Question

In a given tetrahedron ABCD let K and L be the centers of edges AB and CD respectively. Prove that every plane that contains the line KL divides the tetrahedron into two parts of equal volume

Answer 1

Suppose w.l.o.g. that the plane $\alpha$ through KL meets the interiors of edges AC and BD at X and Y. Let $\overrightarrow{AX}=\lambda \overrightarrow{AC}$ and $\overrightarrow{BY}=\mu \overrightarrow{BD}$, for $0\le \lambda ,\mu \le 1$. Then the vector $\overrightarrow{KX}=\lambda \overrightarrow{AC}-\frac{\overrightarrow{AB}}{2},\overrightarrow{KY}=\mu \overrightarrow{BD}+\frac{\overrightarrow{AB}}{2},\overrightarrow{KL}=\frac{\overrightarrow{AC}}{2}+\frac{\overrightarrow{BC}}{2}$ are coplanar; i.e., there exist real numbers a, b, c, not all zero, such that
$\overrightarrow{0}=a\overrightarrow{KX}+b\overrightarrow{KY}+c\overrightarrow{KL}=(\lambda a+\frac{c}{2})\overrightarrow{AC}+(\mu b+\frac{c}{2})\overrightarrow{BC}+\left(\frac{b-a}{2}\right)\overrightarrow{AB}$
Since $\overrightarrow{AC},\overrightarrow{BD},\overrightarrow{AB}$ are linearly independent, we must have $a=b$ and $\lambda =\mu$. We need to prove that the volume of the polyhedron KXLYBC, which is one of the parts of the tetrahedron ABCD partitioned by $\alpha$, equals half of the volume V of ABCD. Indeed, we obtain
$VKXLYBC=VKXLC+VKBYLC=\frac{1}{4}(1-\lambda )V+\frac{1}{4}(1+\mu )V=\frac{1}{2}V$