In $ \triangle ABC$ points $ D$ and $ E$ lie on $ \overline{BC}$ and $ \overline{AC}$, respectively. If $ \overline{AD}$ and $ \overline{BE}$ intersect at $ T$ so that $ AT/DT \equal{} 3$ and $ BT/ET \equal{} 4$, what is $ CD/BD$?
[asy]unitsize(2cm);
defaultpen(linewidth(.8pt));

pair A = (0,0);
pair C = (2,0);
pair B = dir(57.5)*2;
pair E = waypoint(C--A,0.25);
pair D = waypoint(C--B,0.25);
pair T = intersectionpoint(D--A,E--B);

label("$B$",B,NW);label("$A$",A,SW);label("$C$",C,SE);label("$D$",D,NE);label("$E$",E,S);label("$T$",T,2*W+N);

draw(A--B--C--cycle);
draw(A--D);
draw(B--E);[/asy]$ \textbf{(A)}\ \frac {1}{8}\qquad \textbf{(B)}\ \frac {2}{9}\qquad \textbf{(C)}\ \frac {3}{10}\qquad \textbf{(D)}\ \frac {4}{11}\qquad \textbf{(E)}\ \frac {5}{12}$
1. Draw \( \overline{CT} \) and let the area of \( \triangle ECT \) be 1. Then, the area of \( \triangle BTC \) is 4, since \( BT/ET = 4 \).
2. Let \( CD/BD = r \). We need to find \( r \).
3. The area of \( \triangle TDB \) is \( \frac{4}{r+1} \) because the ratio of \( CD \) to \( BD \) is \( r \), and the total area of \( \triangle BTC \) is 4.
4. The area of \( \triangle TDC \) is \( \frac{4r}{r+1} \) because the ratio of \( CD \) to \( BD \) is \( r \), and the total area of \( \triangle BTC \) is 4.
5. The area of \( \triangle ABT \) is \( 3 \times \text{area of } \triangle TDB = 3 \times \frac{4}{r+1} = \frac{12}{r+1} \) because \( AT/DT = 3 \).
6. The area of \( \triangle AET \) is \( \frac{1}{4} \times \text{area of } \triangle ABT = \frac{1}{4} \times \frac{12}{r+1} = \frac{3}{r+1} \) because \( BT/ET = 4 \).
7. Since \( \triangle ATC = 3 \times \text{area of } \triangle TDC \), we have:
   \[
   \frac{3}{r+1} + 1 = \frac{12}{r+1}
   \]
8. Multiplying both sides by \( r+1 \), we get:
   \[
   3 + r + 1 = 12r
   \]
9. Simplifying, we find:
   \[
   4 + r = 12r
   \]
10. Solving for \( r \), we get:
    \[
    11r = 4 \implies r = \frac{4}{11}
    \]

The final answer is \( \boxed{\frac{4}{11}} \).