10,11 In the parallelepiped $A B C D A 1 B 1 C 1 D 1$, points $K$ and $M$ are taken on the lines $A C$ and $B A 1$, respectively, such that $K M \| D B 1$. Find the ratio $K M: D B 1$.
When parallel projecting the parallelepiped ABCDA1B1C1D1 onto the plane of the diagonal section $B B 1 D 1 D$ with the projecting line $A C$, the parallelepiped will transform into the parallelogram $B B 1 D 1 D$, point $K$ will transform into the midpoint $K^{\prime}$ of side $D B$, point $A 1$ - into the midpoint $F$ of segment $B 1 D 1$, segment $B A 1$ - into segment $B F$, line $K M$ - into line $l$, parallel to $K M$ and passing through point $K^{\prime}$, point $M$ - into point $M^{\prime}$ of intersection of lines $B F$ and $l$ (line $K M$ is parallel to the plane $B B 1 D 1 D$, since it is parallel to the line $D B 1$ in this plane). Let $L$ be the point of intersection of $B F$ and $D B 1$. Then $\frac{D L}{L B_{1}}=\frac{B B}{F B_{1}}=2$, and since $K^{\prime} M^{\prime}$ is the midline of triangle $B D L$, we have

$$
\frac{D B_{1}}{K \times M}=\frac{D B_{1}}{K^{\prime} M^{\prime}}=\frac{D L+2 B_{1}}{K^{\prime} M^{\prime}}=\frac{2 K^{\prime} M^{\prime}+K^{\prime} K^{\prime} M^{\prime}}{K^{\prime} M^{\prime}}=3
$$

## Answer

$1: 3$.