Let vec s_1,...,vec s_4 be the outward normals of the faces…
Let $\vec s_1,\dots,\vec s_4$ be the outward normals of the faces of a tetrahedron, with $|\vec s_i|$ equal to the area of the face. Prove that $\vec s_1+\vec s_2+\vec s_3+\vec s_4=\vec 0$.
Explanation of Method
The outward face-area vectors of a tetrahedron sum to zero.
Step 1: The four face vectors
Take $A$ as origin, $\vec p=\overrightarrow{AB}$, $\vec q=\overrightarrow{AC}$, $\vec r=\overrightarrow{AD}$, with $[\vec p,\vec q,\vec r]\gt0$ (otherwise rename $B\leftrightarrow C$). The vector $\vec p\times\vec q$ makes $(\vec p\times\vec q)\cdot\vec r\gt0$, i.e. it points towards $D$, into the solid. The outward vector of face $ABC$ is therefore $-\tfrac12\vec p\times\vec q$, and so on:
| Face | Opposite vertex | Outward area vector |
|---|---|---|
| $BCD$ | $A$ | $\vec s_1=\tfrac12(\vec q-\vec p)\times(\vec r-\vec p)=\tfrac12(\vec q\times\vec r+\vec r\times\vec p+\vec p\times\vec q)$ |
| $ACD$ | $B$ | $\vec s_2=-\tfrac12\,\vec q\times\vec r$ |
| $ABD$ | $C$ | $\vec s_3=-\tfrac12\,\vec r\times\vec p$ |
| $ABC$ | $D$ | $\vec s_4=-\tfrac12\,\vec p\times\vec q$ |
Check for $\vec s_1$: the direction from face $BCD$ towards $A$ is $-\vec p$, and $\vec s_1\cdot(-\vec p)=-\tfrac12[\vec q,\vec r,\vec p]\lt0$. So $\vec s_1$ points away from $A$, i.e. outward $\checkmark$
Step 2: Sum
\[ \vec s_1+\vec s_2+\vec s_3+\vec s_4=\tfrac12(\vec q\times\vec r+\vec r\times\vec p+\vec p\times\vec q)-\tfrac12(\vec q\times\vec r+\vec r\times\vec p+\vec p\times\vec q)=\vec 0. \]
\[\vec s_1+\vec s_2+\vec s_3+\vec s_4=\vec 0\qquad\blacksquare\]
Step 3: Projection argument
For any unit vector $\hat e$, $\hat e\cdot\vec s_i$ is the signed area of the projection of face $i$ onto a plane perpendicular to $\hat e$. The faces seen "from the front" and "from the back" cover the same projected region of the closed solid, so these signed areas cancel. Since $\hat e$ is arbitrary, the sum is $\vec 0$.
Result
\[\boxed{\,\vec s_1+\vec s_2+\vec s_3+\vec s_4=\vec 0\qquad\blacksquare\,}\]
Remark
Correction of signs. The source takes $+\tfrac12\vec p\times\vec q$ etc. as outward for the three faces through $A$ and calls $(\vec q-\vec p)\times(\vec r-\vec p)$ inward for $BCD$. For a right-handed $\vec p,\vec q,\vec r$ all four of these orientations are reversed. The total is still $\vec 0$, because flipping all four vectors does not change a zero sum, so the conclusion stands.