[Equilateral (Equiangular) Triangle]

[Properties and Characteristics of a Parallelogram]

[Auxiliary Congruent Triangles]

[Rotation Helps Solve the Problem.

On the sides $AB, BC$, and $CA$ of an equilateral triangle $ABC$, points $D, E$, and $F$ are chosen respectively such that $DE \parallel AC$ and $DF \parallel BC$.

Find the angle between the lines $AE$ and $BF$.

#
Let $A E$ and $B F$ intersect at point $N$. Since $C E D F$ is a parallelogram and triangle $B D E$ is equilateral, then $B E = C F$ (see the left figure). We can reason in different ways.
![](https://cdn.mathpix.com/cropped/2024_05_06_7adbd1f2e5e559c8684fg-19.jpg?height=596&width=1422&top_left_y=342&top_left_x=316)

First method. Triangles $B A E$ and $C B F$ are congruent (by two sides and the included angle). Angle $A N F$ is an exterior angle for triangle $A B N$, hence,

$\angle A N F = \angle B A E + \angle A B F = \angle C B F + \angle A B F = 60^{\circ}$.

Second method. Consider a rotation around the center $O$ of triangle $A B C$ by an angle of $120^{\circ}$ counterclockwise (see the right figure). Under this rotation, vertex $A$ will move to vertex $B$, side $B C$ will move to side $C A$, and since $B E = C F$, point $E$ will move to point $F$. Therefore, ray $A E$ will move to ray $B F$. The angle between these rays is equal to the angle of rotation, that is, $\angle E N F = 120^{\circ}$. Hence, $\angle A N F = 60^{\circ}$.

## Answer

$60^{\circ}$.