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.