4. On the base AC of the isosceles triangle $\mathrm{ABC}$, a point $\mathrm{E}$ is taken, and on the lateral sides $\mathrm{AB}$ and $\mathrm{BC}$ - points K and M, such that KE is parallel to BC and EM is parallel to AB. What part of the area of triangle ABC does the area of triangle $\mathrm{KEM}$ occupy, if $\mathrm{BM}: \mathrm{EM}=2: 3 ?$
4. Quadrilateral VKEM is a parallelogram, triangles AKE and EMC are similar to triangle ABC with similarity coefficients equal to

$$
\begin{aligned}
& \frac{\mathrm{KE}}{\mathrm{BC}}=\frac{2 \mathrm{x}}{2 \mathrm{x}+3 \mathrm{x}}=\frac{2}{5} \text { and } \frac{\mathrm{MC}}{\mathrm{BC}}=\frac{3 \mathrm{x}}{2 \mathrm{x}+3 \mathrm{x}}=\frac{3}{5} . \text { Therefore } \mathrm{S}_{\text {KME }}=\frac{1}{2} \mathrm{~S}_{\text {VKEM }}=\frac{1}{2}\left(\mathrm{~S}_{\mathrm{ABC}}-\right. \\
& \left.\mathrm{S}_{\mathrm{AKE}}-\quad \mathrm{S}_{\mathrm{EMC}}\right)=\frac{1}{2}\left(1-\frac{4}{25}-\frac{9}{25}\right) \mathrm{S}_{\mathrm{ABC}}=\frac{6}{25} \mathrm{~S}_{\mathrm{ABC}}
\end{aligned}
$$