Ans: C
$\because AB//AD$,
$\because AB//AD$,
$\therefore \angle ADB = \angle DBC$.
$\because ADCB$ are points on the circle,
$\therefore \angle ADB = \angle ACB$.
In $\Delta BCE$,
$\begin{array}{rcl}
\angle EBC + \angle ECB & = & 74^\circ \\
\angle EBC & = & 37^\circ
\end{array}$
$\therefore \angle ADB = \angle ACB = \angle EBC = 37^\circ$.
In $\Delta ABD$,
$\because AB = AD$,
$\therefore \angle ABD = \angle ADB = 37^\circ$.
In $\Delta ABC$,
$\begin{array}{rcl}
\angle BAE & = & 180^\circ – 37^\circ – 37^\circ – 37^\circ \\
& = & 69^\circ
\end{array}$