Let $A, B, C$ and $D$ be four points not lying in the same plane. In what ratio does the plane passing through the points of intersection of the medians of triangles $A B C, A B D$ and $B C D$ divide the segment $B D$?
Let $K, L, M$ be the points of intersection of the medians of triangles $ABC, ABD$, and $BCD$ respectively, $P$ be the midpoint of $BC$, and $Q$ be the midpoint of $AC$. In triangle $ADP$, it is known that

$$
\frac{PM}{MD}=\frac{PK}{KA}=\frac{1}{2}
$$

so $KM \parallel AD$. Similarly, $KL \parallel CD$. Therefore, the plane $KLM$ is parallel to the plane $ACD$. Let $F$ be the point of intersection of line $BD$ with the plane $KLM$. The parallel planes $ACD$ and $KLM$ are intersected by the plane $DQB$ along parallel lines $DQ$ and $KF$. Therefore,

$$
\frac{DF}{FB}=\frac{QK}{KB}=\frac{1}{2}
$$

## Answer

$1: 2$.