Let $A, B, C$ and $D$ be four points not lying in the same plane. A plane is drawn through the centroid of triangle $ABC$ and is parallel to the lines $AB$ and $CD$. In what ratio does this plane divide the median drawn to the side $CD$ of triangle $ACD$?
Let $O$ be the point of intersection of the medians of triangle $ABC$, and $P$ be the midpoint of $CD$. The plane $ABC$ passes through the line $AB$, which is parallel to the intersecting plane, and has a common point $O$ with the intersecting plane. Therefore, the line $l$ of intersection of these planes is parallel to the line $AB$. Let the line $l$ intersect the segment $AC$ at point $M$.

The plane $ACD$ passes through the line $CD$, which is parallel to the intersecting plane, and has a common point $M$ with the intersecting plane. Therefore, the line $m$ of intersection of these planes is parallel to the line $CD$. Let the line $m$ intersect the segment $AD$ at point $L$, and point $E$ be the midpoint of segment $AB$, and $Q$ be the point of intersection of segments $LM$ and $AP$ (and thus, $Q$ is the point of intersection of the intersecting plane with the specified median $AP$ of triangle $ACD$). Then

$$
\frac{AM}{QP}=\frac{MM}{MC}=\frac{EO}{OC}=\frac{1}{2}.
$$

## Answer

$1: 2$.