Answer: $(\sqrt{5} + 1)/2$.
Let $\lambda = AX : BX$ be the required ratio. On one hand, by Thales' theorem
$$\begin{aligned} \lambda &= \frac{AX}{BX} = \frac{YR}{RB} = \\ &= [YR = CD, \text{ since } CDYR \text{ is a parallelogram}] = \\ &= \frac{CD}{RB}. \end{aligned} \quad (1)$$

On the other hand,
$$\begin{aligned} \lambda &= \frac{AX}{BX} = [AX = CY, \text{ since } CYAX \text{ is a parallelogram}] = \frac{CY}{BX} = \\ &= [BX = CP, \text{ since } CPXB \text{ is a parallelogram}] = \frac{CY}{CP} = \\ &= [\text{Thales' theorem for } \angle CYB, PQ \parallel BC] = \frac{BY}{BQ} = \\ &= [BQ = CD, \text{ since } BCDQ \text{ is a parallelogram}] = \frac{BY}{CD} = \frac{YR + BR}{CD} = \\ &= [YR = CD, \text{ since } RCDY \text{ is a parallelogram}] = \frac{CD + BR}{CD} = \\ &= \frac{1 + BR/CD}{1} = [\text{see (1)}] = 1 + \frac{1}{\lambda}. \end{aligned}$$
So, $\lambda = 1 + \frac{1}{\lambda}$ or $\lambda^2 - \lambda - 1 = 0$. From this quadratic equation, taking into account that $\lambda > 0$, we obtain $\lambda = (\sqrt{5} + 1)/2$.