8.3. On the side $BC$ of the rhombus $ABCD$, take a point $M$. Draw perpendiculars from $M$ to the diagonals $BD$ and $AC$, intersecting the line $AD$ at points $P$ and $Q$. If the lines $PB$ and $QC$ intersect $AM$ at the same point, find the ratio $\frac{BM}{MC}$.
8.3. As shown in Figure 1, let the intersection of lines $P B$, $Q C$, and $A M$ be $R$.

From the given conditions, $P M \parallel A C$ and $M Q \parallel B D$, thus quadrilaterals $P M C A$ and $Q M B D$ are both parallelograms. Therefore,
$$
\begin{aligned}
M C &= P A, B M = D Q, \text{ and } \\
P Q &= P A + A D + D Q = M C + A D + B M = 2 B C .
\end{aligned}
$$

Since $B C \parallel P Q$ and $B C = \frac{1}{2} P Q$, $B C$ is the midline of $\triangle P R Q$.
This implies that $B M$ is the midline of $\triangle A R P$.
$$
\text{Thus, } M C = P A = 2 B M \text{. }
$$