## An Easy Approach to the Theorem of Pythagoras

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.

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.

Last modified: 16
May 2005