subdivide a square of side c, into 4 equal parts by the two diagonals. note this gives a simple case of the theorem, when a=b, since the area of the square is c^2 and the areas of the 4 triangles is 2 a^2. i.e. a^2 + a^2 = c^2.
now change the angle of the lines subdividing the square, until they leave a small square in the center, with 4 right triangles around it of sides a,b,c.
then the small square in the center has area (b-a)^2, and the 4 triangles have total area 2ab, so the sum of 2ab and (b-a)^2 must equal c^2.
QED.
i found this proof sitting on the plane for a while with a small scrap of paper, (and not playing with a calculator).