Okay, so this is an example of the proof i found on a site:

This is another proof by Floor van Lamoen; Floor has been led to the proof via Bottema's theorem. However, the theorem is not actually needed to carry out the proof.

In the figure, M is the center of square ABA'B'. Triangle AB'C' is a rotation of triangle ABC. So we see that B' lies on C'B''. Similarly, A' lies on A''C''. Both AA'' and BB'' equal a + b. Thus the distance from M to AC' as well as to B'C' is equal to (a + b)/2. This gives

Area(AMB'C') = Area(MAC') + Area(MBC')
= (a + b)/2 · b/2 + (a + b)/2 · a/2
= a2/4 + ab/2 + b2/4.

But also:

Area(AMB'C') = Area(AMB') + Area(ABC')
= c2/4 + ab/2.

This yields a2/4 + b2/4 = c2/4 and the Pythagorean theorem.

our problem is with the first area equation
Area(AMB'C') = Area(MAC') + Area(MBC')
= (a + b)/2 · b/2 + (a + b)/2 · a/2
= a2/4 + ab/2 + b2/4.
we can see which triangles they are most likely talking about, but they would be MAC' and MB'C' or MBC"

So if anyone could please clarify this and possibly explain where the second part of the equation is coming from it would be a great help.

Since i was unable to upload the picture here, i am including a link of the site i found it on

http://www.cut-the-knot.org/pythagoras/index.shtml

it's proof #63

Thanks!
Reb and Sab

There have been two typos. Many thanks for bringing this up.

Area(AMB'C') = Area(MAC') + Area(MBC')

should have been

Area(AMB'C') = Area(MAC') + Area(MB'C')

and

Area(AMB'C') = Area(AMB') + Area(ABC')

should have been

Area(AMB'C') = Area(AMB') + Area(AB'C')

Both are fixed now.