4.4. In triangle $A B C$, a point $D$ is marked on side $B C$ such that $B D: D C=1: 3$, and on side $A C$ - points $E$ and $K$, with point $E$ lying between points $A$ and $K$. Segment $A D$ intersects segments $B E$ and $B K$ at points $M$ and $N$ respectively, with $B M: M E=7: 5$, $B N: N K=2: 3$. Find the ratio $M N: A D$.
Answer: $11: 45$. Solution. We use Menelaus' theorem 1)

\[
\begin{aligned}
& \frac{1}{3} \frac{C A}{A K} \frac{3}{2}=1 \\
\frac{C A}{A K}= & 2 \Rightarrow \frac{C K}{K A}=1
\end{aligned}
\]

\[
\frac{A N}{N D}=\frac{4}{1}
\]

4) 

\[
\begin{gathered}
\frac{A M}{M D} \frac{D B}{B C} \frac{C E}{E A}=1 \\
\frac{A M}{M D} \frac{x}{4 x} \frac{C E}{E A}=1 \\
\frac{B D}{D C} \frac{C A}{A E} \frac{E M}{M B}=1 \\
\frac{x}{3 x} \frac{C A}{A E} \frac{E M}{M B}=1 \\
\frac{1}{3} \frac{C A}{A E} \frac{5}{7}=1 \\
\frac{C A}{A E}=\frac{21}{5} \quad \Rightarrow \quad \frac{C E}{E A}=\frac{16}{5}
\end{gathered}
\]

\[
\begin{gathered}
\frac{A M}{M D} \frac{1}{4} \frac{16}{5}=1 \\
\frac{A M}{M D}=\frac{5}{4}
\end{gathered}
\]

From the obtained ratios, denoting

\[
\begin{gathered}
\frac{A N}{N D} \frac{D B}{D C} \frac{C K}{K A}=1 \\
\frac{A N}{N D} \frac{1}{4} \frac{C K}{K A}=1
\end{gathered}
\]

\[
\begin{gathered}
A N=4 p, N D=p, A N=5 \kappa, M D=4 \kappa: \\
A D=5 \kappa+4 \kappa=5 p \\
\kappa=\frac{5}{9} p \\
\frac{M N}{A D}=\frac{4 \kappa-p}{5 p}=\frac{4 \frac{5}{9} p-p}{5 p}=\frac{11}{45}
\end{gathered}
\]

\[
\frac{B D}{D C} \frac{C A}{A K} \frac{K N}{N B}=1
\]