The Rearrangement Proof of the Pythagorean Theorem

The rearrangement proof shows that a² + b² = c² by packing four copies of a right triangle into a square of side a + b in two different ways. One arrangement leaves a tilted square of area c² uncovered; the other leaves two squares of area a² and b². Same square, same triangles, so the uncovered areas must be equal.

It is the classic visual proof of the Pythagorean theorem, often called a proof without words because the picture carries the whole argument. This page adds the words: what each figure shows, why each step is allowed, and a complete written proof you could hand in.

The proof, figure by figure

Start with any right triangle. Call its legs a and b and its hypotenuse c (if you need a refresher on the parts, see right triangles). Cut out four identical copies.

1. Two equal squares. Draw two squares with side a + b. They have the same area, (a + b)². You never need to work that number out.
2. Arrangement 1. In the first square, put one triangle in each corner with its right angle in the corner, so that each side of the square is split into a and then b as you go round. The four hypotenuses enclose a tilted shape whose sides all have length c. The next section proves it is a square, so its area is c².
3. Arrangement 2. In the second square, join the triangles in pairs along their hypotenuses to make two rectangles, each a by b, and push them into opposite corners. What is left uncovered is a square of side a and a square of side b, with total area a² + b².
4. Compare. In both cases, big square = four triangles + uncovered part. The big squares are equal and the triangles are identical, so the uncovered parts have equal areas: c² = a² + b².

The animation does steps 2 to 4 inside one square. Triangle 1 slides diagonally down and to the left, triangle 2 slides right by a, triangle 3 slides up by b, and triangle 4 stays put. The order matters only for neatness: in this order no triangle ever passes over another. Sliding a shape doesn't change its area, and that is the entire point.

Why the inner shape is a square

Four equal sides are not enough to make a square. A rhombus, a "pushed-over" square, also has four equal sides, and its area is less than side × side. So the proof has to show that the tilted shape has right angles. This is the step most explanations skip.

Call the two acute angles of the triangle α (opposite side a) and β (opposite side b). Look at the point on the bottom edge where two triangles touch. Three angles sit there side by side along the straight edge: β from the left triangle, the corner θ of the tilted shape, and α from the right triangle. Angles that make a straight line add to 180°:

The three angles of any triangle also add to 180°. In a right triangle one of them is 90°, so the other two add to 90°: α + β = 90°. Subtracting gives θ = 90°. The same argument works at all four corners, so the tilted shape has four equal sides and four right angles. It is a square, and its area is c².

Arrangement 2 needs the same kind of check. Two copies of the triangle joined along their hypotenuse form a four-sided shape with sides a, b, a, b. Two of its corners are the triangles' right angles, and each of the other two corners is made of α + β = 90°. Four right angles: it really is an a × b rectangle.

The complete written proof

Notice that the proof never expands (a + b)². If you do expand it in step 3 and simplify, you get the algebraic proof, which uses only the first arrangement. Cut the first arrangement in half along a diagonal of the tilted square and you get Garfield's trapezoid proof.

What the picture leaves unsaid

A proof without words is convincing at a glance, which is also its danger: the eye fills in things the logic must earn. Before you accept the picture, it should answer three questions.

• Is the tilted shape really a square? Yes, by the angle argument above. It is not obvious, and "it looks square" is not a reason.
• Do the pieces fit with no gaps or overlaps? Yes, because we started from the big square and cut it, rather than pushing triangles together and hoping the edges come out straight.
• Does it work for every right triangle, not just the one drawn? Yes. Nothing in the argument used particular values of a and b. If a = b, the two small squares in arrangement 2 are simply equal, and the proof still works.

Pictures like this one are a long-running tradition: the Mathematical Association of America has published "proofs without words" for decades, and Roger Nelsen's collections of them include versions of this one.

Common mistakes

• Splitting the edges the wrong way. Each edge must be split a, b, a, b, … going round the square. If you put two a pieces next to each other at a corner, the triangles won't fit and the middle shape isn't a square.
• Calling the middle shape a square without a reason. In a written proof you need the 180° argument, or your teacher will (rightly) ask for it.
• Using the theorem inside its own proof. The side of the tilted square is c because it is a hypotenuse, not because √a² + b² = c. Using the formula there would be circular reasoning.

Who discovered it?

Nobody knows. The figure of four triangles around a tilted square is ancient: a closely related diagram appears in the Chinese commentary tradition of the 3rd century CE (see the Chinese proof), and the same pieces reappear in Bhāskara's proof. In the 19th century, the German scholars Carl Anton Bretschneider and Hermann Hankel suggested that Pythagoras himself might have proved the theorem this way. It is an attractive guess, but it can't be checked, because nothing written by Pythagoras survives. Thomas Heath discusses the suggestion in his notes on Euclid's Proposition I.47, whose own proof is quite different (see Euclid's proof).

Make it with paper

This proof works beautifully as a hands-on activity because students physically move the pieces. Allow about 20 minutes.

1. Choose the triangle. Each group picks its own legs, for example a = 5 cm and b = 9 cm. Different groups using different triangles makes the point that the proof doesn't depend on the numbers.
2. Cut the triangles. On card, draw two rectangles 5 cm by 9 cm and cut each along a diagonal. That gives four identical right triangles. Number them 1 to 4.
3. Make two boards. On paper, draw two squares with side a + b = 14 cm. On each side of the first board, mark the point 5 cm from the corner, going round clockwise.
4. Arrangement 1. Place the triangles in the corners of board 1 so their tips touch the marks. Shade the uncovered tilted square and measure its side. It should match the hypotenuse of the triangles, √106 ≈ 10.3 cm.
5. Arrangement 2. Move the same triangles to board 2 as two rectangles in opposite corners. Shade the two uncovered squares, 5 cm and 9 cm on a side.
6. Explain. Ask each group: what changed between the two boards, and what stayed the same? Their answer, in words, is the proof.

With a = 3 cm and b = 4 cm on centimeter grid paper, students can also count 9, 16 and 25 squares. That is a nice check, but make the distinction explicit: counting confirms one triangle, while moving the pieces proves it for all of them (see what counts as a proof). More classroom ideas are on the activities page.