$\begin{aligned} & {[\text { Ratios of areas }]} \\ & {[\underline{\text { Area of a quadrilateral }}]}\end{aligned}$

The vertices of a rhombus are located on the sides of a parallelogram, and the sides of the rhombus are parallel to the diagonals of the parallelogram. Find the ratio of the areas of the rhombus and the parallelogram, if the ratio of the diagonals of the parallelogram is $k$.
Prove that the center of a rhombus coincides with the center of a parallelogram and consider similar triangles.

## Solution

Let the vertices $M, N, K$, and $L$ of the rhombus $MNKL$ be located on the sides $AB, BC, CD$, and $AD$ of the parallelogram $ABCD$, respectively, and the sides $MN$ and $KN$ of the rhombus are parallel to the diagonals $AC$ and $BD$ of the parallelogram, with $\frac{AC}{BD}=k$. If $\alpha$ is the angle between the diagonals of the parallelogram, then

$$
S_{\mathrm{ABCD}}=\frac{1}{2} AC \cdot BD \sin \alpha, \quad S_{\mathrm{KLMN}}=MN \cdot KN \sin \alpha=MN^2 \sin \alpha
$$

Therefore,

$$
\frac{S_{KLMN}}{S_{ABCD}}=\frac{2 MN^2}{AC \cdot BD}
$$

Note that the center of the rhombus coincides with the center $O$ of the parallelogram. Since $ON$ is the angle bisector of triangle $BOC$, we have

$$
\frac{BN}{CN}=\frac{OB}{OC}=\frac{BD}{AC}
$$

Thus,

$$
\frac{BN}{BC}=\frac{BD}{BD+AC}=\frac{1}{1+k}
$$

From the similarity of triangles $BMN$ and $BAC$, we find that

$$
MN=AC \cdot \frac{BN}{BC}=\frac{AC}{1+k}
$$

Therefore,

$$
\frac{S_{KLMN}}{S_{ABCD}}=\frac{2 MN^2}{AC \cdot BD}=\frac{\frac{2 AC^2}{(1+k)^2}}{AC \cdot BD}=2 \cdot \frac{AC}{BD} \cdot \frac{1}{(1+k)^2}=\frac{2 k}{(1+k)^2}
$$

## Answer

$\frac{2 k}{(1+k)^2}$