17.95 As shown in the figure, triangle $ABC$ is a right triangle, where $\angle BCA=90^{\circ}$. The median $CM$ is perpendicular to the median $BN$, and $BC=s$. Then the length of $BN$ is
(A) $s \sqrt{2}$.
(B) $\frac{3}{2} s \sqrt{2}$.
(C) $2 s \sqrt{2}$.
(D) $\frac{s \sqrt{5}}{2}$.
(E) $\frac{s \sqrt{6}}{2}$.
(33rd American High School Mathematics Examination, 1982)
[Solution] In $\triangle A B C$, since $C M$ and $B N$ intersect at point $D$, if we set $x = D N$, then $2 x = B D$.
Also, in the right triangle $\triangle C B N$, $C D$ is the altitude on side $B N$, so
$$
B C^{2} = B D \cdot B N \text {, }
$$

which means $s^{2} = 2 x \cdot 3 x = 6 x^{2}$, or $x = \frac{s}{\sqrt{6}}$.
Therefore, $B N = 3 x = \frac{3 s}{\sqrt{6}} = \frac{\sqrt{6}}{2} s$.
Hence, the answer is $(E)$.