12.306. In a right prism $A B C A_{1} B_{1} C_{1}\left(A A_{1}\left\|B B_{1}\right\| C C_{1}\right)$, the sides of the base $A B$ and $B C$ are equal to $a$ and $b$ respectively, and the angle between them is $\alpha$. A plane is drawn through the bisector of this angle and vertex $A_{1}$, forming an acute angle $\beta$ with the plane of the base. Find the area of the section.
Solution.

Let $B D$ be the bisector of $\triangle A B C, \triangle A_{1} B D$ - the section of the given

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1060.jpg?height=604&width=622&top_left_y=1087&top_left_x=644)
prism (Fig. 12.168) $\frac{A C}{A D}=$

$$
=\frac{A D+D C}{A D}=1+\frac{D C}{A D}=1+\frac{B C}{A B}=1+\frac{b}{a}=\frac{a+b}{a} .
$$

Fig. 12.168

Since $\triangle A B D$ and $\triangle A B C$ have the same height dropped from vertex $B$, then $\frac{S_{\triangle A B D}}{S_{\triangle A B C}}=\frac{A D}{A C}=\frac{a}{a+b}$. Therefore, $S_{\triangle A B D}=\frac{a}{a+b} \cdot S_{\triangle A B C}=\frac{a}{a+b} \times$ $\times \frac{1}{2} a b \sin \alpha=\frac{a^{2} b \sin \alpha}{2(a+b)}, \triangle A B D$ is the projection of $\triangle A_{1} B D$ on the base plane of the prism, i.e., $S_{\triangle A_{1} B D}=\frac{S_{\triangle A B D}}{\cos \beta}=\frac{a^{2} b \sin \alpha}{2(a+b) \cos \beta}$.

Answer: $\frac{a^{2} b \sin \alpha}{2(a+b) \cos \beta}$.