Area of a triangle and its projections

Let S be the area of triangle T in three-dimensional space. Let AB, and C be area of the projections of T to the xyyz, and xz planes respectively. Then


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 ab, and c are the locations of the three corners of T. Then

S2 = v2/2,


v = (a – b) × (c – b)

and by v2 we mean the dot product of v with itself.

Write out the components of v2 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 v2 is A2. A similar argument shows that the other two terms are B2 and C2.