27. As shown in the figure, $AB$ is the diameter of a circle, $CD$ is a chord parallel to $AB$, and $AC$ intersects $BD$ at $E$, with $\angle AED = a$. The ratio of the area of $\triangle CDE$ to the area of $\triangle ABE$ is
(A) $\cos a$; (B) $\sin a$; (C) $\cos^2 a$;
(D) $\sin^2 a$;
(E) $1-\sin a$.
$\frac{27}{C}$
27. In the figure, $A B / / D C, A \overarc{D}=\overarc{C B}, \triangle C D E$ is similar to $\triangle A B E$, thus
$$
\frac{\text { area of } \triangle C D E}{\text { area of } \triangle A B E}=\left(\frac{D E}{A E}\right)^{2}.
$$

Connecting $A D$, since $A B$ is a diameter, $\angle A D B=30^{\circ}$. Considering the right $\triangle A D E$, we have $D E=A E \cos \alpha$ and $\left(\frac{D E}{A E}\right)^{2}=\cos ^{2} \alpha$.