In a rectangle $A B C D, A B=9 \mathrm{~cm}$ and $B C=6 \mathrm{~cm}$. $P$ and $Q$ are two points on $B C$ such that the areas of the figures $A B P, A P Q$, and $A Q C D$ are in geometric progression. If the area of the figure $A Q C D$ is four times the area of triangle $A B P$, then $B P: P Q: Q C$ is
- $1: 1: 2$
- $1: 2: 1$
- $1: 2: 4$
- $2: 4: 1$