In the adjoining figure, $AB$ is a diameter of the circle, $CD$ is a chord parallel to $AB$, and $AC$ intersects $BD$ at $E$, with $\angle AED = \alpha$.  The ratio of the area of $\triangle CDE$ to that of $\triangle ABE$ is

[asy]
size(200);
defaultpen(fontsize(10pt)+linewidth(.8pt));
pair A=(-1,0), B=(1,0), E=(0,-.4), C=(.6,-.8), D=(-.6,-.8), E=(0,-.8/(1.6));
draw(unitcircle);
draw(A--B--D--C--A);
draw(Arc(E,.2,155,205));
label("$A$",A,W);
label("$B$",B,C);
label("$C$",C,C);
label("$D$",D,W);
label("$\alpha$",E-(.2,0),W);
label("$E$",E,N);[/asy]

$ \textbf{(A)}\ \cos\ \alpha\qquad\textbf{(B)}\ \sin\ \alpha\qquad\textbf{(C)}\ \cos^2\alpha\qquad\textbf{(D)}\ \sin^2\alpha\qquad\textbf{(E)}\ 1 - \sin\ \alpha $
1. **Identify Similar Triangles**: 
   Since \(AB\) is a diameter of the circle, \(\angle ADB = 90^\circ\) by the Inscribed Angle Theorem. Given that \(CD\) is parallel to \(AB\), \(\angle ACD = \angle ABD\) and \(\angle CAD = \angle BDA\). Therefore, \(\triangle ABE \sim \triangle CDE\) by AA similarity criterion.

2. **Area Ratio of Similar Triangles**:
   The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. Thus, 
   \[
   \frac{[CDE]}{[ABE]} = \left(\frac{DE}{AE}\right)^2
   \]

3. **Using Trigonometric Relationships**:
   Since \(\angle ADB = 90^\circ\), we can use the definition of cosine in the right triangle \(\triangle ADE\):
   \[
   \cos \alpha = \frac{DE}{AE}
   \]

4. **Substitute and Simplify**:
   Substituting \(\cos \alpha\) into the area ratio expression, we get:
   \[
   \frac{[CDE]}{[ABE]} = \left(\frac{DE}{AE}\right)^2 = (\cos \alpha)^2 = \cos^2 \alpha
   \]

Conclusion:
\[
\boxed{\cos^2 \alpha}
\]