Ans: C
Let $\angle CBD = x$. Since $A$, $B$, $C$ and $D$ are four points on the circumference, then
Let $\angle CBD = x$. Since $A$, $B$, $C$ and $D$ are four points on the circumference, then
$\begin{array}{rcl}
\angle ACD & = & \angle ABD \\
& = & 40^\circ
\end{array}$
Since $AB//DC$, then
$\begin{array}{rcl}
\angle BAC & = & \angle ACD \\
& = & 40^\circ
\end{array}$
In $\Delta ABC$, since $AB = AC$, then
$\begin{array}{rcl}
\angle ACB & = & \angle ABC \\
& = & 40^\circ + x
\end{array}$
Hence, we have
$\begin{array}{rcl}
40^\circ + (40^\circ+x) \times 2 & = & 180^\circ \\
120^\circ + 2x & = & 180^\circ \\
2x & = & 60^\circ \\
x & = & 30^\circ
\end{array}$
Therefore, $\angle CBD = 30^\circ$.
