6. Given that $\odot O$ is the circumcircle of acute $\triangle A B C$, $N$ is the midpoint of $B C$, and point $M$ is on $O A$, and $\angle A B C=$ $4 \angle O N M, \angle A C B=6 \angle O N M$. Then the size relationship between $\angle O M N$ and $\angle O N M$ is ( ).
(A) $\angle O M N>\angle O N M$
(B) $\angle O M N=\angle O N M$
(C) $\angle O M N<\angle O N M$
(D) cannot be determined
6. B.

As shown in Figure 3, connect $O B$, extend $O N$ to intersect $\odot O$ at point $D$, and connect $A D$. Let $\angle O N M=x$, then
$$
\begin{array}{l}
\angle A B C=4 x, \\
\angle A C B=6 x . \\
\text { Hence } \angle A O B=2 \angle A C B \\
=12 x, \\
\angle B A C=180^{\circ}-10 x .
\end{array}
$$

By the converse of the perpendicular diameter theorem, point $D$ bisects $\overparen{B C}$.
Therefore, $\angle B A D=\frac{1}{2} \angle B A C=90^{\circ}-5 x$.
Also, $\angle B A O=\angle A B O=\frac{180^{\circ}-\angle A O B}{2}=90^{\circ}-6 x$, so $\angle O A D=\angle B A D-\angle B A O=x$.
In $\triangle O A D$, $O A=O D$, thus $\angle D=\angle O A D=x$.
Therefore, $\angle O N M=\angle D, M N \parallel A D$.
Thus, $\angle O M N=\angle O A D=x=\angle O N M$.