Lets draw a right angled triangle with sides a, b and c (where c is the longest side):
Using four copies of that triangle we can make a square, like this:
This square has the length [a + b], also there is a small square inside the larger one with the area c2, this is the area the triangles did not occupy.
Now lets rearrange the triangles in the square so they look like this:
Now we have four triangles and two small squares, one with area a2 and the other with area b2:
As you can see, both of these squares have length [a + b], the area of the large square and the area of the triangles didn't change, so the remaining area must be equal for both squares, in other words, c2 must be equal to [a2 + b2].
