12.393 In triangle $A B C$, the acute angles $\alpha$ and $\gamma (\alpha>\gamma)$ at the base $A C$ are given. From vertex $B$, the altitude $B D$ and the angle bisector $B E$ are drawn. Find the area of triangle $B D E$, if the area of triangle $A B C$ is $S$.
Solution.

$$
\begin{aligned}
& \text { B } \\
& \text { Let } \alpha=\angle A, \gamma=\angle C \text {. Then } \\
& \text { } \\
& \angle A B D<\angle D B C \Rightarrow \text { the bisector } B E \text { lies inside } \\
& \triangle D B C . \angle D B E=\angle A B E-\angle A B D= \\
& =\frac{180^{\circ}-(\alpha+\gamma)}{2}-\left(90^{\circ}-\alpha\right)=\frac{\alpha-\gamma}{2} \\
& S_{B D E}=\frac{1}{2} B D \cdot D E=\frac{1}{2} B D \cdot A C \cdot \frac{D E}{A C}= \\
& =S \cdot \frac{B D \cdot \tan(\angle D B E)}{A C}=S \cdot \frac{A B \cdot \sin \alpha \cdot \tan(\angle D B E)}{A C}=\frac{S \cdot \sin \alpha \cdot \sin \gamma \cdot \tan \frac{\alpha-\gamma}{2}}{\sin (\alpha+\gamma)} \\
& \text { (by the sine rule for } \triangle A B C \frac{A B}{A C}=\frac{\sin \gamma}{\sin (\alpha+\gamma)} \text { ). }
\end{aligned}
$$

Answer: $S \cdot \tan\left(\frac{\alpha-\gamma}{2}\right) \cdot \frac{\sin \alpha \cdot \sin \gamma}{\sin (\alpha+\gamma)}$.