4. $\mathrm{ABCD}$ is a rhombus. Points $E$ and $F$ lie on sides $AB$ and $BC$ respectively, such that $\frac{AE}{BE}=\frac{BF}{CF}=5$. Triangle $DEF$ is equilateral. Find the angles of the rhombus. Solution.

The notation is shown in the figure. The situation where angle $D$ of the rhombus is acute is impossible (see below). Mark point $P$ on side $AB$ such that $AP=EB=FC$. Triangles $CDF$ and $ADP$ are congruent, as they have equal angles $\angle A=\angle C$ and corresponding equal sides $AD=DC, AP=FC$. Considering that triangle $FED$ is equilateral by the condition, we get that triangle $EDP$ is isosceles $(ED=FD=PD)$. (At this point, note that if angle $\angle A$ of the rhombus is obtuse, then angle $\angle APD$ is acute, and in the isosceles triangle $EDP$, the angle at the base is obtuse,

![](https://cdn.mathpix.com/cropped/2024_05_06_aa449fffebf3e1dc9820g-3.jpg?height=499&width=1037&top_left_y=116&top_left_x=521)

which is impossible). Mark the equal angles $x=\angle A=\angle C$ and $y=\angle ADP=\angle CDF$. We find $\angle EPD=x+y$ (the exterior angle of triangle $ADP$) and $\angle PDE=180^{\circ}-2(x+y)$. Summing all the angles at vertex $D$ of the rhombus ($\left.\angle D=180^{\circ}-x\right)$, we get the equation

$$
y+180^{\circ}-2(x+y)+60^{\circ}+y=180^{\circ}-x
$$

From this, $x=60^{\circ}$, the angles of the rhombus are $60^{\circ}$ and $120^{\circ}$.

The end of the solution can be different. After proving that triangle $EDP$ is isosceles, note that its exterior angles are equal: $\angle APD=\angle BED$. From this, triangles $ADP$ and $BED$ are congruent by the two indicated angles and sides $(AP=EB, PD=ED)$. Therefore, $BD=AD$, and triangle $ABD$ is equilateral. Thus, $\angle BAD=60^{\circ}$.
Answer. The angles of the rhombus are $60^{\circ}$ and $120^{\circ}$.