Solution.

According to the condition $2 B C+2 C D=90 \text{ cm}, \angle B C D=60^{\circ}, \frac{\angle A B D}{\angle D B C}=\frac{3}{1}$. We have $B C+C D=45, \angle A B C+\angle B C D=180^{\circ}$. Let $\angle C B D=x, \angle A B D=3 x$, therefore, $3 x+x+60^{\circ}=180^{\circ}$ and $x=30^{\circ}, \angle A B D=90^{\circ}$. In $\triangle A B D$ $A B=A D \cdot \cos 60^{\circ}, A B=\frac{A D}{2}=\frac{B C}{2}, A B=C D ; B C+\frac{B C}{2}=45,3 B C=90$; $B C=30$ cm. $A D=D C=15 \text{ cm}$.

Answer: 15 and 30 cm.
10.038. The perimeter of a parallelogram is 90 cm and the acute angle is $60^{\circ}$ (Fig. 10.38). A diagonal of the parallelogram divides its obtuse angle in the ratio $1: 3$. Find the sides of the parallelogram.