AMC 10 2017b Problem 22

The diameter $\overline{AB}$ 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 $\overline{AE}$ intersects the circle at a point $C$ between $A$ and $E$. What is the area of $\triangle ABC$? $\textbf{(A)}\ \frac{120}{37} \textbf{(B)}\ \frac{140}{39} \textbf{(C)}\ \frac{145}{39} \textbf{(D)}\ \frac{140}{37} \textbf{(E)}\ \frac{120}{31}$