# Triangle equilateral proof

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!!

• Are you sure that you haven't missed something? Under the given conditions it's possible to construct $ABC$ which isn't equilateral – Stefan4024 Aug 10 '18 at 17:10
• @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. – Batominovski Aug 11 '18 at 9:02
• I'm voting to close this question as off-topic because the OP made a late change to the question – Stefan4024 Aug 11 '18 at 9:23
• I have undone the edit. Actually I'm new here so didn't know the rules m Sorry. I'll create a new thread – 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.