4.1. On the side $A C$ of triangle $A B C$, points $E$ and $K$ are taken, with point $E$ lying between points $A$ and $K$ and $A E: E K: K C=3: 5: 4$. The median $A D$ intersects segments $B E$ and $B K$ at points $L$ and $M$ respectively. Find the ratio of the areas of triangles $B L M$ and $A B C$. Answer: $\frac{1}{5}$.
Solution. Since the median of a triangle divides its area in half, $S(ABD) = S(ADC)$ and $S(BMD) = S(MDC)$, which means $S(ABM) = S(AMC) = \frac{3}{2} S(AMK)$. At the same time, $\frac{S(ABM)}{S(AMK)} = \frac{BM}{MK}$, so $\frac{BM}{MK} = \frac{3}{2}$. Therefore, $\frac{S(BLM)}{S(LMK)} = \frac{3}{2}$, which means $S(ABL) = \frac{3}{2} SALK = \frac{3}{2} \cdot \frac{8}{3} S(ALE) = 4 S(ALE)$. Thus, $\frac{BL}{LE} = \frac{S(ABL)}{S(ALE)} = 4$. According to the theorem on the ratio of areas of triangles with a common angle, $\frac{S(BLM)}{S(BEK)} = \frac{BL}{BE} \cdot \frac{BM}{BK} = \frac{4}{5} \cdot \frac{3}{5} = \frac{12}{25}$. At the same time, $\frac{S(BEK)}{S(ABC)} = \frac{EK}{AC} = \frac{5}{12}$. Finally, we get that $\frac{S(BLM)}{S(ABC)} = \frac{S(BLM)}{S(BEK)} \cdot \frac{S(BEK)}{S(ABC)} = \frac{12}{25} \cdot \frac{5}{12} = \frac{1}{5}$.