3.63. 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 $ABC$.
3.63. By the condition, $\angle B A C=\alpha, \angle B C A=\gamma(\alpha>\gamma), A D=D C, B E$ is the bisector (Fig. 3.65); it is required to find $S_{\triangle B D E}: S_{\triangle A B C}$. Draw $B F \perp A C$; then $A B=\frac{B F}{\sin \alpha} ; B C=\frac{B F}{\sin \gamma} \cdot$ We have

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-540.jpg?height=304&width=618&top_left_y=777&top_left_x=291)

Fig. 3.65

$$
\begin{gathered}
S_{\triangle B D E}=S_{\triangle A B D}-S_{\triangle A B E}=\frac{1}{2} S_{\triangle A B C}-\frac{1}{2} A E \cdot B F \\
\frac{S_{\triangle B D E}}{S_{\triangle A B C}}=\frac{1}{2}-\frac{0.5 A E \cdot B F}{0.5 A C \cdot B F}=\frac{1}{2}-\frac{A E}{A C}
\end{gathered}
$$

But $\frac{A E}{E C}=\frac{A B}{B C}=\frac{B F}{\sin \alpha}: \frac{B F}{\sin \gamma}=\frac{\sin \gamma}{\sin \alpha}$; let's now form the derived proportion

$$
\frac{A E}{A E+E C}=\frac{\sin \gamma}{\sin \alpha+\sin \gamma}, \text { or } \frac{A E}{A C}=\frac{\sin \gamma}{\sin \alpha+\sin \gamma}
$$

Thus, we finally find

$$
\frac{S_{\triangle B D E}}{S_{\triangle A B C}}=\frac{1}{2}-\frac{\sin \gamma}{\sin \alpha+\sin \gamma}=\frac{\sin \alpha-\sin \gamma}{2(\sin \alpha+\sin \gamma)}=
$$

$$
=\frac{\cos \frac{\alpha+\gamma}{2} \sin \frac{\alpha-\gamma}{2}}{2 \cos \frac{\alpha-\gamma}{2} \sin \frac{\alpha+\gamma}{2}}=\frac{\operatorname{tg} \frac{\alpha-\gamma}{2}}{2 \operatorname{tg} \frac{\alpha+\gamma}{2}}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-541.jpg?height=292&width=404&top_left_y=400&top_left_x=46)

Fig. 3.66