A geometry problem by Sanyam Garg

The completed diagram looks as follows:

Now, we see that $ABCD$ forms a parallelogram, thus $B=E \implies \boxed{\bar{BE} = 0}$

From the given conditions of the problem, $\triangle {ABC}$ is equilateral. Hence $ABCD$ is a parallelogram, and $\overline {DE}$ is a diagonal. So $E$ coincides with $B$ , and $|\overline {BE}|=\boxed 0$ .