At the base of the quadrangular pyramid $SABCD$ lies the parallelogram $ABCD$. Prove that for any point $O$ inside the pyramid, the sum of the volumes of the tetrahedra $OSAB$ and $OSCD$ is equal to the sum of the volumes of the tetrahedra $OSBC$ and $OSDA$ .
parmenides51 wrote:
At the base of the quadrangular pyramid $SABCD$ lies the parallelogram $ABCD$. Prove that for any point $O$ inside the pyramid, the sum of the volumes of the tetrahedra $OSAB$ and $OSCD$ is equal to the sum of the volumes of the tetrahedra $OSBC$ and $OSDA$ .
Translated from here.
**Solution.** Let \( X \) be the point where the ray \( SO \) intersects the plane \( ABCD \) (see Fig. 3). Since the point \( O \) lies inside the pyramid, the point \( X \) lies within its base. In this case, \( S_{XAB} + S_{XCD} = S_{XBC} + S_{XDA} \) (one of the possible proofs of this fact is illustrated in Fig. 4 – each of the sums is equal to half of the area of the parallelogram \( ABCD \)). Consequently,
\[
V_{XSAB} + V_{XSCD} = V_{XSBC} + V_{XSDA},
\]
since the height of these pyramids, drawn from vertex \( S \), is the same. Similarly,
\[
V_{XOAB} + V_{XOCD} = V_{XOBC} + V_{XODA}.
\]
Subtracting equation (1) from equation (2), we obtain the desired result.
