6. In $\square A B C$, points $P$ and $Q$ are on sides $A B$ and $A C$ respectively, and $\frac{B P}{A P}+\frac{C Q}{Q A}=1$. Then $P Q$ must pass through the ( ) of the triangle.
(A) orthocenter (B) centroid (C) incenter (D) circumcenter
6.(B).

As shown in Figure 4, let the median $AD$ intersect $PQ$ at point $G$. Draw perpendiculars from $A, B, C, D$ to the line $PQ$, with the feet of the perpendiculars being $A', B', C', D'$, respectively. Then,
$$
\begin{array}{c}
1=\frac{BP}{AP_{\top}}+\frac{CQ}{QA_{4+0}} \\
=\frac{BB'+CC'}{AA'}=\frac{2DD'}{AA'}=\frac{2DG}{AG},
\end{array}
$$

i.e., $AG=2DG$, so point $G$ is the centroid of $\triangle ABC$.