### Instruction:
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 $\triangle  ABC$?
$\textbf{(A)}\ \frac{120}{37}\qquad\textbf{(B)}\ \frac{140}{39}\qquad\textbf{(C)}\ \frac{145}{39}\qquad\textbf{(D)}\ \frac{140}{37}\qquad\textbf{(E)}\ \frac{120}{31}$


### Response:

[asy]  /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(8.865514650638614cm);  real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */  pen dotstyle = black; /* point style */  real xmin = -6.36927122464312, xmax = 11.361758076634109, ymin = -3.789601803155515, ymax = 7.420015026296013;  /* image dimensions */   draw((-2.,0.)--(0.6486486486486486,1.8918918918918919)--(2.,0.)--cycle);   /* draw figures */ draw(circle((0.,0.), 2.));  draw((-2.,0.)--(5.,5.));  draw((5.,5.)--(5.,0.));  draw((5.,0.)--(-2.,0.));  draw((-2.,0.)--(0.6486486486486486,1.8918918918918919));  draw((0.6486486486486486,1.8918918918918919)--(2.,0.));  draw((2.,0.)--(-2.,0.));  draw((2.,0.)--(5.,5.));  draw((0.,0.)--(5.,5.));   /* dots and labels */ dot((0.,0.),dotstyle);  label("$O$", (-0.10330578512396349,-0.39365890308038826), NE * labelscalefactor);  dot((-2.,0.),dotstyle);  label("$A$", (-2.2370398196844437,-0.42371149511645134), NE * labelscalefactor);  dot((2.,0.),dotstyle);  label("$B$", (2.045454545454548,-0.36360631104432517), NE * labelscalefactor);  dot((5.,0.),dotstyle);  label("$D$", (4.900450788880542,-0.42371149511645134), NE * labelscalefactor);  dot((5.,5.),dotstyle);  label("$E$", (5.06574004507889,5.15104432757325), NE * labelscalefactor);  dot((0.6486486486486486,1.8918918918918919),linewidth(3.pt) + dotstyle);  label("$C$", (0.48271975957926694,2.100706235912847), NE * labelscalefactor);  clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);   /* end of picture */ [/asy]
Let $O$ be the center of the circle. Note that $EC + CA = EA = \sqrt{AD^2 + DE^2} = \sqrt{(2+2+3)^2 + 5^2} = \sqrt{74}$. However, by Power of a Point, $(EC)(EC + CA) = EO^2 - R^2 = (2+3)^2 + 5^2 - 2^2 = 25 + 25 - 4 = 46 \implies EC = \frac{46}{\sqrt{74}}$, so $AC = \sqrt{74} - \frac{46}{\sqrt{74}} = \frac{28}{\sqrt{74}}$. Now $BC = \sqrt{AB^2 - AC^2} = \sqrt{4^2 - \frac{28^2}{74}} = \sqrt{\frac{16 \cdot 74 - 28^2}{74}} = \sqrt{\frac{1184 - 784}{74}} = \frac{20}{\sqrt{74}}$. Since $\angle ACB = 90^{\circ}, [ABC] = \frac{1}{2} \cdot BC \cdot AC = \frac{1}{2} \cdot \frac{20}{\sqrt{74}} \cdot \frac{28}{\sqrt{74}} = \frac{140}{37}$.