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  

$\textbf{(A)}\ \cos\ \alpha\qquad \textbf{(B)}\ \sin\ \alpha\qquad \textbf{(C)}\ \cos^2\alpha\qquad \textbf{(D)}\ \sin^2\alpha\qquad \textbf{(E)}\ 1-\sin\ \alpha$
$ABE$ and $DCE$ are similar isosceles triangles. It remains to find the square of the ratio of their sides. Draw in $AD$. Because $AB$ is a diameter, $\angle ADB=\angle ADE=90^{\circ}$. Thus, \[\frac{DE}{AE}=\cos\alpha\] So \[\frac{DE^2}{AE^2}=\cos^2\alpha\] The answer is thus $\fbox{(C)}$.