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 $\angle{DBC}=70^\circ$,

in $\bigtriangleup$ABC AB=AC and $\angle$A=20. SO $\angle$B=$\angle$C=80


*

*NOW WE DRAW $\angle$BAO=60 with AO=AB.AND WE join
OD and OB

*now we consider $\bigtriangleup$ABC and $\bigtriangleup$ADO. AD=BC; AO=AB and $\angle$DAO=$\angle$B=80

*SO $\bigtriangleup$ABC$\cong$$\bigtriangleup$ADO. SO $\angle$AOD=20 and DO=AB=AC

*furthur we have AO= OB and $\angle$BAO=60.SO $\bigtriangleup$BAO IS equilateral.SO $\angle$AOB =60.FROM THIS we have $\angle$DOB=40.aiso OB=AB=DO

*OB=DO and $\angle$DOB=40 SO WE HAVE $\angle$OBD=$\angle$ODB=70 and $\angle$OBA=60.SO $\angle$ABD=10

*$\angle$B=80 so $\angle$DBC=70
