Algebraic Proof of the Pythagorean Theorem

The algebraic proof of the Pythagorean theorem finds the area of one square in two ways. A square of side a + b contains four right triangles and a tilted square of side c, so (a + b)² = 4 · 1 over 2ab + c². Expanding gives a² + 2ab + b² = 2ab + c², and subtracting 2ab leaves a² + b² = c².

The (a + b)² proof, line by line

Take a right triangle with legs a and b and hypotenuse c. Arrange four copies inside a square of side a + b, one in each corner, so that every side of the big square is split into a piece a and a piece b. The region left in the middle is a square of side c; the next section proves it.

Area from the side. The big square has side a + b, so

Area from the pieces. Each triangle has area 1 over 2 × base × height = 1 over 2ab, and the middle square has area c². The pieces fill the big square with no gaps or overlaps, so

Set them equal and simplify. Both expressions describe the same region, so they are equal. Expand the left side, then subtract 2ab from both sides:

That's the whole proof. Every line is either an area fact or a standard algebra step, and nothing depends on the particular lengths, so it holds for every right triangle. This is the reason the result is often called the "(a + b)² proof".

Why (a + b)² = a² + 2ab + b²

The expansion is the distributive property: (a + b)² = (a + b)(a + b) = a² + ab + ba + b² = a² + 2ab + b². The picture shows the same thing with areas. It also shows why (a + b)² = a² + b² is wrong: that forgets the two rectangles. (This picture is, in fact, the second arrangement of the rearrangement proof with the rectangles left uncut.)

Check it with numbers

Try the 5-12-13 triangle. The big square has side 17, the four triangles have total area 4 · 1 over 2 · 5 · 12, and what's left should be 13²:

A numerical check is not a proof, since it covers only one triangle, but it is a quick way to catch a slip in the algebra.

Why the middle shape really is a square

The proof uses the area c² for the middle shape, so you have to know it is a square. Its four sides are all hypotenuses of identical triangles, so they all have length c. That alone only makes it a rhombus. For the angles, look at any point where two triangles meet on an edge of the big square. Three angles sit there along a straight line: one acute angle of the first triangle, the corner of the middle shape, and the other acute angle of the second triangle. Straight-line angles add to 180°, and the two acute angles of a right triangle add to 90°, so the corner of the middle shape is 180° − 90° = 90°.

Four equal sides and four right angles make a square. There is a labeled diagram of this argument on the rearrangement proof page.

The (b − a)² proof

A second algebraic proof puts the four triangles inside the square on the hypotenuse instead of around it. Place them so that each hypotenuse is one side of a square of side c, with each triangle a quarter turn from the last. They leave a small square hole in the middle. If b is the longer leg, the hole has side b − a.

Add up the pieces and expand (b − a)² = b² − 2ab + a²:

Why is the hole a square of side b − a? At each corner of the big square, two triangles meet: one with its acute angle β there, the other with α. Since α + β = 90°, the corner of the square, the short leg of one triangle lies exactly along the long leg of the next. So each long leg b is made of the neighbor's short leg a plus a side of the hole, which therefore has length b − a. Each corner of the hole lies on a straight leg next to a triangle's right angle, so it measures 180° − 90° = 90°.

This is the figure in Bhāskara's proof, and in the Chinese hypotenuse diagram of the gougu proof, where the triangles were colored red and the central square yellow. If a = b, the hole shrinks to a point and the formula becomes c² = 2a², which is still correct.

Two-column proof for homework

Many geometry classes ask for proofs in two columns: each statement on the left, the reason that justifies it on the right. Here is the (a + b)² proof in that format.

If your class writes paragraph proofs instead, turn each row into a sentence: "Since the corners of a square are right angles, each corner triangle has legs a and b with a right angle between them, so by SAS it is congruent to the original triangle." The logic is identical.

Algebraic proof vs. rearrangement proof

The two proofs start from the same picture, which is why they're often confused. They differ in how they finish.

President Garfield's proof is a third member of the family: it uses half of the same square, a trapezoid, and the same algebra. Proofs that use no areas at all, like the similar triangles proof, work very differently; the guide to all the proofs compares them.

Common mistakes

• Expanding (a + b)² as a² + b². The middle term 2ab is the heart of the proof. See common mistakes for more on squaring sums.
• Using ab for the triangle's area. A right triangle is half of an a × b rectangle, so its area is 1 over 2ab. Four of them make 2ab, not 4ab.
• Skipping the "why is it a square?" step. Without it, the middle area might not be c².
• Assuming what you're proving. Don't find the side of the middle square as √a² + b². It is c because it is the hypotenuse of a triangle congruent to the original, and that is all the proof needs.