In parallelogram $ABCD, AB = 1, BC = 4$, and $\angle ABC = 60^{\circ}$ . Suppose that $AC$ is extended from $A$ to a point $E$ beyond $C$ so that triangle $ADE$ has the same area as the parallelogram. Find the length of $DE$.
I have already gotten the area of the parallelogram using its $2$ diagonals and the cosine law.
$\sqrt{13} \sqrt{21} = \sqrt{273}/2 $
I also see that $CDE$ must be half of the are of the parallelogram, but I cant seem to find any angles and side lengths to compute for $CD$. Some hints or maybe I overlooked some properties?