Goal
Show that for a right triangle with leg lengths a and b and hypotenuse c, we have c² = a² + b², using a tiling (rearrangement) argument — a proof that is largely visual (a "proof without words"). This is the same basic idea used by An‑Nayrizi (circa 900 AD) and by modern visual proofs such as Roger Nelson’s.
What to draw (easy recipe)
- Draw a right triangle and label the two legs a and b, and the hypotenuse c.
- Make four identical copies of that triangle.
- First arrangement: Put the four triangles inside a large square whose side length is (a + b). Place one triangle at each corner so that the right angles sit at the corners of the big square and each triangle’s legs lie along the sides of the big square. The four hypotenuses then form a smaller tilted square in the center — that central square has side length c.
Picture in your head (or draw): a big square of side a + b. Each corner contains one right triangle (legs along the edges), and the slanted hypotenuses touch and form the small central square.
Compute areas (first arrangement)
Area of the big square = (a + b)².
Area of the four triangles together = 4 × (1/2 × a × b) = 2ab.
So the area left over in the middle (the small tilted square) equals:
(a + b)² − 2ab = a² + 2ab + b² − 2ab = a² + b².
But that middle shape is a square whose side is c, so its area is c². Therefore c² = a² + b².
Why this is a rearrangement (another way to see it)
You can think of the same big square (side a + b) arranged in a different way so that the four triangles plus two small squares of side a and side b fill exactly the same big square. In that second arrangement the portion that is not covered by triangles splits into the two separate squares of areas a² and b². Since both arrangements use the same four triangles and the same big square, the uncovered area must be equal. In the first arrangement that uncovered area is the square of side c, so c² = a² + b².
Quick numeric check
Take a = 3 and b = 4. Then the big square side is 7, so its area is 49. The four triangles area is 2ab = 2×3×4 = 24. The middle area is 49 − 24 = 25, which is 5², so c = 5 and indeed 5² = 3² + 4².
Historical note
An‑Nayrizi (often written Al‑Nayrizi) around the year 900 gave a geometric rearrangement argument like this. Modern versions appear as short visual proofs called "proofs without words" (Roger Nelson and others have produced striking tiling pictures that make the same identity obvious at a glance).
What to remember
- The proof uses area and rearrangement: two different ways to fill the same big square give two expressions for the same leftover area.
- Because the leftover piece in one arrangement is a square of side c and in the other arrangement it equals the combined areas a² + b², we get c² = a² + b².
- This is visual, simple, and elegant — hence the name "proof without words." Try drawing it yourself: making the two arrangements reinforces the idea strongly.
If you want, I can give step-by-step drawing instructions (measurements and rotations) so you can construct the two arrangements exactly on paper or using a geometry app.