$[$ Varignon's Parallelogram $]$

In a convex quadrilateral $A B C D$, the segments connecting the midpoints of opposite sides intersect at an angle of $60^{\circ}$, and their lengths are in the ratio $1: 3$. What is the length of the smaller diagonal of the quadrilateral $A B C D$, if the larger one is $\sqrt{39}$?
The midpoints of the sides of any quadrilateral are the vertices of a parallelogram, the sides of which are parallel to the diagonals of the quadrilateral and are respectively equal to their halves.

## Solution

The midpoints of the sides of any quadrilateral are the vertices of a parallelogram, the sides of which are parallel to the diagonals of the quadrilateral and are respectively equal to their halves. Let $x$ and $3x$ be the halves of the diagonals of the parallelogram. Since the angle between them is $60^{\circ}$, by the cosine theorem, the squares of the sides of the parallelogram are

$$
x^{2}+9 x^{2}-3 x^{2}=7 x^{2}, x^{2}+9 x^{2}+3 x^{2}=13 x^{2}
$$

Since the larger diagonal of the quadrilateral is $\sqrt{39}$, the larger side of the parallelogram is $\frac{\sqrt{39}}{2}$, i.e.,

$13 x^{2}=\frac{39}{4}$, from which $x=\frac{\sqrt{3}}{2}$. Then the smaller side of the parallelogram is $x \sqrt{7}=\frac{\sqrt{21}}{2}$. Therefore, the smaller diagonal of the given quadrilateral is $\sqrt{21}$.

## Answer

$\sqrt{21}$.