20.69 As shown in the figure, $CE$ and $DE$ are two equal chords of circle $O$, $\widehat{AB}$ is $\frac{1}{4}$ of the circumference, then the ratio of the area of $\triangle CED$ to the area of $\triangle AOB$ is
(A) $\sqrt{2}: 1$.
(B) $\sqrt{3}: 1$.
(C) $4: 1$.
(D) $3: 1$.
(E) $2: 1$.
(9th American High School Mathematics Examination, 1958)
[Solution] Given that $\angle C E D=\angle A O B=90^{\circ}$, we have $\frac{S_{\triangle C E D}}{S_{\triangle A O B}}=\frac{C E \cdot D E}{A O \cdot O B}=\frac{C E^{2}}{A O^{2}}=\frac{\frac{1}{2} C D^{2}}{\frac{1}{4} C D^{2}}=2$. Therefore, the answer is $(E)$.