 Closure of outward face area vectors

For a closed polyhedron, let $A_F$ be a face area and ${\mathbf{n}}_F$ its outward unit normal vector. The sum of the face area vectors is zero. To prove it, fix any constant vector $\mathbf{q}$ and apply the divergence theorem to the constant vector field $\mathbf{q}$: its divergence vanishes, while its outward flux is $\mathbf{q}\cdot {\sum }_F{A}_F{\mathbf{n}}_F$. This is zero for every $\mathbf{q}$, so the sum vanishes. For a tetrahedron, the same identity follows directly by expanding the four oriented face cross products; multiplying every face vector by two leaves the identity intact.