2019-II-39

答案：B
考慮圓內接四邊形 $ACDE$。

$\begin{array}{rcll}
\angle DAE & = & \angle DCE & \text{(同弓形內的圓周角)} \\
\angle DAE & = & 22^\circ
\end{array}$

考慮 $\Delta ABD$。

$\begin{array}{rcll}
\angle ABD & = & \angle DAT & \text{(內錯弓形的圓周角)} \\
\angle ABD & = & \angle DAE + \angle EAT \\
\angle ABD & = & 22^\circ + 38^\circ \\
\angle ABD & = & 60^\circ
\end{array}$

由此，可得

$\begin{array}{rcll}
\angle ADB & = & 180^\circ – \angle ABD – \angle BAD & \text{($\Delta$ 的內角和)} \\
\angle ADB & = & 180^\circ – 60^\circ – 64^\circ \\
\angle ADB & = & 56^\circ
\end{array}$