In a triangle $ABC$ with side $AB=AC$ and $\measuredangle BAC=20 ^\circ $. $D $ is a point on side $AC$ and $BC = AD$. Find $ \measuredangle DBC$ Problem : In a triangle $ABC$ with side $AB=AC$ and $\measuredangle BAC=20 ^\circ $.
$D $ is a point on side $AC$ and $BC = AD$. Find $ \measuredangle DBC$
Solution: 
$AB =AC$
So  $ \measuredangle ACB = \measuredangle ABC$
$ \measuredangle ACB = \measuredangle ABC=80^\circ$ (Using angle sum property)
I am unable to continue from here.
Any assistance is appreciated.

Let $ABC$ be the triangle with $O$ being the circumcentre. $AO$ is extended to meet $BC$ at $D$. $BO$ is extended to meet $AC$ at $E$. Let $F$ 
be the point on $AC$ such that $AF= BC$. 
Triangle $OAB$ is congruent to triangle $OAC$. 
Therefore $\angle BAO = \angle CAO = 10^{\circ}$.
Let $BD = a$, so $CD$ also equals $a$ 
therefore $AB = \frac{a}{\sin 10^{\circ}}$ 
Now applying sine rule in triangle $AEB$
$$ 
\frac{AE}{\sin 10^{\circ}} = \frac{AB}{\sin 150^{\circ}}
$$ 
Putting the value of $AB$ as $a/\sin 10^{\circ}$ 
We get 
$$
AE = \frac{a}{\sin 30^{\circ}} = 2a.
$$ 
We have $AF= BC = 2a$ 
Therefore $F$ and $E$ are the same point . 
Therefore 
$$
\angle ABF = \angle ABE = 10^{\circ}
$$ 
and hence
$$
\angle FBC = 80 - 10 = 70^{\circ}.
$$
