The proof credited to Leonardo da Vinci builds two hexagons from a right triangle. The first is made of the two leg squares plus two copies of the triangle; the second is the hypotenuse square plus two copies of the triangle. A quarter turn shows that the hexagons have equal areas, so taking away the triangles leaves a2 + b2 = c2.
The proof is elegant and worth knowing, but its link to Leonardo is doubtful. No manuscript of his containing it has been identified, and the earliest known source dates from 1772. The story of the name is at the end of this page.
Building the two hexagons
Start with a right triangle ABC with the right angle at C, legs a = BC and b = AC, and hypotenuse c = AB. Draw a square outward on each side: CADE on b, CBGF on a and ABHI on c.
- Hexagon ABGFED. Join F to E. This closes a second triangle, CEF, so the hexagon is made of the two leg squares and two triangles.
- Hexagon CAIJHB. On the far side HI of the hypotenuse square, attach a copy of the triangle turned through 180°, with its right angle at a new point J. This hexagon is made of the hypotenuse square and two triangles.
Triangle CEF really is a copy of ABC. Its sides CE and CF are sides of the leg squares, so CE = b and CF = a. The angles around C add up to 360°, and three of them are right angles (the triangle's corner and one corner of each square), so the angle ECF is also 90°. By side-angle-side, △CEF ≅ △ACB, and in particular EF = c.
Each hexagon splits into two equal halves
The first hexagon is symmetric. The diagonal of a square makes 45° with its sides, so the angles at C along the line from G to D are
∠GCB + ∠BCA + ∠ACD = 45° + 90° + 45° = 180°That means G, C and D lie on one straight line. This line cuts the right angle ACE exactly in half, and CA = CE, so reflecting in it swaps A and E. In the same way it swaps B and F. The reflection therefore maps the hexagon onto itself, and the two halves GBAD and DEFG are mirror images with equal areas.
The second hexagon has a center. The added triangle HJI is the original triangle turned through 180° about the center of the hypotenuse square. That half turn swaps C with J, A with H and B with I, so it maps the hexagon onto itself. The line CJ passes through the center and cuts the hexagon into two halves, CAIJ and JHBC, that the half turn swaps.
The quarter turn that links them
Now compare a half of one hexagon with a half of the other. Turn the quadrilateral CAIJ a quarter turn about the point A:
- C goes to D, because AC and AD are two sides of the square on b meeting at a right angle at A.
- I goes to B, because AI and AB are two sides of the square on c meeting at a right angle at A.
- J goes to G. The side IJ of the added triangle is parallel to CB and has length a. After the quarter turn it starts at B, has length a and points at right angles to CB, away from the triangle. That is exactly the side BG of the square on a.
So the quarter turn carries CAIJ onto DABG, which is one half of the first hexagon.
Adding up the areas
All four halves have the same area. Each hexagon is two copies of that quadrilateral, so the two hexagons have equal areas. Write T for the area of the triangle, T = 1 over 2ab, and count what each hexagon contains:
a2 + b2 + 2T= c2 + 2Ta2 + b2= c2Example: check the areas with a 3-4-5 triangle
With legs 3 and 4 and hypotenuse 5 (the 3-4-5 triangle), each triangle has area T = 6.
first hexagon= 9 + 16 + 2 · 6 = 37second hexagon= 25 + 2 · 6 = 37Both hexagons have area 37, and each half has area 18.5.
Key ideaInstead of cutting the squares up, the proof adds the same two triangles to both sides, shows the two resulting shapes are equal because they are built from congruent halves, and then takes the triangles away again.
The complete proof, written out
| Statement | Reason |
|---|---|
| 1. ∠ACB = 90°; squares CADE, CBGF and ABHI are drawn outward. | Given; construction |
| 2. △CEF ≅ △ACB | SAS: CE = CA, CF = CB, ∠ECF = 360° − 3 · 90° = 90° |
| 3. △HJI is △ABC turned 180° about the center M of square ABHI. | Construction |
| 4. G, C and D are collinear. | 45° + 90° + 45° = 180° |
| 5. Area of GBAD = area of DEFG = half of hexagon ABGFED. | Reflection in line GD maps the hexagon to itself |
| 6. Area of CAIJ = area of JHBC = half of hexagon CAIJHB. | The half turn about M maps the hexagon to itself |
| 7. CAIJ ≅ DABG | A quarter turn about A maps C to D, I to B and J to G |
| 8. The two hexagons have equal areas. | From 5, 6 and 7 |
| 9. a2 + b2 + 2T = c2 + 2T, so a2 + b2 = c2 | Add up the pieces of each hexagon; subtract 2T |
Step 4 matters more than it looks. Without it, GBAD and DEFG would not be the two halves of the hexagon, and step 5 would fail. It is also the step most often skipped in quick versions of this proof.
Did Leonardo da Vinci really find it?
Many books and websites call this "Leonardo da Vinci's proof", after the Italian artist and engineer (1452–1519). The evidence for that is thin.
- Elisha Loomis's classic collection The Pythagorean Proposition (1940 edition) credits Leonardo on the authority of a 1924 book by F. C. Boon. Thomas Heath's 1908 edition of Euclid's Elements says the proof "appears to come from" Leonardo's papers, without giving a reference.
- The mathematician Franz Lemmermeyer traced these claims back (arXiv, 2013). The first mention of Leonardo he found is in a 1906 report by Max Simon, apparently through a mix-up with another reference. No one has pointed to a page in Leonardo's notebooks that contains the proof.
- The earliest known source is the Göttingen mathematician Johann Tobias Mayer, who found the proof in 1772, according to a compilation of proofs published in 1826. It appeared without any author's name in a German geometry textbook of 1790, and the French mathematician Terquem published the same proof as new in 1838.
Lemmermeyer concludes that the proof is Mayer's, while admitting that he cannot prove Leonardo never thought of it. Leonardo did study geometric problems about shapes of equal area, which may be why the story seemed believable. The fair summary is that "da Vinci's proof" is a traditional name, like the windmill of Euclid's proof, not an established historical fact. For more on how proofs and legends get attached to famous names, see the history of the Pythagorean theorem.
How it compares with other proofs
Leonardo's proof belongs to a family of proofs that compare areas by adding and removing congruent pieces. The rearrangement proof puts four triangles inside two equal squares and compares what is left. President Garfield's proof computes the area of a trapezoid made of three triangles in two ways. Perigal's dissection goes the other way and cuts a leg square into pieces that fill the hypotenuse square. What makes the hexagon proof distinctive is the symmetry: one reflection, one half turn and one quarter turn carry all the logic. Browse the rest on the Pythagorean theorem proofs page.
Frequently asked questions
Did Leonardo da Vinci prove the Pythagorean theorem?
There is no good evidence that he did. No manuscript of Leonardo's containing this proof has been identified, and the earliest known source is Johann Tobias Mayer in 1772. The name "da Vinci's proof" seems to date from the early 20th century.
Why do the two hexagons have the same area?
Each hexagon is made of two copies of the same four-sided shape. A quarter turn about one corner of the triangle carries half of one hexagon exactly onto half of the other, and turning a shape never changes its area.
Why does the line through the outer corners of the leg squares pass through C?
Each square's diagonal makes a 45° angle with its sides. At C the angles along that line are 45°, then the right angle of the triangle, then 45°, which add up to 180°, a straight line.
What is the area of each hexagon?
Each hexagon has area a2 + b2 + ab, which also equals c2 + ab, because the two triangles together have area ab. For legs 5 and 12 that is 25 + 144 + 60 = 229 = 169 + 60.
Is this a dissection proof?
Not exactly. A dissection cuts one figure into pieces that rearrange into another. This proof adds the same triangles to both sides, shows the two larger shapes are equal, then subtracts the triangles.
Sources and further reading
- Franz Lemmermeyer, On Leonardo da Vinci's proof of the Theorem of Pythagoras, arXiv:1311.0816, 2013. Traces the attribution and the proof's earlier appearances.
- Leonardo's proof of Pythagoras's theorem, MacTutor History of Mathematics, University of St Andrews.
- Elisha S. Loomis, The Pythagorean Proposition, 2nd ed., 1940 (reissued by the National Council of Teachers of Mathematics, 1968).
- T. L. Heath, The Thirteen Books of Euclid's Elements, vol. 1, Cambridge University Press, 1908.
- Alexander Bogomolny, Pythagorean Theorem and its many proofs, Cut the Knot.
Keep exploring
- ProofsEuclid's Proof of the Pythagorean TheoremEuclid's windmill proof of the Pythagorean theorem, Book I Proposition 47, step by step: congruent triangles, an animated shear and the converse, I.48.
- ProofsThe Rearrangement Proof of the Pythagorean TheoremThe rearrangement proof of the Pythagorean theorem, animated: four triangles in two squares, why the middle is a square, a full write-up and a paper activity.
- ProofsPerigal's Dissection Proof of the Pythagorean TheoremHenry Perigal's dissection proof of the Pythagorean theorem: cut the larger leg square into four equal pieces, then slide them and the small square into c².
- ProofsProofs of the Pythagorean TheoremProofs of the Pythagorean theorem: an animated proof, how a proof differs from checking examples, all 12 proofs on this site by type, and where to start.
- HistoryHistory of the Pythagorean TheoremThe history of the Pythagorean theorem from Babylon (1800 BCE) to a 2023 proof: who discovered it, who proved it first, and why it bears Pythagoras' name.
- ProofsPresident Garfield's Trapezoid ProofJames Garfield's 1876 trapezoid proof of the Pythagorean theorem: an animated build, the area found two ways, why angle ABD is 90°, and the story behind it.