9.21. (England, 70). On the sides $B C$ and $A C$ of triangle $A B C$, points $D$ and $E$ are chosen respectively such that $\angle B A D=50^{\circ}, \angle A B E=30^{\circ}$. Find $\angle B E D$, if
$\angle A B C=\angle A C B=50^{\circ}$
9.21. Let $O$ be the point of intersection of lines $A D$ and $B E$ (Fig. 17), then $\angle A O B=180^{\circ}-30^{\circ}-50^{\circ}=100^{\circ}$, $\angle B D A=180^{\circ}-50^{\circ}-50^{\circ}=80^{\circ}$, $\angle C B E=50^{\circ}-30^{\circ}=20^{\circ}$, $\angle A E B=\angle C B E+\angle E C B=70^{\circ}$, $\angle C A D=180^{\circ}-\angle A C B$ -

$-\angle A B C-\angle B A D=30^{\circ}$

By the Law of Sines, we have

$$
\begin{gathered}
\frac{O D}{O B}=\frac{\sin 20^{\circ}}{\sin 80^{\circ}}, \quad \frac{O B}{O A}=\frac{\sin 50^{\circ}}{\sin 30^{\circ}}, \\
\frac{O A}{O E}=\frac{\sin 70^{\circ}}{\sin 30^{\circ}}
\end{gathered}
$$

from which we obtain

![](https://cdn.mathpix.com/cropped/2024_05_21_03a501298610b72904feg-184.jpg?height=570&width=519&top_left_y=442&top_left_x=483)

Fig. 17

$$
\begin{aligned}
& \frac{O D}{O E}=\frac{O D}{O B} \cdot \frac{O B}{O A} \cdot \frac{O A}{O E}=\frac{\sin 20^{\circ} \sin 50^{\circ} \sin 70^{\circ}}{\sin 80^{\circ} \sin ^{2} 30^{\circ}}= \\
& \quad=\frac{4 \sin 20^{\circ} \cos 40^{\circ} \cos 20^{\circ}}{\sin 80^{\circ}}=\frac{2 \sin 40^{\circ} \cos 40^{\circ}}{\sin 80^{\circ}}=\frac{\sin 80^{\circ}}{\sin 80^{\circ}}=1
\end{aligned}
$$

i.e., $O D=O E$ and

$$
\angle B E D=\angle O D E=\frac{1}{2}\left(180^{\circ}-\angle E O D\right)=\frac{1}{2}\left(180^{\circ}-100^{\circ}\right)=40^{\circ}
$$