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 $E$ be a point inside the triangle such that $EB=EC=BC$.
$\qquad\qquad\qquad$
Since $\measuredangle{ABE}=\frac{180^\circ-20^\circ}{2}-60^\circ=20^\circ$, we can see that $\triangle{ABD}$ and $\triangle{BAE}$ are congruent.
Hence, $$\measuredangle{DBC}=\measuredangle{ABC}-\measuredangle{ABD}=\measuredangle{ABC}-\measuredangle{BAE}=80^\circ-10^\circ=\color{red}{70^\circ}.$$
