One of the most fundamental truths of Euclidean geometry – and, indeed, of the geometry of the real world, for all that its precision here is limited by the scale of the triangle in relation to the local curvature of space-time – describes a relationship among the sides of a right-angled triangle. It has been known since antiquity – it was known in ancient Indian cultures and (probably subsequently) in Mesopotamian cultures. More recently yet, the ancient Greek school of thought known as the Pythagoreans (notable for also sharing, with some Indian cultures, both vegetarianism and a belief in re-incarnation) attributed it to their (possibly mythical) founder, Pythagoras. It asserts that

The sum of the areas of the squares on the two orthogonal sides of a right-angle triangle is the area of the square on the third side.

The third side, facing the right angle, is called
the hypotenuse

of the right-angle triangle. (The addition
of an extra term in the cosine of the angle between two
sides generalises this rule to the case of
non-right angles.)

Given a right-angle triangle, I need to show you how you can satisfy yourself that it has the claimed property. I'll explain what to do with the right-angle triangle you've got, illustrating each step with a right-angle triangle of my own. Hopefully, that'll help you see how to apply what I say to yours; and, thereby, lead you to see that the same is true of yours.

So take your right-angle triangle and rotate it through a quarter turn about its right-angle corner (in either sense). You now have two copies of it, with each side of either perpendicular to the corresponding side of the other; and the perpendicular sides all meet in the corner about which you rotated. In particular, for each of the perpendicular sides of your triangle, this gives you two perpendicular copies of that side: complete these to form a square, in each case.

I'm now going to show that, if we translate the combination of those two squares through the displacements along the two copies of our hypotenuse, its images fit around the original – abutting it, with no overlap and no gaps. From this it follows that, using these two translations, this figure tessellates the whole plane. If that's immediately apparent to you, you can skip all the diagrams and reasoning below. Since the two translations used are the same two that tessellate the plane with a square on either hypotenuse, this figure's area must in fact be equal to the area of that square, which is exactly the result to be proven.

To show that the tessellation works, let's first translate our pair of squares through one hypotenuse repeatedly (using alternating colours for each copy, to help tell them apart):

I trust it's clear that you can keep doing this indefinitely, in each direction, with each copy's concave corner at one end of the chosen hypotenuse surrounding the next copy's corner at the other end of that hypotenuse. Now let's take that train of pieces and translate it parallel to the other hypotenuse:

As you can see, and as I'm sure you'll find if you carefully perform the same operations with any right-angle triangle, the copies of the figure never overlap or leave gaps; and we can clearly continue repeating the pattern in each direction indefinitely, thereby tessellating the plane.

Now, indeed, I'm appealing to an intuition that figures tessellated by the same pair of translations must have equal area; but we can equally use our tessellated pair of squares and cut out the square on one hypotenuse, to look at the fragments of each square that it gives us. The tessellation property inescapably implies that these fragments, translated through various combinations of the two hypotenuse-displacements, can be reassembled into the two squares on the perpendicular sides – as can readily enough be verified. (Of course, this in turn relies on a different intuition, that cutting a figure up and rearranging its pieces doesn't change area. The two intuitions are, of course, equivalent.)

There are many other proofs, but this is
my favourite – I don't know if anyone else ever thought of it before me,
but I dreamed it up for myself (albeit I can't rule out having been prompted
thereto by some long-forgotten encounter with it, or something similar; and
it *is* equivalent to one of the
standard rearrangement proofs), which gives me a natural bias in its favour.
I do, that notwithstanding, consider it
to be more elegant and to touch more closely the reasons for the truth of the
result it proves. Finally – in case you thought I played a trick by my
choice of which right-angle triangle to use, or suspect that the variants you've
tried only worked because you didn't pick the quirky special case that'd break
the reasoning – consider this
animation (or a line-only
variant), that varies the ratio of the perpendicular sides of the triangle
through the full range of relative values possible:

Pythagoras's theorem enables us to define an addition on squares, pairwise, by using a side of each as a the perpendicular sides of a right-angled triangle, with the square on the hypotenuse serving as sum of the two squares. The construction is inherently symmetric, so the addition is commutative (a.k.a. Abelian). Because two sides and an angle of a triangle determine its remaining side and angles, the addition is cancellable. Suppose we add three squares in this wah: does it matter what order we do the additions ? (i.e., is the addition associative ?)

- Discussion of right-angle triangles (and now simplices), whose sides are in whole-number ratios to one another.
- Taking square roots of areas, i.e. constructing a square whose area is equal to those of any given polygon; one of the techniques for this applies Pythagoras's theorem to a difference of two squares to turn a long thin rectangle into a square.