Pythagorean theorem

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Consider a right triangle, with short “leg”

a

, long “leg”

b

and hypotenuse

c

, and have it stand on its short “leg”

a

;
Add a congruent triangle standing on its long “leg”

b

, with each hypotenuse meeting at a common vertex;

By joining the top vertices of both triangles, you get a trapezium made up of three triangles;

The trapezium has base

a + b

and parallel sides

a

and

b

, with area

(a + b) a + b 2

;
The three triangles are our two former congruent triangles, each with area

a b 2

;

And since the two acute angles of a right triangle are complementary;

The third triangle is a right triangle with both sides

c

, and area

c 2 2

;

Equating the areas, we have

(a + b) a + b 2 = 2 a b 2 + c 2 2

;

or

a 2 + 2 a b + b 2 2 = 2 a b + c 2 2

;

yielding

Pythagoras theorem a 2 + b 2 = c 2

The Pythagorean theorem provides a way to calculate the length of the longest side of a right triangle (one with a right angle for one of the vertices) in the Euclidean plane if we only know the lengths of the two shorter sides.

A simple proof

A B C D a b c α β α β

Almost as simple as Einstein’s proof of the Pythagorean theorem, this proof does not require the concept of area! This simple proof of the Pythagorean theorem is based on the linear proportionality of the sides of similar triangles. Consider a right triangle

ABC

, where

A

is the vertex at angle

α

(with opposite side

a

),

B

is the vertex at angle

β

(with opposite side

b

) and

C

is the vertex at the right angle (with opposite side

c

, the hypothenuse). Draw the height from

C

down to (thus perpendicularly to) the hypotenuse

c

at point

D

. The right triangles

ACD

and

CBD

are similar to the right triangle

ABC

(all with angles

α, β, π  2 ,

since

α + β = π  2

). With

d   :=   | AD |

and

e   :=   | DB |

, we thus have

α:  a c  =  e a  ⟹  a 2  =  c e,

β:  b c  =  d b  ⟹  b 2  =  c d.

Thus

a 2 + b 2  =  c (d + e)  =  c 2,

since

d + e = c

.

Proofs by dissection

Proofs by dissection without rearrangement

Einstein’s proof by dissection without rearrangement

Einstein’s proof of the Pythagorean theorem is as simple as possible... This simple proof of the Pythagorean theorem is based on the quadratic proportionality of the areas of similar triangles. Consider the figure in the above simple proof. Since the right triangles

ACD

and

CBD

are similar to the right triangle

ABC

, we have

A CBD A ABC  =  a 2 c 2  ,

A ACD A ABC  =  b 2 c 2  .

Since

A CBD + AACD  =  AABC ,

we get

a 2 c 2   AABC +   b 2 c 2   AABC  =  AABC

which simplifies to

a 2 + b 2  =  c 2.

Garfield’s proof of the Pythagorean theorem

See (a stylized rendition of!) Garfield's proof at the top of this page. James Garfield, the 20^th president of the United States, gave this proof (via trapezium) of the Pythagorean Theorem in 1876.

Proofs by dissection with rearrangement

A classic proof

The classic proof by Euclid is Proposition 47 of Book I of The Elements. It requires results from earlier in the book, like the triangle side-angle equality. Since we want to get to the number-theoretic relevance of the theorem more quickly, we will instead present a proof that relies on much fewer geometrical axioms and theorems.

Theorem (Pythagorean theorem). (Pythagoras)

Given a right triangle with the two shorter sides (the “legs”) being

a

and

b

, and the longest side (the hypotenuse) being

c

, the equality

c 2 = a 2 + b 2

holds. In other words, if

a

and

b

are known, then

c

can be computed as

c = 2√  a 2 + b 2

.

Proof. [by dissection] (Bhāskara).

1. Draw a right triangle of whatever size you wish, the only constraint being that one of the angles must be precisely 90° (

π  2

radians). Label the shorter sides

a

and

b

, and the longest side

c

.

a b c

2.

a

is a line segment: extend it

b

units. (This can be accomplished with straightedge and compass by drawing a circle centered where

a

meets

b

with radius

b

).

3. Next, use the segment you just extended to construct a square with each side measuring

a + b

, thus having area

(a + b) 2

; the triangle’s area should be within the square’s area.

4. Draw a line perpendicular to

b

that touches

b

where

b

meets

c

.

5. Extend

b

by

a

units (this is easy now thanks to the square).

a b c

6. Observe that the square now contains two smaller squares, as well as two non-square rectangles. The area of one square is

a 2

, the area of the other is

b 2

. Both rectangles have an area of

a b

. Therefore,

(a + b) 2 = a 2 + b 2 + 2 a b

.

7. Since the area of the triangle is

a b 2

and the combined area of the two rectangles is

2 a b

, this means we can easily break both rectangles into four triangles of the same size as our original triangle. While maintaining the size and shape of the larger square, rearrange the four triangles so that the shorter sides of each triangle are all on the perimeter of the larger square.

a b c

8. Note that this rearrangement has created a new square inside the larger square, and each side of the new square is the hypotenuse of one of the four triangles. Therefore,

(a + b) 2 = c 2 + 4   a b 2

. Using the equality obtained in Step 6, we have

a 2 + b 2 + 2 a b = c 2 + 2 a b

, which simplifies to

a 2 + b 2 = c 2

as specified by the theorem. □

There are many other proofs of this theorem, and even a President of the United States has done one, and there is an entire book of different proofs.

Pythagorean triples

One aspect of the Pythagorean theorem that is of particular interest to number theorists is those cases of

c ∈ ℤ+

where the equation has integer solutions; the three integers are then called “Pythagorean triples” (see A046083, A046084 and A009000).

Euclidean 2-space and Euclidean distance

In Euclidean 2-space, the Euclidean distance between two points with Cartesian coordinates

p1 = (x1, y1)

and

p2 = (x2, y2 )

is defined as the length of the hypotenuse of the right triangle with sides (or “legs”) of length equal to

| x2  −  x1 |

and

|   y2  −  y1 |

, i.e.

d12  :=  2√  (x2 − x1) 2 + (  y2 − y1) 2 .

Euclidean 3-space and Euclidean distance

In Euclidean 3-space, we obtain the Euclidean distance between two points with Cartesian coordinates

p1 = (x1, y1, z1)

and

p2 = (x2, y2, z2 )

by applying the Pythagorean theorem twice, giving

d12  :=  2√  (x2 − x1) 2 + (  y2 − y1) 2 + (z2 − z1) 2 .

The definition for Euclidean distance in Euclidean

n

-space follows by applying the Pythagorean theorem

n  −  1

times.

See also

• Fermat’s last theorem (

  a n + b n = c n, n   ≥   1,

  has no solutions in positive integers

  a, b, c,

  for

  n   ≥  3

  )

External links

• A. Bogomolny, Pythagorean Theorem and its many proofs, from Interactive Mathematics Miscellany and Puzzles.