3. In $\triangle A B C$, the midpoints of sides $B C, C A, A B$ are $L, M, N$ respectively, and points $D, E$ are on $B C, A B$ respectively, such that $A D, C E$ both bisect the perimeter of $\triangle A B C$. Points $P, Q$ are the reflections of $D, E$ over $L, N$ respectively, and $P Q$ intersects $L M$ at point $F$. Assuming $A B > A C$, then $A F$ must pass through the (
$(A)$ incenter
$(B)$ circumcenter
$(C)$ centroid
$(D)$ orthocenter
Answer: Choose $A$.
Let the three sides of $\triangle ABC$ and the semi-perimeter be $a, b, c, p$ respectively. Then $B Q = A E = p - b$, and $B P = C D = p - b$, so $B Q = B P$. Since $L M$ is parallel to $A B$, therefore
$$
L F = L P = B P - B L = p - b - \frac{a}{2} = \frac{c - b}{2} \text{, }
$$

Thus, $F M = L M - L F = \frac{b}{2} = A M$, $\angle M A F = \angle A F M = \angle F A B$, so $A F$ is the angle bisector of $\angle A$.