Let $D, E$, and $F$ be points on the sides $BC, CA$, and $AB$ respectively of triangle $ABC$ such that $BD=CE=AF$ and $\angle BDF=\angle AFE$. Prove that triangle $ABC$ is equilateral.

It looks really difficult to me. I tried using sine rule and cevas theorm but doesn't help!!

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    $\begingroup$ Are you sure that you haven't missed something? Under the given conditions it's possible to construct $ABC$ which isn't equilateral $\endgroup$ – Stefan4024 Aug 10 '18 at 17:10
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    $\begingroup$ @DEBOJJAL Please check your problem well before you post it. And it is especially unfair to add or remove essential parts of the problem once an answer has been given. $\endgroup$ – Batominovski Aug 11 '18 at 9:02
  • $\begingroup$ I'm voting to close this question as off-topic because the OP made a late change to the question $\endgroup$ – Stefan4024 Aug 11 '18 at 9:23
  • $\begingroup$ I have undone the edit. Actually I'm new here so didn't know the rules m Sorry. I'll create a new thread $\endgroup$ – DEBOJJAL Aug 14 '18 at 8:18

Here's a picture of the counterexample, where $BD=CE=AF$ and $\angle BDE = \angle AFE$


As you can notice in the picture the angles are slightly different, as I couldn't adjust them manually. However that can be taken care of. As you can notice in the first picture below $\angle BDE < \angle AFE$, while in the second one $\angle BDE > \angle AFE$. As the change in the angle is continuous by Intermediate Value Theorem they have to be same at some point.

Min Max

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    $\begingroup$ I want to warn you that the OP has unfairly changed the problem, making your answer irrelevant. However, this is not your fault, and the OP should revert the question to its original question. The OP can create another thread for the corrected question. $\endgroup$ – Batominovski Aug 11 '18 at 9:07

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