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.