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}.