1.27. A rhombus with a side of $6 \text{ cm}$ is inscribed in a right triangle with an angle of $60^{\circ}$, such that the $60^{\circ}$ angle is common to both and all vertices of the rhombus lie on the sides of the triangle. Find the sides of the triangle.
1.27. In $\triangle F B D$ (Fig. 1.23) the leg $F D=$ $=6 \text{ cm}$ and lies opposite the angle of $30^{\circ}$, from which $F B=12$ cm and, therefore, $A B=18 \text{ cm}$. Further, in $\triangle A B C$ we have $A C=\frac{1}{2} A B=9$ cm. Finally,

$$
B C=\sqrt{A B^{2}-A C^{2}}=\sqrt{243}=9 \sqrt{3}(\text{ cm})
$$

Answer: $9, 9 \sqrt{3}$ and 18 cm.