2019-II-39

Ans: B
Consider cyclic quadrilateral $ACDE$.

$\begin{array}{rcll}
\angle DAE & = & \angle DCE & \text{($\angle$s in the same segment)} \\
\angle DAE & = & 22^\circ
\end{array}$

Consider $\Delta ABD$.

$\begin{array}{rcll}
\angle ABD & = & \angle DAT & \text{($\angle$s in alt. segment)} \\
\angle ABD & = & \angle DAE + \angle EAT \\
\angle ABD & = & 22^\circ + 38^\circ \\
\angle ABD & = & 60^\circ
\end{array}$

Hence, we have

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