16. Let tetrahedron $ABCD$ have $AB=41, AC=7, AD=18, BC=36, BD=27, CD=13$. Let $d$ be the distance between the midpoints of edges $AB$ and $CD$. Find the value of $d^{2}$.
16. Applying the median formula in $\triangle A B C$, we can calculate $C N^{2}=\frac{1}{4}\left(2 A C^{2}+2 B C^{2}-A B^{2}\right)=\frac{1009}{4}$. Similarly, $D N^{2}=$ $\frac{1}{4}\left(2 A D^{2}+2 B D^{2}-A B^{2}\right)=\frac{425}{4}$. In $\triangle N C D$, we have $a^{2}=M N^{2}=\frac{1}{4}\left(2 D N^{2}+2 C N^{2}-C D^{2}\right)=\frac{548}{4}=137$.