TITLE: Prove that $ABC$ is equilateral
QUESTION [5 upvotes]: Let $D,E,F$ be points on the sides BC,CA,AB respectively of a triangle $ABC$  such that $BD=CE=AF$ and $\angle BDF=\angle CED=\angle AFE$.Prove that $\triangle ABC$ is equilateral.
My attempt - Using sine rule in triange $\triangle AFE$ ,$\triangle EDC$ and $\triangle BFD$ respectively, we have $AE \sin\angle AEF=BF \sin\angle BFD=DC\sin\angle FDC $. 
But that does not help much.I am totally unaware what to do.please help.

REPLY [5 votes]: So one has the following settings

WLOG, assume that $A$ is the max angle in $\triangle ABC$. It follows from the picture below that $EF$ is the largest side of $\triangle DEF$.

That means $\angle D2$ is the largest angle in $\triangle DEF$. We have
$$\angle D1=\min\{\angle D1,\angle E1,\angle F1\}.$$
This follows that $\angle C$ is the largest angle of $\triangle ABC$, or that $\angle C=\angle A$. 
P/S: Please excuse my hand-drawing. I was too lazy for computer graphics.