Disclaimer: The solutions we've shared are just one exciting approach, and there are surely many other wonderful methods out there. We’d love to hear your alternative solutions in the community thread below, so let's keep the creativity flowing!

Let $S_a=\frac{AE}{AB} + \frac{FA}{CA}, S_b= \frac{EB}{AB} + \frac{BD}{BC}, S_c=\frac{DC}{BC} + \frac{CF}{CA}$.

Let $M_a=AE/AB \times FA/CA, M_b=EB/AB \times BD/BC, M_c=DC/BC \times CF/CA$.

Now, 

$S_a + S_b + S_c $

$= \frac{AE}{AB} + \frac{FA}{CA} + \frac{EB}{AB} + \frac{BD}{BC} + \frac{DC}{BC} + \frac{CF}{CA} $

$= 3$.

Without loss of generality, we assume that $S_b$ is the lowest among $S_a,S_b,S_c$. By pigeon hole principle, we know that $S_b \leq 1$.

From AM-GM inequality, we know that $\frac{S_b}{2} \geq \sqrt{M_b}$ or $\frac{1}{4} \geq \frac{S_b}{4} \geq M_b$.

Now, the area of $\triangle BED$

$ = \frac{\sin{\angle EBD}}{2}\times BE \times BD $

$= \frac{\sin{\angle ABC}}{2}\times AB \times BC\times M_b $

$= 2016\times M_b \leq 2016 \times \frac{1}{4} $

$= 504$.