在 $\Delta OBD$ 及 $\Delta OAC$ 中，
$\begin{array}{rcll}
OB & = & OA & \text{(半徑)} \\
OD & = & OC & \text{(半徑)} \\
BD & = & AC & \text{(已知 )}
\end{array}$
所以，$\Delta OBD \cong \Delta OAC\ \text{(S.S.S.)}$。由此，可得
$\begin{array}{rcll}
\angle OAC & = & \angle OBD & \text{($\cong \Delta$ 的對應角)}
\end{array}$
在 $\Delta OAC$，
$\begin{array}{rcll}
OA & = & OC & \text{(半徑)} \\
\angle OAC & = & \angle OCA & \text{(等腰 $\Delta$ 的底角)} \\
\angle COB & = & \angle OAC + \angle OCA & \text{($\Delta$ 的外角)} \\
\angle COB & = & 2 \angle OAC & \text{(已證)} \\
\angle COB & = & 2 \angle OBD & \text{(已證)}
\end{array}$
考慮 $\Delta OBD$。
$\begin{array}{rcll}
OD & = & OB & \text{(半徑)} \\
\angle ODB & = & \angle OBD & \text{(等腰 $\Delta$ 的底角)}
\end{array}$
所以，可得
$\begin{array}{rcll}
\angle BOD + \angle ODB + \angle OBD & = & 180^\circ & \text{($\Delta$ 的內角和)} \\
\angle COD + \angle BOC + \angle ABD + \angle ABD & = & 180^\circ &\text{(已證)} \\
48^\circ + 2 \angle ABD + \angle ABD + \angle BAD & = & 180^\circ \\
4 \angle ABD & = & 132^\circ \\
\angle ABD & = & 33^\circ
\end{array}$
