9.3. On the sides $B C$ and $B F$ of parallelogram $A B C D$, points $E$ and $F$ are taken respectively such that $\frac{E B}{E C}=p$ and $\frac{F C}{F D}=q$. Let $M$ be the intersection point of lines $A E$ and $B F$, and $N$ be the intersection point of lines $D E$ and $B F$. Determine the values of the ratios $\frac{A M}{M E}$ and $\frac{D N}{N E}$.
Solution. Draw $C L \| B F (L \in A B)$ and let $K$ be the intersection of lines $C L$ and $A E$.

$$
\begin{aligned}
& B M \| C K \Rightarrow \triangle B E M \sim \triangle C E K \Rightarrow \frac{K E}{E M}=\frac{C E}{E B}=\frac{1}{p} \\
& \frac{A M}{M E}=\frac{A M}{M K} \cdot \frac{M K}{M E} \\
& B M \| K L \Rightarrow \triangle B M A \sim \triangle L K A \Rightarrow \quad \frac{A M}{M K}=\frac{A B}{B L}=\frac{D C}{F C}=\frac{F D+F C}{F C}=1+\frac{F D}{F C}=1+\frac{1}{q}=\frac{q+1}{q} . \\
& \frac{M K}{M E}=\frac{M E+K E}{M E}=1+\frac{K E}{M E}=1+\frac{1}{p}=\frac{p+1}{p}
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_06_06_f3ba4c14cf7da9617052g-2.jpg?height=353&width=1039&top_left_y=897&top_left_x=997)

We apply (3) and (4) in (2) and obtain $\quad \frac{A M}{M E}=\frac{q+1}{q} \cdot \frac{p+1}{p}=\frac{(q+1)(p+1)}{q p}$.

We apply Menelaus' Theorem in $\triangle C D E$ and the collinear points $F, N, B$ on the lines $D C, D E, C E$.

$$
\frac{B E}{B C} \cdot \frac{C F}{F D} \cdot \frac{D N}{N E}=1 \Leftrightarrow \frac{D N}{N E}=\frac{B C}{B E} \cdot \frac{F D}{C F}=\frac{B C}{B E} \cdot \frac{1}{q}=\frac{B E+E C}{B E} \cdot \frac{1}{q}=\left(1+\frac{E C}{B E}\right)\left(\frac{1}{q}\right)=\left(1+\frac{1}{p}\right)\left(\frac{1}{q}\right)=\frac{p+1}{p q}
$$