Let $p$ be the plane through $AB$ parallel to $CD$. Any plane $p'$ parallel to $p$ which intersects the segment $KL$ will intersect the tetrahedron in a parallelogram with center on the line $KL$. Any plane $q$ through $KL$ will pass through the center of the parallelogram and hence divide it into two equal parts. So suppose $q$ divides the tetrahedron into two parts $T$ and $T'$. We have established that the area of the intersection of $p'$ and $T$ is equal to the area of the intersection of $p'$ and $T'$. It follows from Cavalieri's principle that $T$ and $T'$ have equal volume. (To see that $PQRS$ is a parallelogram, note that $PQ$ is parallel to $CD$, and $RS$ is parallel to $CD$, so $PQ$ and $RS$ are parallel. Similarly $QR$ and $PS$ are parallel to $AB$ and hence to each other.)

Arbitrary plane through $K,L$ cuts $AD,BC$ at $X,Y.$ It's clear that the locus of the points that bisect the segments with endpoints on $BC,AD,$ is the midplane $\lambda$ parallel to those passing through $BC,AD$ and parallel to each other, thus $KL \in \lambda$ $\Longrightarrow$ $KL$ bisects $\overline{XY}$ $\Longrightarrow$ $X,Y$ are equidistant from the plane $CDK.$ Since $[DLK]=[CLK],$ then tetrahedra $[DLKX]=[CLKY]$ are equivalent. If $V_1$ and $V_2$ denote the volumen of the pentahedra $LXKYBD$ and $LXKYCA,$ we have then $V_1=[DKBYL]+[CLKY]=[DKBC]=\frac{_1}{^2}[ABCD]$ Similarly, $V_2=\frac{_1}{^2}[ABCD]$ $\Longrightarrow$ $V_1=V_2=\frac{_1}{^2}[ABCD]$