Ans: B
$\because AB$ is the diameter,
$\because AB$ is the diameter,
$\therefore \angle ACB=90^\circ$.
$\because ABCD$ are cyclic quadrilateral,
$\begin{array}{rcl}
\therefore \angle BAD & = & 180^\circ-28^\circ-90^\circ \\
& = & 62^\circ
\end{array}$
$\because BC=CD$,
$\therefore \angle CAD = \angle CAB$.
Hence, we have
$\begin{array}{rcl}
\angle CAD & = & \dfrac{1}{2}\times62^\circ \\
& = & 31^\circ
\end{array}$
Therefore, we have
$\begin{array}{rcl}
\angle ADC & = & 180^\circ-28^\circ-31^\circ \\
& = & 121^\circ
\end{array}$
