A plane passing through the midpoints of edges $A B$ and $C D$ of a tetrahedron $A B C D$ divides edge $A D$ in the ratio 3:1, counting from vertex $A$. In what ratio does this plane divide edge $B C$?
Let $K$ and $M$ be the midpoints of edges $AB$ and $CD$, and $P$ and $Q$ be the points of intersection of the secant plane with edges $AD$ and $BC$ respectively (Fig.1). When the pyramid is orthogonally projected onto a plane perpendicular to the line $KM$, points $K$ and $M$ will project to a certain point $O$, and points $A, B, C, D, P$, and $Q$ will project to points $A'$, $B'$, $C'$, $D'$, $P'$, and $Q'$ respectively, such that the quadrilateral $A' C' B' D'$ is a parallelogram with center $O$ (Fig.2), because its diagonals intersect at point $O$ and are bisected by it. Point $P'$ divides the side $A' D'$ of this parallelogram in the ratio 3:1, counting from point $A'$. Therefore, the line $O P'$ divides the opposite side $B' C'$ of the parallelogram also in the ratio 3:1, counting from point $B'$. Consequently, point $Q$ divides the edge $BC$ of the pyramid $ABCD$ in the ratio 3:1, counting from point $B$.

## Answer

$3: 1$, counting from vertex $B$.