Let $AB$ be a diameter of a circle and let $C$ be a point on $AB$ with $2\cdot AC=BC$. Let $D$ and $E$ be points on the circle such that $DC \perp AB$ and $DE$ is a second diameter. What is the ratio of the area of $\triangle DCE$ to the area of $\triangle ABD$? $\mathrm{(A) \ } \frac{1}{6} \qquad$ $ \mathrm{(B) \ } \frac{1}{4} \qquad$ $ \mathrm{(C) \ } \frac{1}{3} \qquad$ $ \mathrm{(D) \ } \frac{1}{2} \qquad$ $ \mathrm{(E) \ } \frac{2}{3}$
Solution
Connect $EF$. Then $[DCE]=[FCE]=\frac12[EDF]=2[DCO]$.
Let $AB$ be a diameter of a circle and $C$ a point on $AB$ with $3\cdot AC=BC$. Let $D,E$ be points on the circle with $DC\perp AB$ and $DE$ a second diameter. What is the ratio of the area of $\triangle DCE$ to the area of $\triangle ABD$?
$\mathrm{(A) \ } \frac{1}{6} \qquad$ $ \mathrm{(B) \ } \frac{1}{4} \qquad$ $ \mathrm{(C) \ } \frac{1}{3} \qquad$ $ \mathrm{(D) \ } \frac{1}{2} \qquad$ $ \mathrm{(E) \ } \frac{2}{3}$ 
