12.147. Find the cosine of the acute angle of the rhombus if a line drawn through its vertex divides the angle in the ratio $1: 3$, and the opposite side - in the ratio $3: 5$.
## Solution.

Let $L$ be the point of intersection of the given line $BL$ with side $AD$ of the rhombus $ABCD, \angle ABL: \angle CBL = 1: 3, AL: LD = 3: 5$ (Fig. 12.20).

$$
\angle ABL = \alpha, AL = 3x
$$

Then $\angle CBL = 3\alpha, LD = 5x, AB = 8x, \angle ABC = 4\alpha, \angle ABO = 2\alpha$.

If $BL$ is the bisector of $\angle ABD$, then from $\triangle ABD: \frac{AL}{LD} = \frac{AB}{BD}$; $BD = \frac{LD \cdot AB}{AL} = \frac{5x \cdot 8x}{3x} = \frac{40x}{3}$.

In $\triangle AOB \left(\angle AOB = 90^{\circ}\right): \cos 2\alpha = \cos \angle ABO = \frac{BO}{AB} = \frac{20x}{3} : 8x = \frac{5}{6}$. $\cos \angle ABC = \cos 4\alpha = 2 \cos^2 2\alpha - 1 = \frac{7}{18}$.

Answer: $\frac{7}{18}$.