Triangle $ABC$ has a right angle at $C$, $AC = 3$ and $BC = 4$.  Triangle $ABD$ has a right angle at $A$ and $AD = 12$.  Points $C$ and $D$ are on opposite sides of $\overline{AB}$.  The line through $D$ parallel to $\overline{AC}$ meets $\overline{CB}$ extended at $E$.  If $\frac{DE}{DB} = \frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers, then $m + n = $
[asy]
size(170);
defaultpen(fontsize(10pt)+linewidth(.8pt));
pair C=origin, A=(0,3), B=(4,0), D=(7.2,12.6), E=(7.2,0);
draw(A--C--B--A--D--B--E--D);
label("$A$",A,W);
label("$B$",B,S);
label("$C$",C,SW);
label("$D$",D,NE);
label("$E$",E,SE);
[/asy]
$ \textbf{(A)}\ 25\qquad\textbf{(B)}\ 128\qquad\textbf{(C)}\ 153\qquad\textbf{(D)}\ 243\qquad\textbf{(E)}\ 256 $
1. **Identify the given information and draw the diagram:**
   - Triangle \(ABC\) is a right triangle with \(\angle C = 90^\circ\).
   - \(AC = 3\) and \(BC = 4\).
   - Triangle \(ABD\) is a right triangle with \(\angle A = 90^\circ\).
   - \(AD = 12\).
   - Points \(C\) and \(D\) are on opposite sides of \(\overline{AB}\).
   - The line through \(D\) parallel to \(\overline{AC}\) meets \(\overline{CB}\) extended at \(E\).

2. **Calculate the hypotenuse \(AB\) of \(\triangle ABC\):**
   \[
   AB = \sqrt{AC^2 + BC^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5
   \]

3. **Determine the coordinates of points \(A\), \(B\), and \(C\):**
   - \(C\) is at the origin \((0, 0)\).
   - \(A\) is at \((0, 3)\).
   - \(B\) is at \((4, 0)\).

4. **Find the coordinates of point \(D\):**
   - Since \(\triangle ABD\) is a right triangle with \(\angle A = 90^\circ\) and \(AD = 12\), we can use the Pythagorean theorem to find \(BD\):
     \[
     BD = \sqrt{AB^2 + AD^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13
     \]
   - Since \(D\) is on the line extended from \(AB\), we can use the slope of \(AB\) to find the coordinates of \(D\). The slope of \(AB\) is:
     \[
     \text{slope of } AB = \frac{3 - 0}{0 - 4} = -\frac{3}{4}
     \]
   - Using the point-slope form of the line equation, the coordinates of \(D\) can be found by extending the line \(AB\) by a distance of 12 units from \(A\).

5. **Find the coordinates of point \(E\):**
   - Since \(DE\) is parallel to \(AC\), the slope of \(DE\) is the same as the slope of \(AC\), which is undefined (vertical line).
   - Therefore, \(E\) lies directly below \(D\) on the extended line \(CB\).

6. **Calculate the ratio \(\frac{DE}{DB}\):**
   - Since \(DE\) is parallel to \(AC\), the length of \(DE\) is the same as the vertical distance from \(D\) to the x-axis.
   - The coordinates of \(D\) are \((7.2, 12.6)\), so the length \(DE = 12.6\).
   - The length \(DB = 13\).

7. **Simplify the ratio \(\frac{DE}{DB}\):**
   \[
   \frac{DE}{DB} = \frac{12.6}{13} = \frac{63}{65}
   \]
   - Since \(63\) and \(65\) are relatively prime, the ratio is already in its simplest form.

8. **Calculate \(m + n\):**
   - Here, \(m = 63\) and \(n = 65\).
   - Therefore, \(m + n = 63 + 65 = 128\).

The final answer is \(\boxed{128}\)