Pythagorean theorem proof: three clear proofs of a² + b² = c², step by step
The Pythagorean theorem says that in a right triangle the square on the hypotenuse equals the sum of the squares on the legs: a² + b² = c². The simplest proof fits four identical right triangles inside a square of side a + b and compares the leftover area in two arrangements: c² in one, a² + b² in the other. With legs 3 and 4 it gives 49 − 24 = 25 = 9 + 16.
A proof is different from a measurement. Measuring a few triangles can only suggest that a² + b² = c²; a proof shows that it must hold for every right triangle, large or small. This guide walks through three classic proofs, each checked on the 3-4-5 triangle, then looks at who proved the theorem first, the converse that lets you test for a right angle, and a living textbook chapter where you can watch the squares grow.
Key ideas
- The theorem a² + b² = c² holds for right triangles only, and so does its converse: if the squares add up, the angle is 90°.
- Proof by four triangles: (a + b)² = c² + 4 · (ab/2), so a² + 2ab + b² = c² + 2ab and therefore a² + b² = c².
- Similar triangles give a² = c·p and b² = c·q with p + q = c; Garfield’s trapezoid reaches the same result with areas.
- Pythagoras’ own part is unclear: the tablet Plimpton 322 (about 1800 BC) lists right-triangle triples, Euclid’s Elements (about 300 BC) holds the oldest surviving proof, and a 1940 book collects 370 proofs.
What the theorem says, and what a proof has to show
In a triangle with a square corner, the two sides that meet at the right angle are the legs, a and b, and the long side opposite it is the hypotenuse, c. The theorem says that a² + b² = c². In words, a square drawn on each leg has an area, and the two areas together equal exactly the area of the square on the hypotenuse. For the 3-4-5 triangle: 3² + 4² = 9 + 16 = 25 = 5².
Checking 3-4-5 proves nothing about 5-12-13, or about a triangle with legs 1 and 1, where c = √2. A proof starts from facts already accepted, such as the area of a square and of a triangle and the rule that the angles of a triangle add up to 180°, and reaches the claim in steps that work for any a and b. The three proofs below do it with areas.
Proof 1: four triangles inside a square
Take four identical right triangles with legs a and b and hypotenuse c, and a big square of side a + b, whose area is (a + b)². In the first arrangement, place the triangles in the four corners so that their hypotenuses form a tilted square of side c in the middle. In the second, place them so that they leave a square of side a and a square of side b. The big square and the four triangles are the same in both pictures, so the empty area must be the same too: c² in the first, a² + b² in the second.
In symbols, the four triangles cover 4 × (ab/2) = 2ab. So (a + b)² − 2ab = c², and since (a + b)² = a² + 2ab + b², this gives a² + b² = c². Check it with 3 and 4: the big square has side 7 and area 49, the four triangles cover 4 × 6 = 24, and 49 − 24 = 25 is left, which is both 5² and 9 + 16.
One detail makes the proof complete: why is the middle shape of the first arrangement really a square? Its four sides are all c, and at each corner the angle is 180° minus the two acute angles of the triangle. In a right triangle those two angles add up to 90°, so each corner measures 90°, and the shape is a square.
Proofs 2 and 3: similar triangles and Garfield’s trapezoid
Proof 2 uses similar triangles. In a right triangle with legs a and b, draw a line from the right-angle corner to the hypotenuse, meeting it at 90°. It splits the hypotenuse into two pieces, p next to leg a and q next to leg b, with p + q = c, and cuts the triangle into two smaller right triangles that have the same angles as the whole one, so they are similar to it.
Similar triangles have equal ratios of sides. From the first small triangle, a/c = p/a, so a² = c·p. From the second, b/c = q/b, so b² = c·q. Add them: a² + b² = c·p + c·q = c·(p + q) = c². In the 3-4-5 triangle, p = 9/5 = 1.8 and q = 16/5 = 3.2, which add up to 5, and indeed 5 × 1.8 = 9 and 5 × 3.2 = 16.
Proof 3 was published in 1876 by James A. Garfield, then a member of the US House of Representatives and later the twentieth president. Put two copies of the triangle together with a right isosceles triangle whose two legs are c, so that they form a trapezoid whose parallel sides are a and b and whose height is a + b. Its area is (a + b)²/2, and it is also the sum of the three triangles: ab/2 + ab/2 + c²/2. So (a + b)²/2 = ab + c²/2, which simplifies to a² + b² = c². For 3 and 4: 49/2 = 24.5, and 6 + 6 + 12.5 = 24.5.
Who proved it first? Plimpton 322, Euclid and the converse
The theorem carries Pythagoras’ name, but he lived about 570 to 495 BC, left no writings, and historians cannot say who first proved it. The Babylonian clay tablet Plimpton 322, from roughly 1800 BC, already lists right-triangle triples: one row pairs 119 and 169, the leg and hypotenuse of a triangle whose other leg is 120, and indeed 119² + 120² = 14,161 + 14,400 = 28,561 = 169². Indian Sulba Sutras from about 600 BC, and the Chinese Zhoubi Suanjing in texts from about the first century BC, also use the rule.
The oldest proof that survives in a deductive system is Euclid’s, in Book I, Proposition 47 of the Elements, written about 300 BC. The theorem may have more known proofs than any other: Elisha Scott Loomis’s book The Pythagorean Proposition, in its second edition of 1940, contains 370 of them.
The converse is true as well, and it is how builders check a corner: if a² + b² = c², the angle between a and b is a right angle. The triples 5, 12, 13 (25 + 144 = 169), 8, 15, 17 (64 + 225 = 289) and 7, 24, 25 (49 + 576 = 625) all pass. The sides 6, 7, 10 do not: 36 + 49 = 85, which is less than 100, so the angle opposite the side of 10 is larger than 90°.
Watch the proof move in the VJOURNAL SMART BOOK
The “Pythagoras and right triangles” chapter of the VJOURNAL SMART BOOK maths textbook on VITON13 SCHOOL shows the four-triangle proof as a moving picture. Squares grow on the sides of a 3-4-5 triangle, four triangles slide inside a square of side a + b and fill it in two different ways, and a wireframe box turns slowly while its long diagonal lights up. Under the picture, a sandbox has sliders for the two legs: the squares are redrawn to scale, and you can compare c with your own calculation.
Use it as a test of this guide: set the legs to 5 and 12 and check that the hypotenuse comes out as 13, then try 1 and 1 and watch c settle at √2 ≈ 1.414. The free scientific calculator has the square-root key you need for triangles of your own.
In everyday language
Cut four copies of a right triangle out of paper. Lay them in a square frame one way and the gap is one big square; lay them another way and the gap is two smaller squares. Same frame, same four triangles, so the two gaps must hold the same amount of room.
Try it yourself
On squared paper draw a square 7 units wide and cut out four right triangles with legs 3 and 4. Lay them in the square so that two squares remain, one 3 wide and one 4 wide, and count the unit squares in them. Then subtract the four triangles’ area, 4 × 6 = 24, from 49 to find the tilted gap of the other arrangement.
Expected result
9 + 16 = 25 unit squares in the two small squares, and 49 − 24 = 25 for the tilted square, so its side is 5.
Check your answer: Use the four-triangle method for legs 5 and 12. What is the area of the big square, how much do the four triangles cover, and how long is c?
The big square has side 17 and area 289. Four triangles cover 4 × (5 × 12 ÷ 2) = 120. What is left is 289 − 120 = 169 = c², so c = 13.
Questions
What is the easiest proof of the Pythagorean theorem?
The four-triangle proof: it needs only the area of a square and of a triangle and the identity (a + b)² = a² + 2ab + b². It can also be done with paper cut-outs and no algebra at all.
Did Pythagoras prove the Pythagorean theorem?
Nobody knows. He left no writings, and the Babylonian tablet Plimpton 322 shows right-triangle triples from about 1800 BC. The oldest proof that survives in a deductive system is Euclid’s, from about 300 BC.
How many proofs of the Pythagorean theorem are there?
Hundreds. The 1940 edition of Elisha Scott Loomis’s book collects 370 proofs, and the theorem may have more known proofs than any other.
Does a² + b² = c² work for triangles without a right angle?
No. Other triangles need the law of cosines, c² = a² + b² − 2ab·cos C, where C is the angle between a and b. At C = 90°, cos C = 0 and it turns back into the Pythagorean theorem.
How can I test whether a triangle has a right angle?
Square the two shorter sides, add them, and compare with the square of the longest side. If they are equal, the angle is 90°. For 6, 7, 10 the sum is 85, not 100, so it is not a right triangle.
Try the four-triangle proof on paper with legs 3 and 4, then open the Pythagoras chapter and drag the sliders to see the squares grow to scale beside the triangle.
Sources and further reading
- Wikipedia: Pythagorean theorem ↗Sources checked:
- Wolfram MathWorld: Pythagorean theorem ↗Sources checked:
- MacTutor, University of St Andrews: Pythagoras’s theorem in Babylonian mathematics ↗Sources checked:
- Cut-the-knot: Pythagorean theorem and its many proofs ↗Sources checked: