Geometry study question I am currently studying geometry and there is a question which I simply cannot figure out. I have attached the image.

The question asks: In the figure, equilateral triangle ABC is inscribed in circle O, whose radius is 4. Altitude BD is extended until it intersects the circle at E. What is the length of DE?
The answer states the following: To get DE, subtract OD from radius OE, which is 4. Draw AO, Since triangle AOD is a 30-60-90 triangle, OD is 2.
But I simply cannot figure out how we know that triangle AOD is a 30-60-90 triangle. What gets us there?
Any help is appreciated.
Thank you,
Ilya
 A: $O$ is center of gravity of the triangle, it implies that :
$$\vec{OA}+ \vec{OB} + \vec{OC} = \vec{0} = \vec{OD}+\vec{DA}+\vec{OB}+\vec{OD}+\vec{DC} $$
Since ABC is equilateral : $\vec{DA}+\vec{DC}=\vec{0}$
And then you have $OD = \frac{OB}{2} = 2$
A: Extend $AO$ to meet $BC$ at $X$. By symmetry of an equilateral triangle, $\angle AXC=90^o$. $\angle XCA=60^o$, so $\angle CAO=30^o$.
A: In an equilateral triangle the angle bisector, the perpendicular bisector and the height coincide. 
The three perpendicular bisectors intersect in the circumcenter (O in the figure). 
Tthe line AO is actually the perpendicular bisector of side BC, the height of the triangle and the bisector of the angle $\hat{A}$ . Therefore, $\hat{A}=30º$.
On the other hand, the line BO is perpendicular to the side AC, that is $\hat{ADO}=90º$. Thus the remaining angle of the triangle must measure 60º.
Hope that helps.
A: AO=OB=4 so  triangle AOB is isosceles so  $30^o$= (1/2)angle ABC= angle ABO= angle BAO. Therefore angle OAD= $(60^o)$-(angle BAO)=$30^o$. 
