4. On the diagonal $B D$ of parallelogram $A B C D$, points $P$ and $Q$ are chosen such that $B P=P Q=Q D$. Lines $\boldsymbol{C P}$ and $\boldsymbol{C} \boldsymbol{Q}$ intersect sides $\boldsymbol{A B}$ and $\boldsymbol{A D}$ at points $M$ and $N$, respectively. Find the ratio $M N: B D$.
Solution. Draw the diagonal $AC$, then the intersection point of the diagonals $O$ - divides both diagonals in half (Fig. 2). Then for $\triangle ABC$, the segment $BO$ is a median, and the point $P$ satisfies the condition $BP: PO=2: 1$, which means $P$ is the centroid of $\triangle ABC$. Therefore, $CM$ is also a median of $\triangle ABC$, so $M$ is the midpoint of side $AB$, and similarly, $N$ is the midpoint of side $AD$. Thus, $CN$ is the midline of $\triangle ABD$, from which it follows that $MN: BD=1: 2$.

Answer: $MN: BD=1: 2$.

![](https://cdn.mathpix.com/cropped/2024_05_06_7440f25d146e96f0abdbg-2.jpg?height=437&width=594&top_left_y=935&top_left_x=731)

Fig. 2