1. The picture proof: one square, four triangles
Take four copies of one right triangle, with legs a and b and long side (hypotenuse) c. Lay them in a big square whose side is a + b, first one way and then another. Watch what is left uncovered.
How this picture stays honest. The triangles never stretch, shrink, fade or turn into anything else. Each one slides as a solid piece, and you can see every bit of its trip. The numbers under the picture are worked out from the shapes on the screen at every moment. Change a and b and it still works, because the argument works for every right triangle.
2. Euclid's proof: Elements Book I, Proposition 47
Here is the proof in Euclid's Elements. In our figure the right angle is at A, the legs are AB = a and AC = b, and the hypotenuse is BC = c. Squares stand on all three sides. The square on the legs ABFG is moved, one honest step at a time, onto one part of the big square, and the square ACKH onto the other part.
You can also drag point A (or focus it and use the arrow keys). A stays on the half-circle over BC, so the angle at A stays a right angle.
A word about shears. A shear slides one side of a shape along its own line while the opposite side stays put. The base and the height stay the same, so the area stays the same (Euclid I.35 and I.41). Every shear here is labelled. While a piece moves, its area is measured from the shape on the screen and shown under the picture. Euclid himself argues with the hatched triangles; the shears and the turn are a way to see his argument. While a piece turns, it may pass over the other piece or the triangle. Each piece is still one solid shape, so its area does not change.
A note on the name. The theorem is named for Pythagoras of Samos. The proof on this page is the one written down in Euclid's Elements, Book I, Proposition 47, which closes Book I. No writing by Pythagoras himself has come down to us.
3. Try it
1. A rectangular garden bed is 9 feet wide and 12 feet long. A straight stepping-stone path runs from one corner to the opposite corner. How long is the path?
2. Build the picture proof with a = 5 and b = 12. What is the area of the big square? What do the four triangles cover together? What area is left uncovered in the first arrangement, and so how long is c?
3. In Euclid's figure, AB = 6, AC = 8, and the right angle is at A. Find BC. Then find BM, the short side of rectangle BMLD. (Hint: the rectangle has the same area as the square on AB.)
Worked answers
- The path is the hypotenuse: 9² + 12² = 81 + 144 = 225, and √225 = 15. The path is 15 feet.
- The big square has side 5 + 12 = 17, so its area is 17² = 289. Each triangle is ½ · 5 · 12 = 30, so the four cover 120. Uncovered: 289 − 120 = 169, and that is the tilted square c². So c = 13. Check: 5² + 12² = 25 + 144 = 169.
- BC² = 6² + 8² = 36 + 64 = 100, so BC = 10. Rectangle BMLD has long side BD = BC = 10 and must have area 6² = 36, so BM = 36 ÷ 10 = 3.6. Likewise MC = 64 ÷ 10 = 6.4, and 3.6 + 6.4 = 10. ✓