Let $P$ and $Q$ be the centers of the faces $BCC_1B_1$ and $DCC_1D_1$ and let $\alpha = (APQ)$ (Fig. 1). Since $PQ \parallel BD$ the plane $\alpha$ meets the plane $(ABCD)$ at the line through $A$ which is parallel to $BD$. We denote by $T$ and $S$ the intersection points of this line with the lines $CB$ and $CD$, respectively.

The lines $TP$ and $SQ$ meet the edge $CC_1$ at the intersection point $R$ of $\alpha$ and $CC_1$. Let $M = TR \cap BB_1$ and $N = SR \cap DD_1$. Then the intersection of $\alpha$ and the surface of the cube is the quadrilateral $AMRN$. (It is easy to see that it is a rhombus.) It is clear that $BT = BA = 1$. Hence $B$ and $M$ are the midpoints of $TC$ and $TR$, respectively. Then $\frac{BM}{RC} = \frac{1}{2}$ and since $BM = RC_1 = 1 - RC$, we find $BM = \frac{1}{3}$. Analogously $DN = \frac{1}{3}$.

Denote by $V$ the volume of the polytope cut from the cube by the planes $(ABCD)$ and $(AMRN)$ (Fig. 2). Let $A_2$ and $C_2$ be the intersection points of $AA_1$ and $CC_1$ with the plane through $MN$ and parallel to $(ABCD)$. Then $RC_2 = RC - CC_2 = MB = AA_2 = \frac{1}{3}$ and hence the tetrahedra $NMC_2R$ and $NMA_2A$ have equal volumes. This shows that $V = V_{ABCD A_2 MC_2 N} = \frac{1}{3}$. Hence the required ratio is $1:2$.

Fig. 1

Fig. 2