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!!
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.