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> In parallelogram $ABCD, A=1, B=4$, 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 $DE$.
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To make $\triangle{ADE}$ have the same area as the parallelogram, the diagonal should be extended to double its length. Using Law of Cosines, you can find the length of the diagonal to be $\sqrt{13}$. 

So, the two sides of $\triangle{ADE}$ would be $4$ and $2 \sqrt{13}$. How do you find $DE$? Are you able to do so using the lengths of the two other sides?