imc

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