The diameter $AB$ of a circle of radius $2$ is extended to a point $D$ outside the circle so that $BD=3$. Point $E$ is chosen so that $ED=5$ and line $ED$ is perpendicular to line $AD$. Segment $AE$ intersects the circle at a point $C$ between $A$ and $E$. What is the area of $\triangle ABC$? $\textbf{(A)}\ \frac{120}{37}\qquad$ $\textbf{(B)}\ \frac{140}{39}\qquad$ $\textbf{(C)}\ \frac{145}{39}\qquad$ $\textbf{(D)}\ \frac{140}{37}\qquad$ $\textbf{(E)}\ \frac{120}{31}$
Solution
$\triangle ABC\sim\triangle AED$. Since $AE=\sqrt{49+25}=\sqrt{74}$, the similarity ratio is $4:\sqrt{74}$.
The diameter $AB$ of a circle of radius $3$ is extended to $D$ outside the circle so $BD=2$. Point $E$ satisfies $ED=6$, $ED\perp AD$. Segment $AE$ meets the circle at $C$ between $A,E$. What is the area of $\triangle ABC$?
