Here is one more of the 145 proofs of this theorem.

The reason I particularly like this proof is that you can actually get a piece of paper and a pair scissors and do it. (My old maths teacher had a set of pretty wooden blocks to demonstrate it.)

Ok here goes:

- Draw a square.
- Call the vertices A, B, C, D.
**Inside** the square draw two identical right angle triangles with the lines AB and AD as their hypotenuses. The shorter side of one should lie on the longer side of the other.

You should end up with something looking like this:

A_
/|`-._
/ | `-._
/ | `-._
/ |______________`-.___ B
/ |F /
/c | /
/ |b /
/ | /
__/__________| /
D`-._ a E /
`-._ /
`-._ /
`-._ /
`'
C

The area of the square is **c**^{2}

Now rotate triangle ADE 90^{o} clockwise about point D. This should map line AD onto line DC.

Now rotate triangle ABF 90^{o} anticlockwise about point B mapping AB onto BC.

The resulting shape should look like this:

b
_________________
| /|
| / |
| / |
a | / |
_________| / | b
|`-. c / |
| `-. / |
a | c `-. / |
| `-. / |
|_________________`'___________|

This shape consists of two squares with sides of length a and b. Since all the operations performed on the original square are area preserving the area of the new shape must have the same as the area of the square.

Therefore **a**^{2}=b^{2}+c^{2}