(3) As shown in the figure, given that $L, M, N$ are the midpoints of the three sides $BC, CA, AB$ of $\triangle ABC$, respectively, and $D, E$ are points on $BC, AB$, respectively, such that $AD, CE$ both bisect the perimeter of $\triangle ABC$. $P, Q$ are the points symmetric to $D, E$ with respect to $L, N$, respectively. $PQ$ 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) Let the three sides and the semi-perimeter of $\triangle A B C$ be $a, b, c, p$, respectively, then
$$
B Q=A E=p-b, B P=C D=p-b,
$$

Therefore, $B Q=B P$. Since $L M$ is parallel to $A B$, we have
$$
L F=L P=B P-B L=p-b-\frac{a}{2}=\frac{c-b}{2},
$$

Thus,
$$
\begin{array}{l}
F M=L M-L F=\frac{b}{2}=A M, \\
\angle M A F=\angle A F M=\angle F A B,
\end{array}
$$

Therefore, $A F$ is the angle bisector of $\angle A$. Hence, the answer is $\mathrm{A}$.