8,9
[Trigonometric ratios in a right triangle]

In an isosceles triangle $ABC$, the angle at vertex $B$ is $\alpha$. A tangent to the circumcircle of this triangle is drawn at point $C$, intersecting the extension of the angle bisector $BD$ of angle $B$ at point $E$. Find the ratio of the area of triangle $CDE$ to the area of triangle $ABC$.
Let $C D=A D=a$. By the theorem on the angle between a tangent and a chord, $\angle D C E=\angle A C E=\alpha$.

From the right triangles $C D E$ and $B C D$, we find that $D E=a \operatorname{tg} \alpha, B D=a \operatorname{ctg} \alpha / 2$.

Therefore, $\frac{S_{C D E}}{S_{A B C}}=\frac{D E}{2 B D}=\frac{\operatorname{tg} \alpha}{2 \operatorname{ctg} \frac{\alpha}{2}}=\frac{1}{2} \operatorname{tg} \alpha \operatorname{tg} \frac{\alpha}{2}$.

## Answer

$1 / 2 \operatorname{tg} \alpha \operatorname{tg} \alpha / 2$.