A circle crosses the sides of a triangle, dividing each of them into three equal parts. Prove that the triangle is equilateral.
I think that the best way is to show that $\angle BAC = \angle ABC$, and then $\angle ABC = \angle ACB$. Something like that and I would be very grateful if you could help me because I don't see how this can be done.
By the power of a point theorem, $$AD\cdot AE=AK\cdot AF$$
$$2 AD^2 = 2 AK^2 $$
That means that $ AC = 3 AD = 3 AK = AB$
Analogically $BA=BC$