A parallelogram has points $M$ and $N$ on sides $AB$ and $CD$ respectively, such that $AM: MB = CN: ND = m$. The diagonal $AC$ intersects lines $DM$ and $BN$ at points $R$ and $S$ respectively. What is the ratio $\frac{AC}{RS}$?
From the ratio $A M: M B = C N: N D = m$, and from the fact that the quadrilateral is a parallelogram, it follows that $D M \| N B$.

![](https://cdn.mathpix.com/cropped/2024_05_02_f3230ec2cab52b05fbe2g-1.jpg?height=303&width=542&top_left_y=263&top_left_x=781)

Applying the theorem of parallel intercepts to the sides of angle $C A B$,

$$
\begin{gathered}
A R: R S = A M: M B = m, \text { hence } \\
A R = m R S .
\end{gathered}
$$

Applying the theorem of parallel intercepts to the sides of angle $A C D$,

$$
\begin{gathered}
C S: S R = C N: N D = m, \text { hence } \\
S C = m R S .
\end{gathered}
$$

Using (1) and (2),

$$
A C = A R + R S + S C = m R S + R S + m R S = (2 m + 1) R S,
$$

from which the desired ratio is

$$
\frac{A C}{R S} = 2 m + 1.
$$