Ratio of segment lengths in rectangle

Area of $\triangle ABP = \frac{1}{2} \times AB \times BP$. Area of $\triangle APQ = \frac{1}{2} \times AB \times PQ$.

Let Area($\triangle ABP$) $= A$. Given Area($AQCD$) $= 4A$.

Since Area($ABP$), Area($APQ$), and Area($AQCD$) form a geometric progression with first term $A$ and third term $4A$, the second term Area($APQ$) is: $\text{Area}(APQ) = \sqrt{A \times 4A} = 2A$

Thus, the areas are in the ratio $1 : 2 : 4$.

Since triangles $ABP$, $APQ$, and trapezium $AQCD$ all share the same height perpendicular to $BC$ (or can be seen via bases on $BC$), their areas are directly proportional to the base lengths $BP$, $PQ$, and $QC$ respectively.

Therefore, $BP : PQ : QC = 1 : 2 : 4$.