12.170. 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 median $B E$ are drawn.

Find the area of triangle $B D E$, if the area of triangle $A C B$ is $S$.
Solution.

In $\triangle ABC, \angle A=\alpha, \angle C=\gamma$ (Fig. 12.40).

Since $\alpha>\gamma$, point $D$ is located between points $A$ and $E$.

$A E=A D+D E, C D=C E+D E$. Since $C E=A E$, then $C D=(A D+D E)+D E$,

$D E=\frac{C D-A D}{2}$.

Let $B D=H$. Then from $\triangle A B D$ we get $A D=H \operatorname{ctg} \alpha$, and from $\triangle C D B - C D=H \operatorname{ctg} \gamma$.

$\triangle A B C$ and $\triangle D B E$ have a common height $B D$, hence

$\frac{S_{\triangle D B E}}{S_{\triangle A B C}}=\frac{D E}{A C}=\frac{C D-A D}{2(C D+A D)}=\frac{H \operatorname{ctg} \gamma-H \operatorname{ctg} \alpha}{2 H(\operatorname{ctg} \gamma+\operatorname{ctg} \alpha)}=\frac{\sin (\alpha-\gamma)}{2 \sin (\alpha+\gamma)} ;$

$S_{\triangle D B E}=\frac{S \sin (\alpha-\gamma)}{2 \sin (\alpha+\gamma)}$.

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