4. $A B$ is a chord of a circle with centre $O . C D$ is the diameter perpendicular to the chord $A B$, with $A B$ closer to $C$ than to $D$. Given that $\angle A O B=90^{\circ}$, then the quotient
$$
\frac{\text { area of } \triangle A B C}{\text { area of } \triangle A O D}=
$$
$\qquad$
(A) $\sqrt{2}-1$;
(B) $2-\sqrt{2}$;
(C) $\frac{\sqrt{2}}{2}$;
(D) $\frac{1+\sqrt{2}}{2}$;
(E) $\frac{1}{2}$.
4. Answer: (B).
Let the radius of the circle be 1 . Then
$$
\frac{S_{\triangle A B C}}{S_{\triangle A O D}}=\frac{A E \times C E}{S_{\triangle A O D}}=\frac{A E \times C E}{\frac{1}{2} \times A E \times O D}=\frac{2 \times C E}{O D}=\frac{2(1-1 \sqrt{2})}{1}=2-\sqrt{2}
$$