Points $E$ and $F$ lie on the sides $A B$ and $B C$ of the rhombus $A B C D$, respectively, such that $A E=5 B E, B F=5 C F$. It is known that triangle $D E F$ is equilateral. Find the angle $B A D$.
On side $AB$, we lay off the segment $AK = CF = BE$. From the equality of triangles $AKD$ and $CFD$ (by two sides and the included angle), it follows that

$DK = DF = ED$. The angles at the base $KE$ of the isosceles triangle $DKE$ are equal, so the adjacent angles $AKD$ and $BED$ are also equal, and triangles $AKD$ and $BED$ are equal by two sides and the included angle. Therefore, $BD = AD = AB$, which means triangle $ABD$ is equilateral. Hence, $\angle BAD = 60^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_cdc67dfeb844af03d94fg-50.jpg?height=397&width=646&top_left_y=516&top_left_x=732)

## Answer

$60^{\circ}$.