2. Based on $AC$ of the isosceles triangle $ABC$, a point $E$ is taken, and on the lateral sides $AB$ and $BC$ - points $K$ and $M$ such that $KE$ is parallel to $BC$ and $EM$ is parallel to $AB$. What part of the area of triangle $\mathrm{ABC}$ is occupied by the area of triangle $KEM$, if $\mathbf{BM}: \mathbf{EM}=2: 3 ?$
Answer: $\frac{6}{25}$.

Solution.

Quadrilateral BKEM is a parallelogram. Triangles AKE and EMC are isosceles. Each of them is similar to triangle $\mathrm{ABC}$ with similarity coefficients $\frac{\mathrm{KE}}{\mathrm{BC}}=\frac{2 x}{2 x+3 x}=\frac{2}{5} \quad$ and $\quad \frac{\mathrm{MC}}{\mathrm{BC}}=\frac{3 x}{2 x+3 x}=\frac{3}{5} . \quad$ Therefore, $S_{K M E}=\frac{1}{2} S_{B K E M}=\frac{1}{2}\left(S_{A B C}-S_{A K E}-S_{E M C}\right)=\frac{1}{2}\left(1-\frac{4}{25}-\frac{9}{25}\right) S_{A B C}=\frac{6}{25} S_{A B C}$.