Find an angle of an isosceles triangle $\triangle ABC$ is an isosceles triangle such that $AB=AC$ and $\angle  BAC$=$20^\circ$. And a point D is on $\overline{AC}$ so that AD=BC, , How to find $\angle{DBC}$?

I could not get how to use the condition $AD=BC$ , How do I use the condition to find $\angle{DBC}$?
EDIT 1: With MvG's observation, we can prove the following fact.

If we set on a point $O$ in $\triangle{ABC}$ such that $\triangle{OBC}$ is a regular triangle, then $O$ is the circumcenter of $\triangle{BCD}$.

First, we will show if we set a point $E$ on the segment $AC$ such that $OE=OB=OC=BC$, then $D=E$.
Becuase $\triangle{ABC}$ is a isosceles triangle, the point $O$ is on the bisecting line of $\angle{BAC}$. $\angle{OAE}=20^\circ/2=10^\circ$. 
And because $OE=OC$, $\angle{OCE}=\angle{OEC}=20^\circ$, $\angle{EOA}=20^\circ-10^\circ=10^\circ=\angle{EAO}$. 
Therefore $\triangle{AOE}$ is an isosceles triangle such that $EA=EO$. so $AD=BC=AE$, $D=E$.
Now we can see the point $O$ is a circumcenter of the $\triangle{DBC}$ because $OB=OC=OD.$
By using this fact, we can find $\angle{DBC}=70^\circ$,

I saw the following solution may years ago:
On side $AD$ construct in exterior equilateral triangle $ADE$. Connect $BE$.

Then $AB=AC, AE=BC, \angle BAE=\angle ABC$ gives $\Delta BAE =\Delta ABC$ and hence $AB=BE$.
But then 
$$AB=BE, BD=BD, DA=DE \Rightarrow ADB =EDB$$
Hence $\angle ADB=\angle EDB$. Since the two angles add to $300^\circ$ they are each $150^\circ$. Then $\angle ABD + \angle ADB+ \angle BAD=180^\circ$ gives $ABD=10^\circ$.  
