Ans: $18^\circ$
$\begin{array}{rcl}
\angle BAC & = & \angle BDC \\
& = & 36^\circ
\end{array}$
$\begin{array}{rcl}
\angle BAC & = & \angle BDC \\
& = & 36^\circ
\end{array}$
Join $BC$. In $\Delta ABC$,
$\because AB = AC$,
$\angle ABC = \angle ACB$.
$\begin{array}{rcl}
\therefore \angle ACB & = & \dfrac{1}{2}(180^\circ-36^\circ) \\
& = & 72^\circ
\end{array}$
Since $BD$ is a diameter, $\angle BCD = 90^\circ$.
$\begin{array}{rcl}
\angle ACD & = & \angle BCD – \angle ACB \\
& = & 90^ \circ – 72^\circ \\
& = & 18^\circ
\end{array}$
Hence, we have
$\begin{array}{rcl}
\angle ABD & = & \angle ACD \\
& = & 18^\circ
\end{array}$
