China Mathematical Competition (Shaanxi)

As in the diagram, from point $C$ draw a line $CE$ such that it is equal and parallel to $AB$. Construct a prism $ABF$-$ECD$ with $△CDE$ as base and $BC$ as a lateral edge. Denote $V_1$ as the volume of the tetrahedron and $V_2$ the volume of the prism, then $V_1=\frac{1}{3}{V}_2$.



Since $S_{△CDE}=\frac{1}{2}CE\cdot CD\sin \angle ECD$, and the common perpendicular line $MN$ of $AB$ and $CD$ is the height of the prism, then $$V_2=\frac{1}{2}MN\cdot CE\cdot CD\sin \angle ECD=\frac{3}{2}.$$ So $V_1=\frac{1}{3}{V}_2=\frac{1}{2}$.

Answer: B.