Let *S* be the area of triangle *T* in three-dimensional space. Let *A*, *B*, and *C* be area of the projections of *T* to the *xy*, *yz*, and *xz* planes respectively. Then

*S*^{2} = *A*^{2} + *B*^{2} + *C*^{2}.

There’s an elegant proof of this theorem here using differential forms. Below I’ll sketch a less elegant but more elementary proof.

You could prove the identity above by using the fact that the area of a triangle spanned by two vectors is half the length of their cross product. Suppose *a*, *b*, and *c* are the locations of the three corners of *T*. Then

*S ^{2} = v*

^{2}/2,

where

*v* = (*a* − *b*) × (*c* − *b*)

and by *v*^{2} we mean the dot product of *v* with itself.

Write out the components of *v*^{2} and you get three squared terms. Notice that when you set the *x* components to zero, i.e. project onto the *yz* plane, the first of the three terms is unchanged and the other two are zero. In other words, the first of the three terms of *v*^{2} is *A*^{2}. A similar argument shows that the other two terms are *B*^{2} and *C*^{2}.

That’s a neat result!

A special case is visually analogous to Pythagoras’ theorem: If the three vertices of T lie on the three coordinate axes, we get a “right-angle pyramid” whose base is T and whose lateral faces are the three projections of T. The square of the “hypotenuse face” is the sum of squares of the three “leg faces.” Could there be a pure geometric proof in four dimensions?

2S = sqrt(dot(v, v))

(2S)^2 = [sqrt(dot(v, v))]^2

4S^2 = dot(v, v)

S^2 = v^2 / 4

Can we phrase this as a generalization of the Pythagorean Theorem? If C is the length of a line in two-dimensional space, and A and B are the length of its projections onto the x-axis and y-axis, the Pythagorean Theorem tells us that C^2 = A^2 + B^2 . That seems like the sort of thing that isn’t a coincidence.

If we have a tetrahedral volume V in a four-dimensional space, does the analogous equation V^2 = A^2 + B^2 + C^2 + D^2 hold for the projections of V into the xyz, xyw, xzw, and yzw subspaces?

Why stop there? This cries out for an induction proof of k-dimensional volumes projected into the k (k-1) dimensional subspaces.

Mike, we’re talking about (k-1) dimensional volumes in a k-dimensional space being projected into (k-1) dimensional subspaces. A line in 2-space projected onto a line is still a line; a triangle in 3-space projected onto a plane is still a triangle. There’s no loss of dimensionality in the projection, just a distortion.

(Technically, the projection of one line onto another line may be a point, if the lines are orthogonal, and the projection of a triangle onto a plane may be a line, if the plane the triangle lies in is orthogonal to the projection plane. But in general the image of a two-dimensional figure in a two-dimensional space will still be two-dimensional.)

Niiiice. Presumably the theorem works for any planar figure (because it can be decomposed into mutually exclusive triangles). Nonconvex, donuts, non-simply connected, doesn’t matter. Does the theorem hold for planar figures with countable vertices? Uncountable? If you can get all that, PLUS higher dimensions (infinite dimensions?), then you might be looking at something pretty powerful.

In fact, there is so many ways to prove this that you can spend less then all you life trying to find elegant solution. IMHO, finding the simplest solution to be _enough_ to prove the theorem (but not simpler) – the better. So you close to it. If someone wants to generalization theorem – they just changed the point of view for that particular case, 3d, and moved to k-space with no reason. (k can be infinite?)

An aside: Setting v^2 to the dot product of v with itself is actually a very natural choice, more than just notational sugar and convenience. In Clifford (or Geometric) algebra, this is called the contraction axiom, one of the fundamental properties associated with the vector product of that algebra (which leads directly to the anticommutative property of orthogonal vectors, and other important results).