In a right triangle with an angle of $60^{\circ}$, a rhombus with a side length of 6 is inscribed, and the $60^{\circ}$ angle is common to both, with all vertices of the rhombus lying on the sides of the triangle. Find the sides of the triangle.
## Solution

Let vertices $N$ and $M$ of the rhombus $B M K N$ be on the hypotenuse $A B$ and the leg $B C$ of the triangle $A B C$, and vertex $K$ be on the leg $A C$. Since $K N=6, \angle A=30^{\circ}, K N \| B C$, then $A N=2 K N=12$. Therefore, $A B=A N+N B=18, B C=1 / 2$ $A B=9, A C=9 \sqrt{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_9617f685748a8dfcd82ag-08.jpg?height=394&width=575&top_left_y=366&top_left_x=747)

## Answer

$9, 9 \sqrt{3}, 18$.