12.145. 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?
Solution.

In $\triangle ABC$, $AB = BC$, $\angle BAC = \alpha$, $\angle DAC = \beta$ (Fig. 12.18)

Then $\angle BAD = \alpha - \beta$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0905.jpg?height=500&width=664&top_left_y=289&top_left_x=96)

Fig. 12.19

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0905.jpg?height=714&width=406&top_left_y=79&top_left_x=788)

Fig. 12.20

Let $AB = a \Rightarrow AC = 2a \cos \alpha$ and

\[
\begin{aligned}
& \frac{S_{\triangle BAD}}{S_{\triangle CAD}} = \frac{\frac{1}{2} AB \cdot AD \sin \angle BAD}{\frac{1}{2} AC \cdot AD \sin \angle CAD} = \frac{AB \sin \angle BAD}{AC \sin \angle CAD} = \frac{a \sin (\alpha - \beta)}{2a \cos \alpha \sin \beta} = \\
& = \frac{\sin (\alpha - \beta)}{2 \cos \alpha \sin \beta}
\end{aligned}
\]

Answer: $\frac{\sin (\alpha - \beta)}{2 \cos \alpha \sin \beta}$.