一、Fill in the Blanks (10 points each, total 80 points)
1. Given $A B$ is the diameter of $\odot O$, $A B=1$, extend $A B$ to point $C$, such that $B C=1$, $C D$ is the tangent of $\odot O$, $D$ is the point of tangency. Then the area of $\triangle A B D$ is $\qquad$
$-1 . \frac{\sqrt{2}}{6}$.
By the secant-tangent theorem, we know
$$
C D^{2}=C B \cdot C A=2 \Rightarrow C D=\sqrt{2} \text {. }
$$

As shown in Figure 5, draw $D E \perp O C$ at point $E$.
Then $D E \cdot O C=2 S_{\triangle O D C}=O D \cdot C D=\frac{\sqrt{2}}{2}$
$$
\Rightarrow D E=\frac{\sqrt{2}}{3} \Rightarrow S_{\triangle A B D}=\frac{1}{2} A B \cdot D E=\frac{\sqrt{2}}{6} .
$$