How Many Ways Can You Prove Pythagoras’ Theorem?

What is left uncovered

Before you slide, the triangles leave two squares uncovered, one on side a and one on side b. Slid into the corners (never turned), they leave one tilted square on the long side, c.

Try other shapes: 1 and 6, 4.5 and 2, two equal sides. “Uncovered” never changes when you slide. That is the whole proof, once you see why it must be so.

Why it is a proof

The frame is the same before and after, and the four triangles, only moved, cover the same area. So what they leave uncovered is the same area too:

$$a^2 + b^2 = c^2$$

Nothing was measured. The argument never used the numbers, so it works for every right triangle, even one too big for any page.

One thing to check: is the tilted shape really a square? Its four sides are long sides of the triangles, so they are equal. At each of its corners, on the frame’s side, a sharp angle of one triangle, the corner and the other sharp angle of the next triangle make a straight line, 180∘°. A triangle’s angles add up to 180∘°, the right angle takes 90∘°, so its two sharp angles make 90∘°. That leaves 90∘° for the corner: it is a square.

The theorem, and what a proof is

A triangle with a square corner is a right triangle, and its longest side, the one opposite the square corner, is the hypotenuse. You have just shown:

Pythagoras’ theorem: in every right triangle, the square on the hypotenuse equals the squares on the other two sides added together. With the short sides a and b and the hypotenuse c, $a^2 + b^2 = c^2$.

Nepal’s Grade 7 maths books write it $h^2 = p^2 + b^2$: h for the hypotenuse, p for the perpendicular and b for the base. The letters don’t matter; which side is which does.

A theorem is a statement that has been proved. A proof is an argument, each step following from what is already known, that shows a statement is true in every case it talks about. Measuring a hundred triangles is evidence: it makes you believe. A proof makes you certain, because it covers the triangles nobody will ever draw.

A second proof can’t make the theorem truer, but each shows it from another side.

Cut the big square

In the 1100s, the mathematician and astronomer Bhaskara (1114–1185), head of the observatory at Ujjain, wrote the Bijaganita (“seed arithmetic”), his book of algebra. In it he asks: a triangle has short sides 15 and 20, what is its hypotenuse, “and show the demonstration of the received mode of computation”: prove the rule everyone uses.

One of his proofs puts four copies of the triangle inside the square on the hypotenuse, round a small square in the middle. Try it, with his own numbers:

The square on the hypotenuse is made of five pieces. Turn two of them out, and the same five pieces become a square on side 20 beside a square on side 15. Nothing was added and nothing thrown away, so

$$25^2 = 20^2 + 15^2, \qquad 625 = 400 + 225$$

and the same cut works for any right triangle.

Counting the pieces, as Bhaskara did

Bhaskara didn’t leave the picture to speak for itself: he added up the pieces. Each triangle is half of a 20 by 15 rectangle, 150, so the four make 600. The small square in the middle has side 20 − 15 = 5, so its area is 25. Together, 600 + 25 = 625, and the square with area 625 has side 25: the hypotenuse.

The same counting works with letters. The four triangles are $4 \times \tfrac12 ab = 2ab$. The small square’s side is a − b, and Grade 7’s rule for the square of a difference gives its area: $(a - b)^2 = a^2 - 2ab + b^2$. Add the two:

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

The 2ab taken away in the small square is put back by the four triangles. Bhaskara wrote this rule out in words in his other book, the Lilavati (§ 135; H. T. Colebrooke’s English of 1817): “Twice the product of two quantities, added to the square of their difference, will be the sum of their squares.”

“Behold!”

A famous story says Bhaskara drew this figure and wrote under it a single word, “Behold!”. Elisha Loomis repeated the story in 1940 with a warning: the one word could be fair only if Bhaskara also drew the pieces put together the other way; with the first figure alone, he wrote, it “is hardly defensible”. The book itself, in Colebrooke’s translation, has two proofs with numbers, the count above and a proof by triangles of the same shape (later in this topic), and then, after the rule, places “the same portions of figure in another form” with the word “see”, pointing at a drawing. The one-word proof seems to have grown out of that.

In 1876 James Garfield, then a member of the United States Congress, found a proof that was printed in the New-England Journal of Education. Five years later he became President.

It is half of the first proof’s frame. Two copies of the triangle stand side by side (Figure 18.1), their right angles at the ends of a bottom line a + b long. Their long sides meet in the middle at a right angle (the two sharp angles there make 90∘°, leaving 90∘° of the straight line). Join their tops: the gap is half of a square on c.

The whole shape is a trapezium: two parallel sides, a and b, which are a + b apart. Its area (Grade 8) is the average of the parallel sides times the distance between them, $\tfrac12(a + b) \times (a + b)$. Count it by its pieces instead: two triangles of $\tfrac12 ab$ and one of $\tfrac12 c^2$. The same area counted two ways:

$$\tfrac12 (a + b)^2 = ab + \tfrac12 c^2$$

Double both sides: $(a + b)^2 = 2ab + c^2$. Grade 7’s square of a sum is $a^2 + 2ab + b^2$, so $a^2 + 2ab + b^2 = 2ab + c^2$, and taking away 2ab from both leaves $a^2 + b^2 = c^2$.

Cut the triangle itself

The proofs so far moved pieces about. This one moves nothing. It was Bhaskara’s first proof in the Bijaganita, and more than seven hundred years later a boy in Germany found it again for himself.

Stand the triangle on its hypotenuse and draw the line from the right angle straight down to it, meeting it square. The triangle falls into two smaller triangles, and the surprise is that both have the same shape as the whole. Each piece has a right angle (where the line meets the hypotenuse), and each shares one sharp angle with the big triangle. A triangle’s angles add up to 180∘°, so the third angles match too. Three angles the same: the same shape, only a different size. Figures like that are similar, as in Grade 8.

Look at Figure 18.2 again. The long side of the small piece is 3, one of the big triangle’s short sides. The long side of the bigger piece is 4, the other short side. And the long side of the whole is 5.

Same shape, area grows as the square

Double every length of a triangle and four copies of the old one fit inside the new (join the midpoints of its sides and it falls into four): 2 × 2 = 4 times the area. Three times the lengths, 9 times the area. Stretch every length by any number, 2.5 say, and the area grows 2.5 × 2.5 times, just as a square’s does when its side is stretched the same way. So for figures of one shape, the area is always the same fraction of the square on the long side.

For these triangles that fraction is $\tfrac{6}{25}$, since the whole triangle has area 6 and the square on its long side is 25:

Triangle	Long side	Square on it	Area
Small piece	3	9	9 × 6/25 = 2.16
Bigger piece	4	16	16 × 6/25 = 3.84
Whole	5	25	25 × 6/25 = 6

And the two pieces make the whole: 2.16 + 3.84 = 6. Divide each area by the same fraction and what is left is the squares: 9 + 16 = 25. For any right triangle the same holds, with the fraction depending on its shape. Call it k: the pieces have areas $k a^2$ and $k b^2$, the whole $k c^2$, and since the pieces make the whole,

$$k a^2 + k b^2 = k c^2, \quad\text{so}\quad a^2 + b^2 = c^2$$

The argument never used squares, only that the three figures have the same shape. So it works for any three figures of one shape drawn on the sides: half circles, triangles, even three copies of Nepal’s flag, each as wide as its side. The two small ones together always have the area of the big one.

A rope with twelve knots

Pythagoras’ theorem says: if the corner is square, then $a^2 + b^2 = c^2$. Is the other way round true as well? If three sides fit the rule, must the corner be square?

That is what a builder needs. To lay out a square corner, the Journal of Light Construction tells carpenters to mark a point 12 feet from the corner along one wall’s line, then stretch two tapes, 9 feet from the corner and 15 feet from that point, and mark where they meet: 81 + 144 = 225, so the second wall’s line, through that mark, is square to the first. A loop of rope knotted into 12 equal parts, pulled tight as 3, 4 and 5, does the same. Try other lengths:

Why the rule works backwards

You will find a square corner for 3, 4, 5 and 6, 8, 10, and none for 6, 8, 9. Here is why the corner must be square when the sides fit the rule.

Take a triangle whose sides fit, $a^2 + b^2 = c^2$. Now build a second triangle with the sides a and b at a square corner. By the theorem, its third side squared is $a^2 + b^2$, which is $c^2$: its third side is c. So the two triangles have the same three sides, and two triangles with the same three sides are congruent, the same in every way (Grade 8’s SSS). So the first triangle’s corner is a right angle too. This is the converse of the theorem: its if and then swapped. (Not every converse is true: “if it is a dog, it is an animal” is true, but “if it is an animal, it is a dog” isn’t.)

Bigger, smaller, and no triangle at all

The widget showed more. When $c^2$ is less than $a^2 + b^2$, the corner is less than a right angle; when it is more, the corner is more. Think of a and b as two sticks hinged at the corner: open the hinge wider and their ends move apart, so c grows. At exactly 90∘°, $c^2 = a^2 + b^2$; narrower gives a shorter c, wider a longer one.

Every proof here used flat paper: a square frame, triangles slid without stretching, the angles of a triangle adding up to 180∘°. On a curved surface those fail, and so does the theorem.

Walk from the North Pole south to the equator, about 10,000 km. Turn left, a square corner, and walk a quarter of the way round the equator, another 10,000 km. Turn left again and walk north: after 10,000 km you are back at the Pole, arriving at a right angle to the path you left on. A triangle with three right angles, and 10,000² + 10,000² is twice 10,000², not 10,000². On the land a house is built on, though, the Earth is so nearly flat that the theorem is right to far better than any tape measure.

These are from Bhaskara’s Lilavati and China’s Nine Chapters. Each hides a right triangle: find it first.

The lotus in the lake

From the Lilavati (§ 153, in Colebrooke’s English): “In a certain lake swarming with ruddy geese and cranes, the tip of a bud of lotus was seen a span above the surface of the water. Forced by the wind, it gradually advanced, and was submerged at the distance of two cubits. Compute quickly, mathematician, the depth of water.” A span is half a cubit.

The stalk swings about its root on the lake bed, so its length stays the same (Figure 18.3). Upright, it is the depth plus half a cubit. Swung over, it is the hypotenuse of a right triangle whose other sides are the depth and the 2 cubits along the surface.

The broken bamboo

The reed in the pond

Hundreds. In 1940 an American teacher, Elisha Scott Loomis, finished the second edition of The Pythagorean Proposition, every proof he could find: 370, 109 using algebra, 255 geometry, and a few odd ones. “And the end is not yet,” he wrote. One of them, counting this topic’s frame with algebra, was sent to him in 1939 from a high school in Indiana: its author was a schoolboy, Maurice Laisnez.

One kind, he said, could never exist. Trigonometry, the maths of angles you meet in Grade 10, is built on the theorem, so “there are no trigonometric proofs”: they would use the theorem to prove itself. Mathematicians have since found ways round the trap (Jason Zimba published one in 2009). Then in December 2022, two high-school students in New Orleans, Calcea Johnson and Ne’Kiya Jackson, took up a bonus question in their school’s maths contest: prove the theorem with trigonometry. After months of work they had new proofs that avoid the trap. They presented them to the American Mathematical Society in March 2023, and in 2024 their paper, “Five or Ten New Proofs of the Pythagorean Theorem”, led the American Mathematical Monthly.

Wherever a distance is worked out from how far across and how far up, the theorem is inside it: the distance formula of coordinate geometry is Pythagoras’ theorem with the two short sides read off the axes, and it is how a computer finds the distance between two points on a map or a screen.

• Where Did the Missing Square Go? A picture “proof” that lies, and how to catch it.
• Who Is Telling the Truth? Logic and proof with puzzles: how to be sure of an answer without trying every case.
• Is a Pattern That Works Ten Times Always True? Why checking a rule many times, as Bibek did, is not a proof.
• What Is the Shortest Way to the River and Back? A distance puzzle solved with this topic’s triangles.
• How Do You Find Every Right Triangle with Whole Sides? (Grades 9–12) 3, 4, 5 and 20, 21, 29, and a recipe that makes them all.
• Can You Prove √2 Is Not a Fraction? (Grades 9–12) The diagonal of the Babylonians’ square, and why no fraction is exactly it.
• How Does a Calculator Find a Square Root? (Computing, Going Further, Grades 9–12) Every hypotenuse ends in one.
• Grade 10 Math uses the theorem everywhere: the slant height of a pyramid in Area and Volume, and heights and distances in Trigonometry.