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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.

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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 $BA=BC$

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  • $\begingroup$ We haven't studied this Theorem yet. Is it this one: cut-the-knot.org/pythagoras/PPower.shtml? $\endgroup$ – Nikol Dimitrova May 18 '19 at 9:50
  • $\begingroup$ I think it is much simpler to write the second equation as $$\frac29 AB^2=\frac29AC^2.$$ $\endgroup$ – user May 18 '19 at 9:51
  • $\begingroup$ @NikolDimitrova yes. It is proven by similarity of triangles. $\endgroup$ – liaombro May 18 '19 at 9:54

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