Authors: Berlov S.L., Petrov F., Akopyan A.V.

On the side $B C$ of the rhombus $A B C D$, a point $M$ is chosen. Lines drawn through $M$ perpendicular to the diagonals $B D$ and $A C$ intersect the line $A D$ at points $P$ and $Q$ respectively. It turns out that the lines $P B, Q C$ and $A M$ intersect at one point. What can the ratio $B M: M C$ be?
Let $R$ be the point of intersection of $P B, Q C$ and $A M$ (see the figure). Note that $P M\|A C$ and $M Q\| B D$, so quadrilaterals $P M C A$ and $Q M B D$ are parallelograms. Therefore, $M C=P A$, $B M=D Q$, and $P Q=P A+A D+D Q=$ $M C+A D+B M=2 B C$. Since $B C \| P Q$ and $B C=1 / 2 P Q$, $B C$ is the midline of triangle $P R Q$. Thus, $B M$ is also the midline of triangle $A R P$. Therefore, $M C=P A=2 B M$.

## Answer

$1: 2$.