4. As shown in Figure 1, in $\triangle A B C$, it is given that $A B>A C$, points $D$ and $E$ are on $A B$ and $A C$ respectively, and $B D=C E$. Taking the midpoints $M$ and $N$ of $B E$ and $C D$ respectively, line $M N$ intersects $A B$ and $A C$ at points $P$ and $Q$. Then ( ).
(A) $A P>A Q$
(B) $A P<A Q$
(C) $A P=A Q$
(D) Cannot be determined
4. C.

As shown in Figure 3, take the midpoint $K$ of $BC$. Then $KM$ and $KN$ are the midlines of $\triangle BCE$ and $\triangle BCD$, respectively. Hence,
$$
\begin{array}{c}
K N=\frac{1}{2} B D \\
=\frac{1}{2} C E=K M \\
\Rightarrow \angle K M N=\angle K N M .
\end{array}
$$

But from $KM \parallel AC$ and $KN \parallel AB$, we get
$$
\begin{array}{l}
\angle A Q P=\angle K M N, \angle A P Q=\angle K N M \\
\Rightarrow \angle A Q P=\angle A P Q \Rightarrow A P=A Q
\end{array}
$$