In $\Delta OBD$ and $\Delta OAC$,
$\begin{array}{rcll}
OB & = & OA & \text{(radii)} \\
OD & = & OC & \text{(radii)} \\
BD & = & AC & \text{(given)}
\end{array}$
Therefore, $\Delta OBD \cong \Delta OAC\ \text{(S.S.S.)}$. Hence, we have
$\begin{array}{rcll}
\angle OAC & = & \angle OBD & \text{(corr. $\angle$s, $\cong \Delta$)}
\end{array}$
In $\Delta OAC$,
$\begin{array}{rcll}
OA & = & OC & \text{(radii)} \\
\angle OAC & = & \angle OCA & \text{(base $\angle$s, isos. $\Delta$)} \\
\angle COB & = & \angle OAC + \angle OCA & \text{(ext. $\angle$ of $\Delta$)} \\
\angle COB & = & 2 \angle OAC & \text{(proved)} \\
\angle COB & = & 2 \angle OBD & \text{(proved)}
\end{array}$
Consider $\Delta OBD$.
$\begin{array}{rcll}
OD & = & OB & \text{(radii)} \\
\angle ODB & = & \angle OBD & \text{(base $\angle$s, isos. $\Delta$)}
\end{array}$
Therefore, we have
$\begin{array}{rcll}
\angle BOD + \angle ODB + \angle OBD & = & 180^\circ & \text{($\angle$ sum of $\Delta$)} \\
\angle COD + \angle BOC + \angle ABD + \angle ABD & = & 180^\circ &\text{(proved)} \\
48^\circ + 2 \angle ABD + \angle ABD + \angle BAD & = & 180^\circ \\
4 \angle ABD & = & 132^\circ \\
\angle ABD & = & 33^\circ
\end{array}$
