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 the line $ED$ is perpendicular to the line $AD$.  Segment $\overline{AE}$ intersects the circle at point $C$ between $A$ and $E$. What is the area of $\triangle ABC$?

$\textbf{(A) \ } \frac{120}{37}\qquad \textbf{(B) \ } \frac{140}{39}\qquad \textbf{(C) \ } \frac{145}{39}\qquad \textbf{(D) \ } \frac{140}{37}\qquad \textbf{(E) \ } \frac{120}{31}$
1. **Identify the given information and draw the diagram:**
   - The radius of the circle is \(2\), so the diameter \(AB\) is \(4\).
   - \(BD = 3\), so \(AD = AB + BD = 4 + 3 = 7\).
   - \(ED = 5\) and \(ED \perp AD\).

2. **Determine the coordinates of the points:**
   - Place \(A\) at \((0, 0)\) and \(B\) at \((4, 0)\).
   - Since \(D\) is on the extension of \(AB\), \(D\) is at \((7, 0)\).
   - \(E\) is directly above \(D\) by \(5\) units, so \(E\) is at \((7, 5)\).

3. **Find the coordinates of point \(C\):**
   - Since \(C\) is on the circle and on line \(AE\), we need to find the intersection of \(AE\) with the circle.
   - The equation of the circle centered at \((2, 0)\) with radius \(2\) is \((x-2)^2 + y^2 = 4\).
   - The line \(AE\) can be parameterized as \(y = \frac{5}{7}x\) since the slope of \(AE\) is \(\frac{5-0}{7-0} = \frac{5}{7}\).

4. **Solve for the intersection of \(AE\) and the circle:**
   - Substitute \(y = \frac{5}{7}x\) into the circle's equation:
     \[
     (x-2)^2 + \left(\frac{5}{7}x\right)^2 = 4
     \]
     \[
     (x-2)^2 + \frac{25}{49}x^2 = 4
     \]
     \[
     x^2 - 4x + 4 + \frac{25}{49}x^2 = 4
     \]
     \[
     \left(1 + \frac{25}{49}\right)x^2 - 4x + 4 = 4
     \]
     \[
     \frac{74}{49}x^2 - 4x = 0
     \]
     \[
     x\left(\frac{74}{49}x - 4\right) = 0
     \]
     \[
     x = 0 \quad \text{or} \quad x = \frac{4 \cdot 49}{74} = \frac{196}{74} = \frac{98}{37}
     \]
   - Since \(C\) is between \(A\) and \(E\), we discard \(x = 0\) and use \(x = \frac{98}{37}\).

5. **Find the corresponding \(y\)-coordinate of \(C\):**
   - Substitute \(x = \frac{98}{37}\) into \(y = \frac{5}{7}x\):
     \[
     y = \frac{5}{7} \cdot \frac{98}{37} = \frac{490}{259} = \frac{70}{37}
     \]

6. **Calculate the area of \(\triangle ABC\):**
   - Use the coordinates \(A(0, 0)\), \(B(4, 0)\), and \(C\left(\frac{98}{37}, \frac{70}{37}\right)\).
   - The area of \(\triangle ABC\) is given by:
     \[
     \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
     \]
     \[
     = \frac{1}{2} \left| 0(0 - \frac{70}{37}) + 4\left(\frac{70}{37} - 0\right) + \frac{98}{37}(0 - 0) \right|
     \]
     \[
     = \frac{1}{2} \left| 4 \cdot \frac{70}{37} \right|
     \]
     \[
     = \frac{1}{2} \left| \frac{280}{37} \right|
     \]
     \[
     = \frac{140}{37}
     \]

The final answer is \(\boxed{\frac{140}{37}}\)