In the parallelepiped $A B C D A 1 B 1 C 1 D 1$, a segment is drawn connecting vertex $A$ to the midpoint of edge $C C 1$. In what ratio does this segment get divided by the plane $B D A 1$?
When the parallelepiped ABCDA1B1C1D1 is projected in parallel onto the plane of the face AA1B1B with the projecting line $A D$, the parallelepiped will transform into the parallelogram $A A 1 B 1 B$, the midpoint $M$ of the edge $C C 1$ will transform into the midpoint $M^{\prime}$ of the segment $B B 1$, the point $P$ of intersection of the diagonals of the face $A B C D$ will transform into the midpoint $P^{\prime}$ of the side $A B$, and the point $K$ of intersection of the segment $A M$ with the plane $B D A 1$ will transform into the point $K^{\prime}$ of the segment $A 1 P^{\prime}$, connecting the vertex $A 1$ of the parallelogram $A A 1 B 1 B$ with the midpoint of the side $A B$. Let the lines $A 1 P^{\prime}$ and $B B 1$ intersect at the point $F$. Then

$$
B F=A A 1, F M^{\prime}=B F+B M^{\prime}=A A 1+\frac{1}{2} A A 1=\frac{3}{2} A A 1
$$

From the similarity of triangles $A K^{\prime} A 1$ and $M^{\prime} K^{\prime} F$, we find that

$$
A K^{\prime}: K^{\prime} M^{\prime}=A A 1: F M^{\prime}=A A 1: \frac{3}{2} A A 1=2: 3 .
$$

Therefore,

$$
A K: K M=A K^{\prime}: K^{\prime} M^{\prime}=2: 3 .
$$

## Answer

2:3, measured from point $A$.