Problem
1982 AHSME Problem 21
In the adjoining figure, the triangle ABC is a right triangle with \angle BCA=90^\circ. Median CM is perpendicular to median BN, and side BC=s. The length of BN is
![[asy] size(200); defaultpen(linewidth(0.7)+fontsize(10));real r=54.72; pair B=origin, C=dir(r), A=intersectionpoint(B--(9,0), C--C+4*dir(r-90)), M=midpoint(B--A), N=midpoint(A--C), P=intersectionpoint(B--N, C--M); draw(M--C--A--B--C^^B--N); pair point=P; markscalefactor=0.005; draw(rightanglemark(C,P,B)); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$M$", M, S); label("$N$", N, dir(C--A)*dir(90)); label("$s$", B--C, NW);[/asy]](/static/aops/54b9945aa814f7d66e69a2c36ba71be4ce47b8ad.png)
\text {(A)} s\sqrt 2 \qquad \text {(B)} \frac 32s\sqrt2 \qquad \text {(C)} 2s\sqrt2 \qquad \text{(D)}\frac{1}{2}s\sqrt5\qquad \text{(E)}\frac{1}{2}s\sqrt6
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