10.098. A line parallel to the base of a triangle divides it into parts whose areas are in the ratio of $2: 1$. In what ratio, counting from the vertex, does it divide the lateral sides?
Solution.

1. $S_{\triangle A B C}=S_{A M N C}+S_{\triangle M B N}=x+2 x=3 x ; \frac{S_{\triangle A B C}}{S_{\triangle B M N}}=\frac{3}{2}$ (Fig. 10.94). From this, $\left(\frac{M B}{A B}\right)^{2}=\frac{2}{3}, \frac{M B}{A B}=\sqrt{\frac{2}{3}}, A B=\sqrt{3} x, M B=\sqrt{2} x ; A M=(\sqrt{3}-\sqrt{2}) x$, $\frac{M B}{A M}=\frac{\sqrt{2}}{\sqrt{3}-\sqrt{2}}=\sqrt{2}(\sqrt{3}+\sqrt{2})=\frac{\sqrt{6}+2}{1}$.
2. $S_{\triangle A B C}=S_{A M_{1} N_{1} C}+S_{\triangle B M_{1} N_{1}}$ (Fig. 10.94). Similarly, we find $M_{1} B: A M_{1}=(\sqrt{3}+1): 2$.

Answer: $(\sqrt{6}+2): 1$ or $(\sqrt{3}+1): 2$.

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-603.jpg?height=342&width=486&top_left_y=139&top_left_x=91)

Fig. 10.95

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-603.jpg?height=412&width=598&top_left_y=82&top_left_x=653)

Fig. 10.96