What does the Pythagorean theorem say?

In a right triangle, the square of the hypotenuse equals the sum of the squares of the two legs: a² + b² = c². The hypotenuse is the side opposite the right angle.

Why is the middle piece a square?

Its four sides are the four hypotenuses, all of length c. At each corner the straight side of the big square is split into the two acute angles of the triangle and the angle of the middle piece; since the acute angles add up to 90°, so does the remaining angle.

Is this Euclid's proof?

No. Euclid's proof (Elements I.47) draws the squares on the three sides and cuts the largest into two rectangles. The proof shown here is the rearrangement in a square of side a + b, which is older and shorter but needs a line of algebra.

The Pythagorean Theorem

How to use

▶ Play runs the proof, Next → and ← Back step one scene at a time, Reset restores the opening scene and the original figure. Learn more about the controls
Slide $X$ along the bottom side to change $a$ and $b$ , or drag $P$ or $Q$ to move or turn the square; the readout under the canvas keeps $a^2 + b^2$ equal to $c^2$ . Learn more about why it holds
Labels toggles the point names, Measures writes each marked angle's size beside its arc. The Step explanations log keeps every step so far. Learn more about reading a scene

Proof?Put four copies of the right triangle into the corners of a square of side a + b, each turned a quarter turn further than the last. What is left in the middle is a square of side c, because the two acute angles of the triangle add up to 90°. The big square has area (a + b)² = a² + 2ab + b², and also c² + four triangles = c² + 2ab. So a² + b² = c². Drag X to change a and b, or P and Q to move the square.Pythagorean theorem

DisplayLabelsMeasures

Drag X to change a and b, P or Q to move the square

a² + b² = 4.0 + 9.0 = 13.0,  c² = 13.0

Step 1 / 6

Step explanations

1A right triangle with legs a, b and hypotenuse c

Given
The angle at P is right; PX = a, PW = b, WX = c. The two acute angles are α at X and β at W. To prove: a² + b² = c².

The readout shows a² + b² and c² in the frame's units. Learn more about the statement

Getting Started with the Visualizer

The Statement

A Square of Side a Plus b

Four Copies of the Triangle

The Inner Square

Counting the Area Twice

Why It Holds

Dragging the Figure

Consequences

Getting Started with the Visualizer

DemoA run through the proof player

Step 0 of 7

Step through the proof, then drag the point that splits the side into

a

and

b

and watch the squares of the sides keep their sum. The picture below is the player with its parts numbered; the other tabs list what can be dragged, what each control does and what the step log shows.

Opening scene: the controls

1234567

1Proof — the theorem loaded in the player; the ? gives its idea in one line

2Display — Labels shows the point names, Measures writes each marked angle's size beside its arc

3Draggable points — X slides along the bottom side and sets a and b; P and Q move or turn the square; the hollow points follow

4Readout — the line under the figure: a² + b² and c² in the frame's units, live while you drag

5Back, Play, Next, Reset — step through the proof or run it; Reset returns to step 1 and restores the original figure

6Speed and step counter — how fast Play advances, and which of the 6 steps is on screen

7Step explanations — every step reached so far, with its justification tag and a link to the section that explains it; click an entry to jump back to it

Proof complete: step 6 of 6

Proof?Pythagorean theoremDisplayLabelsMeasuresDrag X to change a and b, P or Q to move the square

a² + b² = 4.0 + 9.0 = 13.0,  c² = 13.0

← Back↻ ReplayNext →ResetNormal ▾Step 6 / 6

Step explanations

1A right triangle with legs a, b and hypotenuse c

Given
The angle at P is right; PX = a, PW = b, WX = c. The two acute angles are α at X and β at W. To prove: a² + b² = c².

2A square of side a + b

Construction
Extend PX by b to Q and PW by a to S, and complete the square PQRS of side a + b. Going round it, cut each side at a from one corner and b from the next: the points X, Y, Z, W.

3Four copies of the triangle

SAS congruence
In each corner of the square sits a triangle with legs a and b and a right angle between them. By side-angle-side all four are congruent to the given triangle: each hypotenuse is c, each has area ab/2, and each has the angles α and β.

4The middle piece is a square of side c

Angle sum
XYZW has four sides of length c. At X the straight angle along PQ is made of α, the angle of XYZW, and β; since α + β = 90° in a right triangle, the angle of XYZW at X is 90°. The same holds at Y, Z and W, so XYZW is a square, of area c².

5(a + b)² = c² + 4 · ab/2

Area two ways
The big square has side a + b, so its area is (a + b)². It is also the inner square plus the four triangles: c² + 4 · (ab/2) = c² + 2ab.

6a² + b² = c²

Algebra
Expanding the left side, (a + b)² = a² + 2ab + b². So a² + 2ab + b² = c² + 2ab, and taking 2ab from both sides leaves a² + b² = c². ∎

12345

1The inner square — XYZW, four sides c and four right angles: area c²

2The four triangles — one in each corner, legs a and b, all congruent to the given one: area ab/2 each

3The big square — PQRS, side a + b: its area (a + b)² is counted twice

4Justification tags — Given, Construction, SAS congruence, Angle sum, Area two ways, Algebra

5Replay — at the last step Play becomes Replay and Next is off

The Statement

In a right triangle, with legs

a

and

b

and hypotenuse

c

opposite the right angle,

a^2 + b^2 = c^2

the square on the hypotenuse equals the two squares on the legs together. This is Proposition 47 of Book I of Euclid's Elements, the last substantial result of the book, and the best known theorem in mathematics. The player shows the right angle at

P

, the legs

PX = a

and

PW = b

, the hypotenuse

WX = c

, and the two acute angles

\alpha

and

\beta

.

There are hundreds of proofs. Euclid's own, with the squares drawn on the three sides and a line from the right angle cutting the big square into two rectangles, is the longest page of Book I. The proof on this page is the rearrangement in a square of side

a + b

, found in the ancient Chinese Zhoubi Suanjing and in Bhaskara's work; it is shorter and uses only two facts from the earlier pages, side-angle-side congruence and the angle sum of a triangle. The readout under the canvas shows

a^2 + b^2

and

c^2

side by side, in the frame's own units, live while you drag.

Opening scene, frozen

The right triangle WPX: legs a and b, hypotenuse c, acute angles α and β.

A Square of Side a Plus b

The construction is a square of side

a + b

with the triangle in one corner. Extend the leg

PX = a

b

Q

, extend the leg

PW = b

a

S

, and complete the square

PQRS

. Its side is

a + b

both ways, and its four corners are right angles.

Now go round the square and cut each side the same way: at

a

from one corner and

b

from the next, always turning in the same direction. On

PQ

the cut is

X

itself; on

QR

it is

Y

, with

QY = a

; on

RS

it is

Z

, with

RZ = a

; and on

SP

it is

W

, with

SW = a

, so that

WP = b

, which is where the given triangle's leg already was. In the player these four points are hollow, since they follow

X

: slide

X

and all four move together.

The square of side a + b, frozen

PQRS with each side cut at a from one corner and b from the next: X, Y, Z, W.

Four Copies of the Triangle

In each corner of the big square there is now a triangle:

WPX

XQY

YRZ

ZSW

. Each has a leg

a

, a leg

b

and the right angle of the square between them. Two triangles with two sides and the included angle equal are congruent (side-angle-side), so all four are copies of the given triangle.

Everything else about them follows. Each hypotenuse,

XY

YZ

ZW

and

WX

, has length

c

. Each has area

\frac{1}{2}ab

, half the rectangle on its legs. And each has the same two acute angles,

\alpha

opposite the leg

b

and

\beta

opposite the leg

a

; this last fact is what the next step needs.

Four copies, frozen

A triangle with legs a, b and a right angle in every corner: all four congruent by SAS.

The Inner Square

The piece left in the middle,

XYZW

, has four sides of length

c

. It remains to see that its angles are right angles. Look at

X

. The side

PQ

is a straight line through it, so the three angles at

X

add up to

180°

: the angle

\alpha

of triangle

WPX

, the angle of

XYZW

, and the angle

\beta

of triangle

XQY

.

In a right triangle the two acute angles add up to

90°

, because all three angles add up to

180°

by the angle sum theorem and one of them is the right angle. So the angle of

XYZW

X

180° - \alpha - \beta = 90°

. The same argument at

Y

Z

and

W

shows that every angle of

XYZW

is right. Four equal sides and four right angles:

XYZW

is a square, and its area is

c^2

The inner square, frozen

XYZW: four sides c, and at X the angle between α and β is 180° − 90° = 90°.

Counting the Area Twice

The big square can be measured in two ways. Its side is

a + b

, so its area is

(a + b)^2

. It is also made of exactly five pieces that do not overlap: the inner square, area

c^2

, and the four triangles, area

\frac{1}{2}ab

each. So

(a + b)^2 = c^2 + 4 \cdot \tfrac{1}{2}ab = c^2 + 2ab

This is the whole proof in one line; what remains is to read off

a^2 + b^2

from it.

Area two ways, frozen

(a + b)² for the big square; c² + four halves of ab for its pieces.

Why It Holds

Expand the left side:

(a + b)^2 = a^2 + 2ab + b^2

. Then

a^2 + 2ab + b^2 = c^2 + 2ab

and taking

2ab

from both sides leaves

a^2 + b^2 = c^2

. The end-of-proof mark ∎ in the log closes the argument. The tags name the facts used: the given right triangle, a construction, side-angle-side congruence, the angle sum, area counted two ways, and a line of algebra.

The

2ab

that cancels is the four triangles. That is the idea of every rearrangement proof: the triangles are the same on both sides of the equation, so only the squares matter. Euclid's proof avoids the algebra by cutting the square on

c

into two rectangles and showing each rectangle equal to one of the squares on the legs; the result is the same.

The conclusion, frozen

a² + 2ab + b² = c² + 2ab, so a² + b² = c².

Dragging the Figure

Slide

X

along the bottom side:

a

grows as

b

shrinks, the four triangles change shape together, the inner square turns, and the readout keeps

a^2 + b^2

equal to

c^2

. When

X

is in the middle,

a = b

, the triangles are isosceles right triangles and the inner square is the big square turned through

45°

, with half its area.

Drag

P

Q

to move or turn the whole square; the lengths change but the equation does not. A drag that would push a corner of the square out of the frame, or make the square too small to read, is refused, as is one that would take

X

too close to a corner.

After dragging X, frozen

a longer than b now: the triangles are thinner, the inner square has turned, the equation still holds.

Consequences

• Distances. The distance between two points with coordinates

(x_1, y_1)

and

(x_2, y_2)

\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

: the theorem applied to a right triangle with horizontal and vertical legs.
• Trigonometry. Dividing

a^2 + b^2 = c^2

c^2

gives

\sin^2\theta + \cos^2\theta = 1

, the Pythagorean identity.
• Any triangle. For a triangle that is not right-angled the Law of Cosines,

c^2 = a^2 + b^2 - 2ab\cos\gamma

, adds a correction term that vanishes when

\gamma = 90°

.
• The converse. If

a^2 + b^2 = c^2

then the angle opposite

c

is right (Euclid I.48). This is how a

3

4

5

rope makes a right angle.
• Irrational lengths. The diagonal of a unit square is

\sqrt{2}

, which is not a fraction; the theorem is where irrational numbers first appeared.