8.3 On the side $BC$ of the rhombus $ABCD$, select a point $M$. From $M$, draw perpendiculars to the diagonals $BD$ and $AC$, which intersect the line $AD$ at points $P$ and $Q$, respectively. It is known that the lines $PB$, $QC$, and $AM$ intersect at the same point. What is the ratio $\frac{BM}{MC}$?
Let the intersection of lines $PB$, $QC$, and $AM$ be denoted as $R$ (refer to Figure 1). We note that $PM \parallel AC$ and $MQ \parallel BD$, thus quadrilaterals $PMCA$ and $QMBD$ are both parallelograms. Therefore, $MC = PA$, $BM = DQ$, and $PQ = PA + AD + DQ = MC + AD + BM = 2BC$. Since $BC \parallel PQ$ and $BC = \frac{1}{2} PQ$, $BC$ is the midline of $\triangle PRQ$. This implies that $BM$ is the midline of $\triangle ARP$. Therefore, $MC = PA = 2BM$. Hence, $\frac{BM}{MC} = \frac{1}{2}$.