29 Dec '11 09:55>
The way I show it to my students is to draw two squares with side a + b. Obviously the areas of the two squares are the same. In one square draw the lines
(0,a) (b,0)
(b,0),(a+b,a)
(a+b,a),(a,a+b)
(a,a+b),(0,a)
which cuts four triangles with side a, b, c off the big square and leaves the square with side c at the middle. Area c^2
In the other big square, draw the lines (0,a),(a+b,a) and (b,0),(b,a+b) and join the intersection point to (0,0) and (a+b,a+b). That gives the same four triangles as in the first square, and squares with side a and side b, areas a^2 and b^2;
(a+b)^2 = (a+b)^2
a^2 + b^2 + 4 x (ab)/2 = c^2 + 4 x (ab)/2
a^2 + b^2 = c^2
It's far from the only proof for it, but simple, easy to draw, and makes sense.
(0,a) (b,0)
(b,0),(a+b,a)
(a+b,a),(a,a+b)
(a,a+b),(0,a)
which cuts four triangles with side a, b, c off the big square and leaves the square with side c at the middle. Area c^2
In the other big square, draw the lines (0,a),(a+b,a) and (b,0),(b,a+b) and join the intersection point to (0,0) and (a+b,a+b). That gives the same four triangles as in the first square, and squares with side a and side b, areas a^2 and b^2;
(a+b)^2 = (a+b)^2
a^2 + b^2 + 4 x (ab)/2 = c^2 + 4 x (ab)/2
a^2 + b^2 = c^2
It's far from the only proof for it, but simple, easy to draw, and makes sense.