3. As shown in Figure 1, given that the midpoints of the three sides $BC$, $CA$, and $AB$ of $\triangle ABC$ are $L$, $M$, and $N$ respectively, and points $D$ and $E$ are on $BC$ and $AB$ respectively, satisfying that $AD$ and $CE$ both bisect the perimeter of $\triangle ABC$. Points $P$ and $Q$ are the symmetric points of $D$ and $E$ with respect to $L$ and $N$ respectively, and $PQ$ intersects $LM$ at point $F$. If $AB > AC$, then $AF$ must pass through the ( ) of $\triangle ABC$.
(A) incenter
(B) circumcenter
(C) centroid
(D) orthocenter
3. A.

Let the three sides and the semi-perimeter of $\triangle A B C$ be $a$, $b$, $c$, and $p$ respectively. Then,
$$
B Q=A E=p-b, B P=C D=p-b .
$$

Thus, $B Q=B P$.
Since $L M / / A B$, we have,
$$
L F=L P=B P-B L=p-b-\frac{a}{2}=\frac{c-b}{2} \text {. }
$$

Therefore, $F M=L M-L F=\frac{b}{2}=A M$,
$$
\angle M A F=\angle A F M=\angle F A B .
$$

Hence, $A F$ is the angle bisector of $\angle A$.