Answer: D
When diagonal BD is drawn, it intersects PQ at point O. In triangle \( \triangle ABD \), since \( AD \parallel PO \), ∴ \( DO : OB = AP : PB = 2 : 1 \) Again, in triangle \( \triangle DBC \), since \( BC \parallel OQ \), ∴ \( DQ : QC = DO : OB = 2 : 1 \)
When diagonal BD is drawn, it intersects PQ at point O. In triangle \( \triangle ABD \), since \( AD \parallel PO \), ∴ \( DO : OB = AP : PB = 2 : 1 \) Again, in triangle \( \triangle DBC \), since \( BC \parallel OQ \), ∴ \( DQ : QC = DO : OB = 2 : 1 \)