3.57. A line is drawn through the vertex of angle $\alpha$ at the base of an isosceles triangle, intersecting the opposite lateral side and forming an angle $\beta$ with the base. In what ratio does this line divide the area of the triangle?
3.57. By the condition, \( A B = B C, \angle A C B = \alpha, \angle C A M = \beta \) (Fig. 3.59). Let \( A B = a \); then \( A C = 2 a \cos \alpha \) and, therefore,

\[
\frac{S_{\triangle A M B}}{S_{\triangle A M C}} = \frac{0.5 A B \cdot A M \sin (\alpha - \beta)}{0.5 A C \cdot A M \sin \beta} = \frac{a \sin (\alpha - \beta)}{2 a \cos \alpha \sin \beta} = \frac{\sin (\alpha - \beta)}{2 \cos \alpha \sin \beta}
\]

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-536.jpg?height=388&width=358&top_left_y=102&top_left_x=106)

Fig. 3.59

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-536.jpg?height=321&width=633&top_left_y=105&top_left_x=522)

Fig. 3.60