12.003. In a rhombus, a line is drawn through the vertex of the acute angle, equal to $\alpha$, dividing this angle in the ratio $1: 2$. In what ratio does this line divide the side of the rhombus that it intersects?
Solution.

Let $\angle A B C=\alpha$ and $\frac{\angle A B E}{\angle E B C}=\frac{1}{2}$, then $\angle E B C=\frac{2}{3} \alpha, \angle A B E=\frac{\alpha}{3}$, $\angle D B C=\frac{\angle A B C}{2}=\frac{\alpha}{2}, \angle E B M=\angle E B C-\angle D B C=\frac{2 \alpha}{3}-\frac{\alpha}{2}=\frac{\alpha}{6}$ (Fig. 12.6). From the similarity of $\triangle A D O \sim \triangle E D M$ it follows that $\frac{A D}{E D}=\frac{D O}{D M}$. Since $A D=E D+A E$, then $\frac{A E}{E D}=\frac{D O}{D M}-1$. Let $A B=B C=C D=A D=x$. From $\triangle C O D \angle C O D=90^{\circ}, \angle O D C=\alpha / 2, O D=D C \cdot \cos \frac{\alpha}{2}=x \cdot \cos \frac{\alpha}{2}$, $B D=2 D O=2 x \cdot \cos \frac{\alpha}{2}$. From $\triangle B M E, \angle B M E=90^{\circ}, E M=B M \cdot \operatorname{tg} \frac{\alpha}{6}$. Consider $\triangle E M D, \angle E M D=90^{\circ}, D M=E M \cdot \operatorname{ctg} \frac{\alpha}{2}, D M=B M \cdot \operatorname{tg} \frac{\alpha}{6} \cdot \operatorname{ctg} \frac{\alpha}{2}$, where $B M=B D-D M$, then $D M=(B D-D M) \cdot \operatorname{tg} \frac{\alpha}{6} \operatorname{ctg} \frac{\alpha}{2}$. We get that $\frac{1}{\operatorname{tg} \frac{\alpha}{6} \cdot \operatorname{ctg} \frac{\alpha}{2}}=\frac{B D-D M}{D M}$, where $B D=2 x \cdot \cos \frac{\alpha}{2}$. Then $\frac{1}{\operatorname{tg} \frac{\alpha}{6} \cdot \operatorname{ctg} \frac{\alpha}{2}}=$ $=\frac{2 x \cdot \cos \frac{\alpha}{2}}{D M}-1 ; D M=\frac{2 x \operatorname{tg} \frac{\alpha}{6} \cdot \operatorname{ctg} \frac{\alpha}{2} \cdot \cos \frac{\alpha}{2}}{1+\operatorname{tg} \frac{\alpha}{6} \cdot \operatorname{ctg} \frac{\alpha}{2}} \cdot$ Therefore, $\frac{A E}{E D}=$

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-718.jpg?height=604&width=298&top_left_y=94&top_left_x=219)

Fig. 12.6

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-718.jpg?height=468&width=516&top_left_y=222&top_left_x=718)

Fig. 12.7

$$
\begin{aligned}
& =\frac{x \cdot \cos \frac{\alpha}{2} \cdot\left(1+\operatorname{tg} \frac{\alpha}{6} \cdot \operatorname{ctg} \frac{\alpha}{2}\right)}{2 x \cdot \operatorname{tg} \frac{\alpha}{6} \cdot \operatorname{ctg} \frac{\alpha}{2} \cdot \cos \frac{\alpha}{2}}-1=\frac{1-\operatorname{tg} \frac{\alpha}{6} \operatorname{ctg} \frac{\alpha}{2}}{2 \operatorname{tg} \frac{\alpha}{6} \cdot \operatorname{ctg} \frac{\alpha}{2}}=\frac{\sin \frac{\alpha}{2} \cdot \cos \frac{\alpha}{6}-\sin \frac{\alpha}{6} \cdot \cos \frac{\alpha}{2}}{2 \sin \frac{\alpha}{6} \cdot \cos \frac{\alpha}{2}} \\
& \frac{A E}{E D}=\frac{\cos \frac{\alpha}{6}}{\cos \frac{\alpha}{2}}
\end{aligned}
$$

Answer: $\cos \frac{\alpha}{6}: \cos \frac{\alpha}{2}$.