Ans: C
In $\Delta ABD$,
In $\Delta ABD$,
$\because AB$ is the diameter,
$\therefore \angle ADB=90^\circ$.
$\begin{array}{rcl}
\angle BAD & = & 180^\circ-90^\circ-24^\circ \\
& = & 66^\circ
\end{array}$
In $\Delta OAC$,
$\because OA=OC$,
$\therefore \angle OAC=\angle OCA$.
Also,
$\because AD//OC$,
$\therefore \angle DAC = \angle ACO$.
$\therefore \angle OAC = \angle DAC$.
In $\Delta ADE$,
$\begin{array}{rcl}
\angle AED & = & 180^\circ – 90^\circ – \dfrac{1}{2}(66^\circ) \\
& = & 57^\circ
\end{array}$
