

72

,

72

)

*

D

,

(

30

,

72

)

*

H

,

(

72

,

40

)

*

G

,

(

40

,

-

2

)

*

F

,

(

-

2

,

30

)

*

E

Figure 1

{xy}

,

(

0

,

0

)

;

(

0

,

70

)

*

*

@

-

;

(

70

,

70

)

*

*

@

-

;

(

70

,

0

)

*

*

@

-

;

(

0

,

0

)

*

*

@

-

,

(

0

,

40

)

;

(

70

,

40

)

*

*

@

-

,

(

30

,

0

)

;

(

30

,

70

)

*

*

@

-

,

(

0

,

40

)

;

(

30

,

0

)

*

*

@

-

,

(

30

,

70

)

;

(

70

,

40

)

*

*

@

-

,

(

15

,

72

)

*

a

,

(

50

,

72

)

*

b

,

(

15

,

-

2

)

*

a

,

(

50

,

-

2

)

*

b

,

(

-

2

,

20

)

*

b

,

(

-

2

,

55

)

*

a

,

(

72

,

20

)

*

b

,

(

72

,

55

)

*

a

,

(

28

,

55

)

*

a

,

(

32

,

20

)

*

b

,

(

15

,

42

)

*

a

,

(

50

,

38

)

*

b

,

(

14

,

19

)

*

c

,

(

51

,

56

)

*

c

,

(

-

2

,

72

)

*

A

′

,

(

-

2

,

-

2

)

*

B

′

,

(

72

,

-

2

)

*

C

′

,

(

72

,

72

)

*

D

′

,

(

30

,

72

)

*

K

,

(

72

,

40

)

*

L

,

(

30

,

-

2

)

*

M

,

(

-

2

,

40

)

*

N

,

(

32

,

38

)

*

O

Figure 2

We now discuss the construction of these diagrams and note some facts about them,
starting with the first one. To construct figure 1, we proceed as follows:

•

Construct a square

A

⁢

B

⁢

C

⁢

D

whose sides have length

a

+

b

.

•

Lay point

E

on the

line segment

A

⁢

B

at a

distance

a

from

B

.
Likewise, lay the point

F

on the line segment

B

⁢

C

at a distance

a

from

C

, lay the point

G

on the line segment

C

⁢

D

at a distance

a

from

D

, and lay the point

H

on the line segment

D

⁢

A

at a distance

a

from

A

.

•

Connect the line segments

E

⁢

F

,

F

⁢

G

,

G

⁢

H

, and

H

⁢

E

.

We now note the following facts about figure 1:

•

Since the lengths of the four sides of the square

A

⁢

B

⁢

C

⁢

D

all equal

a

+

b

and the line segments

E

⁢

B

,

F

⁢

C

,

G

⁢

D

, and

H

⁢

A

were constructed to have length

a

, it follows that the lengths of the line segments

A

⁢

E

,

B

⁢

F

,

C

⁢

G

,

H

⁢

D

all
equal

b

, as indicated upon the figure.

•

Since

A

⁢

B

⁢

C

⁢

D

is a square, the angles

A

⁢

B

⁢

C

,

B

⁢

C

⁢

D

,

C

⁢

D

⁢

A

, and

D

⁢

A

⁢

B

are all

right angles

, and hence equal each other.

•

By the side-angle-side

theorem

, the triangles

H

⁢

A

⁢

E

,

E

⁢

B

⁢

F

,

F

⁢

C

⁢

G

, and

G

⁢

D

⁢

H

are all

congruent

to each other.

•

By

definition

of

c

as the length of a hypotenuse, it follows that the
line segments

E

⁢

F

,

F

⁢

G

,

G

⁢

H

, and

H

⁢

E

all have length

c

, as indicated in the
figure.

•

Because the sum of the angles of a triangle equals two right angles and
the angle

E

⁢

A

⁢

H

is a right angle, it follows that the sum of the angles

A

⁢

E

⁢

H

and

A

⁢

H

⁢

E

equals a right angle. Since the triangle

A

⁢

E

⁢

H

is congruent to the triangle

B

⁢

F

⁢

E

, the angles

A

⁢

H

⁢

E

and

B

⁢

E

⁢

F

are equal. Hence the sum of the angles

A

⁢

E

⁢

H

and

B

⁢

E

⁢

F

equals a right angle. Thus, we may conclude that

H

⁢

E

⁢

F

is a right angle.
By

similar

reasoning, we conclude that

E

⁢

F

⁢

G

,

F

⁢

G

⁢

H

, and

G

⁢

H

⁢

E

are also right angles.

•

Since its sides are of equal length and its angles are all right angles, the

quadrilateral

E

⁢

F

⁢

G

⁢

H

is a square.

To construct figure 2, we proceed as follows:

•

Construct a square

A

′

⁢

B

′

⁢

C

′

⁢

D

′

whose sides have length

a

+

b

.

•

Lay point

N

on the line segment

A

′

⁢

B

′

at a distance

a

from

A

′

.
Likewise, lay the point

M

on the line segment

B

′

⁢

C

′

at a distance

a

from

B

′

, lay the point

L

on the line segment

C

′

⁢

D

′

at a distance

a

from

D

′

, and lay the point

K

on the line segment

D

′

⁢

A

′

at a distance

a

from

A

′

.

•

Connect the line segments

K

⁢

M

,

N

⁢

L

,

M

⁢

N

, and

K

⁢

L

.

We now note the following facts about figure 2:

•

Since the lengths of the four sides of the square

A

′

⁢

B

′

⁢

C

′

⁢

D

′

all equal

a

+

b

and the line segments

A

′

⁢

N

,

A

′

⁢

K

,

B

′

⁢

M

, and

D

′

⁢

L

were constructed to have length

a

, it follows that the lengths of the line segments

B

′

⁢

N

,

C

′

⁢

L

,

C

′

⁢

M

,

D

′

⁢

K

all
equal

b

, as indicated upon the figure.

•

Since

A

′

⁢

B

′

⁢

C

′

⁢

D

′

is a square, the angles

A

′

⁢

B

′

⁢

C

′

,

B

′

′

⁢

C

′

⁢

D

′

,

C

′

⁢

D

′

⁢

A

′

, and

D

′

⁢

A

′

⁢

B

′

are all right angles, and hence equal each other.

•

Since

A

′

⁢

B

′

and

C

′

⁢

D

′

, as opposite sides of a square, are

parallel

and their
subsegments

A

′

⁢

N

and

D

′

⁢

L

have equal length, it follows that

N

⁢

L

is parallel to

A

′

⁢

D

′

and

B

′

⁢

C

′

. Likewise, since

A

′

⁢

D

′

and

B

′

⁢

C

′

, as opposite sides of the
same square, are parallel and their subsegments

A

′

⁢

K

and

B

′

⁢

M

have equal length,
it follows that

K

⁢

M

is parallel to

A

′

⁢

B

′

and

C

′

⁢

D

′

.

•

Since

N

⁢

L

is parallel to

B

′

⁢

C

′

and

A

′

⁢

B

′

⁢

C

′

is a right angle, it follows
that

A

′

⁢

N

⁢

L

is a right angle; moreover, since

A

′

⁢

D

′

⁢

C

′

is a right angle,

C

′

⁢

L

⁢

N

is
also a right angle. Likewise, since

M

⁢

K

is parallel to

C

′

⁢

D

′

and

A

′

⁢

D

′

⁢

C

′

is a
right angle, it follows that

A

′

⁢

K

⁢

M

is a right angle; moreover, since

A

′

⁢

B

′

⁢

C

′

is
a right angle,

C

′

⁢

M

⁢

K

is also a right angle. Since

A

′

⁢

B

′

is parallel to

K

⁢

M

and

A

′

⁢

N

⁢

L

is a right angle, it follows that

K

⁢

O

⁢

L

,

L

⁢

O

⁢

M

,

M

⁢

O

⁢

N

, and

N

⁢

O

⁢

K

are all
right angles.

•

Since all the angles of the quadrilateral

A

′

⁢

K

⁢

O

⁢

N

are right angles and two
of its

adjacent sides

,

A

′

⁢

K

and

A

′

⁢

N

, have the same length

a

, this figure is a
square, hence the remaining sides,

O

⁢

K

and

O

⁢

N

, also have length

a

. Likewise,
Since all the angles of the quadrilateral

C

′

⁢

L

⁢

O

⁢

M

are right angles and two
of its adjacent sides,

C

′

⁢

L

and

C

′

⁢

M

, have the same length

b

, this figure is a
square, hence the remaining sides,

O

⁢

L

and

O

⁢

M

, also have length

b

.

•

By the side-angle-side theorem, the traingles

K

⁢

O

⁢

L

,

L

⁢

D

′

⁢

K

,

M

⁢

B

′

⁢

N

, and

N

⁢

O

⁢

M

are congruent to each other and to the triangles

H

⁢

A

⁢

E

,

E

⁢

B

⁢

F

,

F

⁢

C

⁢

G

, and

G

⁢

D

⁢

H

.

Hence, we have shown that a square with sides of length

a

+

b

may be dissected into
either

•

a square

A

′

⁢

K

⁢

O

⁢

N

with sides of length

a

,

•

a square

C

′

⁢

L

⁢

O

⁢

M

with sides of length

b

,

•

four right triangles,

K

⁢

O

⁢

L

,

L

⁢

D

′

⁢

K

,

M

⁢

B

′

⁢

N

, and

N

⁢

O

⁢

M

with sides of length

a

,

b

,

c

or

•

a square

E

⁢

F

⁢

G

⁢

H

with sides of length

c

,

•

four right triangles,

H

⁢

A

⁢

E

,

E

⁢

B

⁢

F

,

F

⁢

C

⁢

G

, and

G

⁢

D

⁢

H

, with sides of length

a

,

b

,

c

.

Since the area of the whole equals the sum of the areas of the parts and the squares

A

⁢

B

⁢

C

⁢

D

and

A

′

⁢

B

′

⁢

C

′

⁢

D

′

are congruent, hence have equal areas, it follows that the
sum of the areas of the figures comprising the former dissection equals the sum of the
areas of the figures comprising the latter dissection. Since the triangles involved in
these dissections all are congruent, hence have equal areas, we may cancel their areas
to conclude that the area of

E

⁢

F

⁢

G

⁢

H

equals the sum of the areas of

A

′

⁢

K

⁢

O

⁢

N

and

C

′

⁢

L

⁢

O

⁢

M

,
or

a

2

+

b

2

=

c

2

.

Title

proof of Pythagorean theorem

Canonical

name

ProofOfPythagoreanTheorem12

Date of creation

2013-03-22 17:09:35

Last modified on

2013-03-22 17:09:35

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

22

Author

rspuzio (6075)

Entry type

Proof

Classification

msc 51-00

Related topic

GeneralizedPythagoreanTheorem

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