在 $\Delta ABE$ 中，
$\begin{array}{rcll}
\angle AEB & = & 180^\circ-90^\circ-\alpha & \text{(三角形的內角和)} \\
\angle AEB & = & 90^\circ -\alpha
\end{array}$
另外，
$\begin{array}{rcl}
\sin \alpha & = & \dfrac{BE}{AE} \\
AE & = & \dfrac{BE}{\sin \alpha}
\end{array}$
在 $\Delta BCE$ 中，
$\begin{array}{rcll}
\angle AEC & = & 180^\circ -\angle BCE & \text{(同旁內角，$AD$//$BC$)} \\
\angle AEB +\angle BEC & = & 180^\circ-90^\circ \\
90^\circ-\alpha+\angle BEC & = & 90^\circ \\
\angle BEC & = & \alpha
\end{array}$
所以，可得
$\begin{array}{rcl}
\cos \angle BEC & = & \dfrac{CE}{BE} \\
CE & = & BE\cos \alpha
\end{array}$
由於 $E$ 為 $AD$ 的中點，可得
$\begin{array}{rcl}
\dfrac{CE}{DE} & = & \dfrac{CE}{AE} \\
\dfrac{CE}{DE} & = & \dfrac{BE\cos \alpha}{\frac{BE}{\sin\alpha}} \\
\dfrac{CE}{DE} & = & \sin \alpha \cos \alpha
\end{array}$
