A rhombus with side $m$ is inscribed in a triangle such that one angle is common to both, and the opposite vertex of the rhombus lies on the side of the triangle, dividing this side into segments of lengths $p$ and $q$. Find the sides of the triangle.
The sides of the rhombus that do not pass through the vertex of the triangle cut off similar triangles from the given triangle.

## Solution

Let the vertices $M, N$ and $K$ of the rhombus $A M K N$ lie on the sides $A B, A C$ and $B C$ of triangle $A B C$ respectively, and $B K = p, C K = q$. From the similarity of triangles $C K N$ and $K B M$, we find that $K N: B M = C K: B K$. Therefore, $B M = \frac{K N}{C K} \cdot B K = \frac{m p}{q}, A B = A M + B M = m + \frac{m p}{q} = \frac{m(p+q)}{q}$.

Similarly, we find $A C$.

## Answer

$p+q, \frac{m(p+q)}{p}, \frac{m(p+q)}{q}$.