Apollonius's theorem: If ${\overline{BE}}$ is a median of ${\triangle}ABC$, then$${\overline{AB}}^2+{\overline{BC}}^2=2{\overline{BE}}^2+{\frac{1}{2}}{\overline{AC}}^2$$
By using the parallelogram law, we get $$\begin{align} {\overline{AC}}^2+{\overline{BD}}^2 &={\overline{AB}}^2+{\overline{BC}}^2+{\overline{CD}}^2+{\overline{DA}}^2 \\&={\overline{AC}}^2+(2{\overline{BE}})^2 \end{align}$$ therefore we have $${\overline{AB}}^2+{\overline{BC}}^2=4{\overline{BE}}^2+{\overline{AC}}^2-{\overline{CD}}^2-{\overline{DA}}^2$$ I'm stuck here. Any hints?
You seem to be using the theorem that you're trying to prove. This is fine for exploratory work, but you'll need to rewrite it in the final version by reversing the steps.
Looking at your equation $$\begin{align} {\overline{AC}}^2+{\overline{BD}}^2 &={\overline{AB}}^2+{\overline{BC}}^2+{\overline{CD}}^2+{\overline{DA}}^2 \\&={\overline{AC}}^2+(2{\overline{BE}})^2 \end{align}$$ notice that we can remove ${\overline{AC}}^2$ from both sides to get ${\overline{BD}}^2 = (2{\overline{BE}})^2$, which is true because ${\overline{BD}} = 2{\overline{BE}}$.
Now start at the end and work backwards.