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 $\angle{DBC}=70^\circ$,
 A: One way to calculate this is to write sin laws for two triangles $ABD$ and $BDC$. Call the angle $\angle ABD=x$. Then we have:
$$
\frac{AD}{\sin x}=\frac{BD}{\sin 20},\frac{BC}{\sin (20+x)}=\frac{BD}{\sin 80}
$$
Using $AD=BC$ and $\sin 80=\cos 10$ we get the following:
$$
\frac{\sin x}{\sin 20}=\frac{\sin (20+x)}{\sin 80}\implies \sin x=2{\sin 10}\sin (20+x)\implies\\
 \tan x=\frac{2\sin 10\sin 20}{1-2\sin 10\cos 20}
$$
Now consider the following identities:
$$
1-2\sin 10\cos 20=1-(\sin 30-\sin 10)=\frac{1}{2}+\sin 10=2\cos 10 \sin 20
$$
Replacing this result in previous equation we get: 
$$
 \tan x=\frac{2\sin 10\sin 20}{2\cos 10\sin 20}=\tan 10 \implies x=10
$$
and hence $\angle DBC=70$.
A: 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 $ABD=10^\circ$.  
A: We construct $\overline{CE}$ such that $E$ lies on $\overline{AB}$ and $\overline{DE}\,||\,\overline{CB}$. We then construct a circle through $A$, $B$ and $C$. Circle $ABC$ has a centre $O$ at the intersection of $\overline{CE}$ and $\overline{BD}$. By the inscribed angle theorem, $\widehat{BOC}=40°$ and since $\triangle BOC$ is isosceles, $\widehat{DBC}=70°$ $\blacksquare$.
A: 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
