## [

 Relationships of Areas In triangle $ABC$, lines parallel to sides $AB$ and $AC$ are drawn through point $M$, which lies on side $BC$. The area of the parallelogram formed by these lines is $\frac{5}{18}$ of the area of triangle $ABC$. Find the ratio $\frac{BM}{MC}$.

#
Let the specified lines intersect sides $AB$ and $AC$ of triangle $ABC$ at points $K$ and $L$ respectively. Denote $\frac{BM}{MC}=k, S_{\triangle ABC}=S$. Triangle $KBM$ is similar to triangle $ABC$ with a similarity coefficient of $\frac{k}{k+1}$, and triangle $LMC$ is similar to triangle $ABC$ with a similarity coefficient of $\frac{1}{k+1}$, therefore

$$
S_{\triangle K B M}=\left(\frac{k}{k+1}\right)^2 \cdot S_{\triangle A B C}=\left(\frac{k}{k+1}\right)^2 \cdot S,
$$

$$
S_{\triangle L M C}=\left(\frac{1}{(k+1)}\right)^2 \cdot S_{\triangle A B C}=\left(\frac{1}{(k+1)}\right)^2 \cdot S
$$

Then, if $S_1$ is the area of parallelogram $AKML$, we have

$$
\begin{gathered}
S_1=S_{\triangle A B C}-S_{\triangle K B M}-S_{\triangle L M C}=S-\left(\frac{k}{(k+1)}\right)^2 \cdot S-\left(\frac{1}{(k+1)}\right)^2 \cdot S= \\
=S\left(1-\frac{k^2}{(k+1)^2}-\frac{1}{(k+1)^2}\right)=\frac{2kS}{(k+1)^2}=\frac{5}{18}S
\end{gathered}
$$

from which we find that $k=\frac{1}{5}$ or $k=5$.

## Answer

$5:1, 1:5$.