6. As shown in Figure 2, point $O$ is the center of the regular hexagon $A B C D E F$, $O M$ $\perp C D$, and $N$ is the midpoint of $O M$. Then $S_{\triangle A B N}: S_{\triangle B C N}$ equals ( ).
(A) $9: 5$
(B) $7: 4$
(C) $5: 3$
(D) $3: 2$
6. (C).

Obviously, $C$, $O$, and $F$ are collinear. As shown in Figure 6, connect $N D$ and $N E$, and draw a line $P Q \perp A B$ through $N$. Since $A B \parallel D E \parallel C F$, then $P Q \perp D E$, $P Q \perp C F$, and $P$, $K$, $Q$ are all feet of the perpendiculars.

By the axial symmetry of the regular hexagon, $\triangle B C N \cong \triangle E D N$.
It is easy to see that $P K=K Q=O M=2 O N$.

Also, $\angle N O K=\angle C O M=30^{\circ}$, $K N=\frac{1}{2} O N$. Let $K N=1$, then we get
$$
\begin{array}{l}
P N=P K+K N=5, N Q=K Q-K N=3 . \\
S_{\triangle M B N}: S_{\triangle \angle D N}=P N: N Q=5: 3 .
\end{array}
$$