In $\Delta ABE$,
$\begin{array}{rcll}
\angle AEB & = & 180^\circ-90^\circ-\alpha & \text{($\angle$ sum of $\Delta$)} \\
\angle AEB & = & 90^\circ -\alpha
\end{array}$
Also,
$\begin{array}{rcl}
\sin \alpha & = & \dfrac{BE}{AE} \\
AE & = & \dfrac{BE}{\sin \alpha}
\end{array}$
In $\Delta BCE$,
$\begin{array}{rcll}
\angle AEC & = & 180^\circ -\angle BCE & \text{(int. $\angle$s, $AD$//$BC$)} \\
\angle AEB +\angle BEC & = & 180^\circ-90^\circ \\
90^\circ-\alpha+\angle BEC & = & 90^\circ \\
\angle BEC & = & \alpha
\end{array}$
Then, we have
$\begin{array}{rcl}
\cos \angle BEC & = & \dfrac{CE}{BE} \\
CE & = & BE\cos \alpha
\end{array}$
Since $E$ is the mid-point of $AD$, we have
$\begin{array}{rcl}
\dfrac{CE}{DE} & = & \dfrac{CE}{AE} \\
\dfrac{CE}{DE} & = & \dfrac{BE\cos \alpha}{\frac{BE}{\sin\alpha}} \\
\dfrac{CE}{DE} & = & \sin \alpha \cos \alpha
\end{array}$
