2. As shown in Figure 1, the two sides $A B C D$ and $A B E F$ of the triangular prism are both rectangles, and $P, Q$ are points on $A E$ and $B D$ respectively. If $P Q \parallel$ plane $B C E$, then, ( ).
(A) It must be that $B Q = D Q$ and $E P = P A$
(B) It must be that $E P = P A$, but it is possible that $B Q \neq D Q$
(C) It must be that $B Q = D Q$, but it is possible that $E P \neq P A$
(D) It is possible that $B Q \neq D Q$ and $E P \neq P A$
2.D.

As shown in Figure 3, draw $PM \perp BE$ at point $M$, $QN \perp BC$ at point $N$, and connect $MN$. Since quadrilateral $ABEF$ is a rectangle, $PM \parallel AB$.
Similarly, $NQ \parallel AB$.
Therefore, $QN \parallel PM$.
Thus, points $P, Q, N, M$ are coplanar.
Since $PQ \parallel$ plane $BCE$, $PQ \subseteq$ plane $PQNM$, and plane $PQNM \cap$ plane $BCE = MN$, it follows that $PQ \parallel MN$.
Therefore, quadrilateral $PQNM$ is a parallelogram.
Hence, $PM = QN$.
Also, $\frac{PM}{AB} = \frac{EP}{EA}$, $\frac{QN}{CD} = \frac{BQ}{BD}$, $AB = CD$, and $PM = QN$, so $\frac{EP}{EA} = \frac{BQ}{BD}$, which means points $P$ and $Q$ divide $EA$ and $BD$ in the same ratio.