Proof involving chords of a circle 

In a circumference with center $O$, three chords $\overline{AB},\overline{AD}$ and $\overline{CB}$ such that the last two intersect in $E$. Show that $AE·AD+BE·BC=AB^2 $.
Added: $O\in\overline{AB}$.

Hi, I have been trying to solve this problem with the power of a point with respect to a circumference and with pythagoras, but it seems that I'm going nowhere:
$2(AB)^2=AD^2+BD^2+BC^2+AC^2$
I hope you could give me a hint. Thanks
Edit: Taking into account what Blue mentioned: 

You were on a good track. You found that
$$2 \cdot AB^2 = AD^2 + BD^2 + BC^2 + AC^2.$$
But you want something with $AD \cdot AE,$ not $AD^2,$ you don't want to see $BD^2,$
and so forth.
But notice that $AC^2 = AE^2 - CE^2$ and $BD^2 = BE^2 - DE^2,$
and also $AD = AE + DE$ and $BC = BE + CE.$
So this suggests you could put everything on the right-hand side in terms of 
$AE,$ $BE,$ $CE,$ and $DE.$
So the hint is, do that, then see if you can put the pieces back together to make
$2 \cdot AD \cdot AE + 2 \cdot BC \cdot BE.$ If you get that far,
you can easily complete the proof.
