$7 \cdot 63$ As shown, $AB$ is the diameter of a circle, $CD$ is a chord parallel to $AB$, and $AC$ intersects $BD$ at $E$. Given $\angle AED = \alpha$, the ratio of the area of $\triangle CDE$ to the area of $\triangle ABE$ is
(A) $\cos \alpha$.
(B) $\sin \alpha$.
(C) $\cos^2 \alpha$.
(D) $\sin^2 \alpha$.
(E) $1 - \sin \alpha$.
[Solution] Given $\triangle C D E \backsim \triangle A B E$,
$$
\therefore S_{\triangle C D E}: S_{\triangle A B E}=E D^{2}: E B^{2} \text {, }
$$

Noting that $E B=A E(\because A B 、 C D$ are parallel chords $)$,
$$
\therefore \quad S_{\triangle C D E}: S_{\triangle A B E}=E D^{2}: A E^{2} \text {. }
$$

Connecting $A D$, we have $\angle A D E=90^{\circ}$,

Thus,
$$
\begin{array}{ll}
\therefore \quad & \frac{D E}{A E}=\cos \alpha, \\
& S_{\triangle C D E}: S_{\triangle A B E}=\cos ^{2} \alpha .
\end{array}
$$

Therefore, the answer is $(C)$.