Pythagoras's theorem says that for a right angled triangle with sides of length a,b,c (with c the length of the hypotenuse) we have c2=a2+b2.

Here is one proof (of many). Start with a square of side length a+b, call it square 1. Put a square of side length c in the middle of square 1, call it square 2. Now rotate square 2 so that each vertex of the square 2 meets one of the edges of square 1.

    ____
   | /\ |
   |/  \|
 a |\c /|
   |_\/_|
    b

At this point our original right angled triangle occurs in each of the four corners of square 1. So the area of square 1 is the area of square 2 + the 4 x the area of our original triangle. That is square 1 has area c2+2ab. On the other hand square 1 has area (a+b)2. The result follows easily.

See also Pythagorean triple.