Pythagoras’ theorem is usually presented as a theorem about triangles: in a right triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides. It is perhaps easier to think of it as a nice result relating rectangles and squares: in a rectangle, the square on the diagonal is equal to the sum of the squares on the long and short sides. There are many proofs of the theorem, most of them not as well-known as they might be. Euclid’s proof in Elements I, 47 (see David Joyce’s excellent website) is an elegant argument in cut-and-paste geometry. If it seems a little daunting at first, it can be tamed by using shaded triangles to keep track of the steps. My purpose here is to give a simple approach to the theorem using metric geometry, that is, geometry where the lengths and areas have values that can be calculated. I make no claims to originality in the approach, but it seems to be one that is not readily accessible. Mathematical arguments are generally easier to follow if you work along with them, and if you want to do that here, all you need is some squared paper and a pencil. Of course, squared paper was exactly what Euclid was avoiding.
Proving Pythagoras’ theorem is one thing. Another is wondering why anyone would come up with it in the first place. The metric geometry approach makes it a natural result that anyone might of thought of in a half-hour of idle doodling, rather than some deep mystery. Begin with a square and bisect it horizontally and vertically, then join the diagonals of the smaller squares to make a diamond, as in Figure 1.
It is clear that the area of the diamond is half the area of the whole square, since it includes half of each of the smaller squares.
On the other hand, half of the area of the large square can be chosen by taking all of the area of half of the squares.
The connection between these two choices can be seen as building squares on the edges of a triangle. The blue squares are formed on two sides of the triangle, and the red diamond on the hypotenuse of the triangle.
An obvious question is whether this result, the equality of the blue and red areas is just a coincidence, or whether there is something more general gong on. What if you divided the original square into a 3x3 grid, or a 4x4 grid? You can try it for yourself, or go on to the next page.