3.14. 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?
3.14. Given that in rhombus $ABCD$ we have: $\angle BAD = \alpha$, $\angle EAD = \frac{\alpha}{3}$, $\angle BAE = \frac{2\alpha}{3}$ (Fig. 3.16). Let $CE = m$, $DE = n$; we need to find $\frac{m}{n}$. In $\triangle AED$, $\angle AED = \frac{2\alpha}{3}$ and by the Law of Sines, $n: \sin \frac{\alpha}{3} = AD: \sin \frac{2\alpha}{3}$, from which

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-508.jpg?height=354&width=456&top_left_y=106&top_left_x=689)

Fig. 3.16

$$
n = \frac{AD \sin \frac{\alpha}{3}}{\sin \frac{2\alpha}{3}} = \frac{AD}{2 \cos \frac{\alpha}{3}}
$$

Next, in $\triangle ACE$, $\angle CAE = \frac{\alpha}{2} - \frac{\alpha}{3} = \frac{\alpha}{6}$, $\angle AEC = 180^\circ - \angle AED = 180^\circ - \frac{2\alpha}{3}$, and by the Law of Sines, $m: \sin \frac{\alpha}{6} = AC: \sin \left(180^\circ - \frac{2\alpha}{3}\right)$, from which

$$
m = \frac{AC \sin \frac{\alpha}{6}}{\sin \frac{2\alpha}{3}} = \frac{AC \sin \frac{\alpha}{6}}{2 \sin \frac{\alpha}{3} \cos \frac{\alpha}{3}}
$$

Thus,

$$
\frac{m}{n} = \frac{AC \sin \frac{\alpha}{6}}{2 \sin \frac{\alpha}{3} \cos \frac{\alpha}{3}} : \frac{AD}{2 \cos \frac{\alpha}{3}} = \frac{AC \sin \frac{\alpha}{6}}{AD \sin \frac{\alpha}{3}} = \frac{AC}{2 AD \cos \frac{\alpha}{6}}
$$

Finally, applying the Law of Sines to $\triangle ACD$, we find

$$
\frac{AC}{AD} = \frac{\sin (180^\circ - \alpha)}{\sin \frac{\alpha}{2}} = \frac{2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}{\sin \frac{\alpha}{2}} = 2 \cos \frac{\alpha}{2}
$$

and ultimately we get $\frac{m}{n} = \frac{2 \cos \frac{\alpha}{2}}{2 \cos \frac{\alpha}{6}} = \frac{\cos \frac{\alpha}{2}}{\cos \frac{\alpha}{6}}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-509.jpg?height=323&width=609&top_left_y=367&top_left_x=55)

Fig. 3.17