2011SP-I-07

答案：$19^\circ$
$\because AB//OC$，

$\begin{array}{rcl}
\angle COD & = & \angle BAD \\
& = & 38^\circ
\end{array}$

連結 $OB$。

$\because OA \mbox{ 及 } OB$ 為半徑，
$\begin{array}{rcl}
\angle OBA & = & \angle OAB \\
& = & 38^\circ
\end{array}$

在 $\Delta OAB$ 中，

$\begin{array}{rcl}
\angle OAB+\angle OBA & = & \angle BOD \\
38^\circ+38^\circ & = & \angle BOC+\angle COD \\
\angle BOC+38^\circ & = & 76^\circ \\
\angle BOC & = & 38^\circ
\end{array}$

由此，可得

$\begin{array}{rcl}
\angle BDC & = & \dfrac{1}{2} \angle BOC \\
& = & \dfrac{1}{2} \times 38^\circ \\
& = & 19^\circ
\end{array}$