12.155. In triangle $ABC$, the acute angles $\alpha$ and $\gamma (\alpha > \gamma)$ adjacent to side $AC$ are given. From vertex $B$, the median $BD$ and the angle bisector $BE$ are drawn. Find the ratio of the area of triangle $BDE$ to the area of triangle $ABC$.
Solution.

In $\triangle ABC$, $\angle A = \alpha$, $\angle C = \gamma$ (Fig. 12.28). Since $\alpha > \gamma$, point $D$ is located between points $A$ and $E$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0912.jpg?height=438&width=608&top_left_y=1061&top_left_x=653)

Fig. 12.28

$\triangle ABC$ and $\triangle BDE$ share a common height, so $\frac{S_{\triangle BDE}}{S_{\triangle ABC}} = \frac{DE}{AC}$. $AC = CE + AE$. $AD = AE + DE$, $CD = CE - DE$. Since $AD = CD$, then $AE + DE = CE - DE$, $DE = \frac{CE - AE}{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0913.jpg?height=440&width=719&top_left_y=77&top_left_x=315)

Fig. 12.29

Since $BE$ is the angle bisector of $\triangle ABC$, then $\frac{CE}{AE} = \frac{BC}{AB} = \frac{\sin \alpha}{\sin \gamma}$.

Thus, $\frac{S_{\triangle BDE}}{S_{\triangle ABC}} = \frac{DE}{AC} = \frac{CE - AE}{2(CE + AE)} = \frac{\frac{CE}{AE} - 1}{2\left(\frac{CE}{AE} + 1\right)} = \frac{\frac{\sin \alpha}{\sin \gamma} - 1}{2\left(\frac{\sin \alpha}{\sin \gamma} + 1\right)} = \frac{\sin \alpha - \sin \gamma}{2(\sin \alpha + \sin \gamma)} = \frac{2 \sin \frac{\alpha - \gamma}{2} \cos \frac{\alpha + \gamma}{2}}{4 \sin \frac{\alpha + \gamma}{2} \cos \frac{\alpha - \gamma}{2}} = \frac{\tan \frac{\alpha - \gamma}{2}}{2 \tan \frac{\alpha + \gamma}{2}}$.

Answer: $\frac{\tan \frac{\alpha - \gamma}{2}}{2 \tan \frac{\alpha + \gamma}{2}}$.