Easier way to calculate this point besides line intersection? 
Given are all points except E, plus |AF| = |DC|. Considering that the lines AB and FE, as well as BC and ED are parallel, is there an easier way to calculate E? Maybe some relation with B? 
I'd like to express the coordinates from E as short as possible and therefore I didn't want to calculate the intersection between the lines FE and ED.
Many thanks in advance!
 A: Here is an answer without explanation.
$\vec{E} = \vec{B}+\frac{1}{1+\frac{\vec{CD}\cdot\vec{AF}}{\vec{AF}\cdot\vec{AF}}}(\vec{AF}+\vec{CD})$
Here is the explanation:
First, translate so that $B$ is at the origin.  This will simplify notation.  At the end, we will translate back, obtaining the "$\vec{B}+$" in the formula.
Sketch the vectors $\vec{AF}$ and $\vec{CD}$ with their heads at $E$.  Sketch direct lines from their tails to $B$.  The right angles and the fact that $|AF|=|CD|$ imply that this quadrilateral is a kite.  Moreover, the diagonal out of $B$ is spanned by the vector $(\vec{AF}+\vec{CD})$.  So $\vec{E}$ is some scalar multiple of this: $\vec{E}=t(\vec{AF}+\vec{CD})$.
The conditions that have been laid out demand that $$proj_{\vec{AF}}\vec{E}=\vec{AF}$$  That means $\frac{\vec{AF}\cdot\left(t(\vec{AF}+\vec{CD})\right)}{\vec{AF}\cdot\vec{AF}}\vec{AF}=\vec{AF}$, or rather $\frac{\vec{AF}\cdot\left(t(\vec{AF}+\vec{CD})\right)}{\vec{AF}\cdot\vec{AF}}=1$.  This implies $$t=\frac{\vec{AF}\cdot\vec{AF}}{\vec{AF}\cdot\vec{AF}+\vec{AF}\cdot\vec{CD}}$$ which simplifies to the coefficient in the solution.
