In $\Delta ABC$, angle bisector of $\angle ABC$ and median on side $BC$ intersect perpendicularly 
In $\Delta ABC$, $BE$ is the angle bisector of $\angle ABC$, $AD$ is the median on side $BC$. $AD$ intersects $BE$ at $O$ perpendicularly. If $AD = BE = 4$, find the lengths of each side of $\Delta ABC$.

What I Tried: At first I was having a hard time trying to make a bit of an accurate picture of the problem, and I made this :-

As solving this, I got no idea. Tried angle-chasing for example, if $\angle ABO = \angle DBO = x$ , then the green angles come to be $(90 - x)$ each, and then you have the brown angle to be $(90 + x)$ . You only get that $\Delta ABO \sim \Delta DBO$ , and that gives me no useful information for now.
I don't think I can use Pythagorean Theorem that much because except $AD = BE = 4$ , I have no other side-lengths to proceed. So right now, I am literally out of ideas.
Can anyone help me do this? Thank You!
 A: In $\triangle ABD$, $BD=AB$. $OA=OD=2$
Let $AB=c$, $AC=b$. $BC=a=2c$.
Also $OE=x$. $OB=4-x$
From $$\dfrac{AE}{CE}=\dfrac{BA}{BC}=\dfrac{1}{2}$$
$$AE = \dfrac{b}{3} , CE= \dfrac{2b}{3}$$
From Apollonius theorem,
$$ b^2 + c^2 = 2(4^2 + c^2)$$
$$ \Rightarrow b^2 - c^2 = 32$$
In right $\triangle BOD$,
$$ 2^2 + (4-x)^2 = c^2$$
In right $\triangle AOE$,
$$ 2^2 + x^2 = \dfrac{b^2}{9}$$
On solving, $x=1$
So $$ ({a,b,c}) = ({2\sqrt{13},3\sqrt{5},\sqrt{13}}) $$
A: Took your diagram and added a line for my solution.

$\triangle ABO \cong \triangle DBO$ (by Angle-Angle-Side)
So, $AO = OD = 2$ and $AB = BD = DC$.
Also, as $BE$ is besector of $\angle B$, $\frac{AE}{CE} = \frac{AB}{BC} = \frac{1}{2}$
Now extend line $BE$ and draw a perpendicular from point $C$ to extended line $BE$. Say it meets line $BE$ at point $F$.
Now $\triangle CEF \sim \triangle AEO$
So $\frac{EF}{CE} = \frac{OE}{AE} \implies EF = 2 OE \implies OF = 3 OE$
Also note that $\triangle BCF \sim \triangle BDO$
So $OB = OF = 3 OE; OB = 3, OE = 1 \,$ as $BE = 4$
$AB = \sqrt{OB^2 + OA^2} = \sqrt {13}$
$BC = 2 AB = 2\sqrt{13}$
$AE = \sqrt{OA^2 + OE^2} = \sqrt {5} \implies AC = 3\sqrt5$
