In $\triangle ABC$ points $D$ and $E$ lie on $BC$ and $AC$, respectively. If $AD$ and $BE$ intersect at $T$ so that $\frac{AT}{DT}=3$ and $\frac{BT}{ET}=4$, what is $\frac{CD}{BD}$?




$\mathrm{(A) \ } \frac{1}{8} \qquad \mathrm{(B) \ } \frac{2}{9} \qquad \mathrm{(C) \ } \frac{3}{10} \qquad \mathrm{(D) \ } \frac{4}{11} \qquad \mathrm{(E) \ } \frac{5}{12}$
We use the square bracket notation $[\cdot]$ to denote area.
WLOG, we can assume $[\triangle BTD] = 1$. Then $[\triangle BTA] = 3$, and $[\triangle ATE] = 3/4$. We have $CD/BD = [\triangle  ACD]/[\triangle ABD]$, so we need to find the area of quadrilateral $TDCE$.
Draw the line segment $TC$ to form the two triangles $\triangle TDC$ and $\triangle TEC$. Let $x = [\triangle TDC]$, and $y = [\triangle TEC]$. By considering triangles $\triangle BTC$ and $\triangle ETC$, we obtain $(1+x)/y=4$, and by considering triangles $\triangle ATC$ and $\triangle DTC$, we obtain $(3/4+y)/x=3$. Solving, we get $x=4/11$, $y=15/44$, so the area of quadrilateral $TDEC$ is $x+y=31/44$.
Therefore $\frac{CD}{BD}=\frac{\frac{3}{4}+\frac{31}{44}}{3+1}=\boxed{\textbf{(D)} \frac{4}{11}}$