4. In $\triangle A B C$, $\angle A B C=12^{\circ}, \angle A C B=$ $132^{\circ}, B M$ and $C N$ are the external angle bisectors of these angles, and points $M, N$ are on lines $A C, A B$ respectively. Then ( ).
(A) $B M>C N$
(B) $B M=C N$
(C) $B M<C N$
(D) The size relationship between $B M$ and $C N$ is uncertain
4. B.

Since $\angle A B C=12^{\circ}$ and $B M$ is the external angle bisector of $\angle A B C$, we have
$$
\angle M B C=\frac{1}{2}\left(180^{\circ}-12^{\circ}\right)=84^{\circ} .
$$

Also, $\angle B C M=180^{\circ}-\angle A C B=48^{\circ}$, thus
$$
\angle B M C=180^{\circ}-84^{\circ}-48^{\circ}=48^{\circ} \text {. }
$$

Therefore, $B M=B C$.
Given $\angle A C N=\frac{1}{2}\left(180^{\circ}-\angle A C B\right)=24^{\circ}$, hence
$$
\begin{array}{l}
\angle B N C=180^{\circ}-\angle A B C-\angle B C N \\
=180^{\circ}-12^{\circ}-(\angle A C B+\angle A C N) \\
=168^{\circ}-\left(132^{\circ}+24^{\circ}\right)=12^{\circ}=\angle A B C .
\end{array}
$$

So, $C N=C B$.
Therefore, $B M=B C=C N$.