Finding the side of a triangle made from extending a diagonal

Question

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 $DE$.

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?

Answer 1

Extend $DC$ to $G$ so that $DC=CG$. Then it is easy to see that the parallelogram defined by the two segments $DG$ and $GE$ has $DG=2,GE=4$ and $\angle DGE=60^\circ$, so we can work out by the law of cosines that $DE=\sqrt{12}$.

Answer 2

Let $C$ be a mid-point of $DF$ and $DE=x$.

Thus, since also $C$ is a mid-point of $AE$, we see that $AFED$ is a parallelogram, which gives: $$AF^2+FE^2+ED^2+DA^2=FD^2+AE^2,$$ which by your work gives: $$x^2+4^2+x^2+4^2=2^2+(2\sqrt13)^2$$ or $$x=2\sqrt3.$$

Answer 3

Picture proof:

So $DE$ is twice the length of the side opposite to the $120^\circ$ angle in the isosceles triangle with equal sides $1$ and $1$. The angle bisector of the $120^\circ$ angle cuts it in two easy triangles with angles $60^\circ$, $90^\circ$, $30^\circ$. So the answer is $4\cos 60^\circ=2\sqrt 3$.