In $\triangle ABC$ points $D$ and $E$ lie on $BC$ and $AC$, respectively. If $AD$ and $BE$ intersect at $T$ so that $AT/DT=3$ and $BT/ET=4$, what is $CD/BD$? $\mathrm{(A) \ } \frac{1}{8} \qquad$ $ \mathrm{(B) \ } \frac{2}{9} \qquad$ $ \mathrm{(C) \ } \frac{3}{10} \qquad$ $ \mathrm{(D) \ } \frac{4}{11} \qquad$ $ \mathrm{(E) \ } \frac{5}{12}$
Solution
Draw $EF\parallel AD$ with $F$ on $BC$.
Since $TD\parallel EF$, $\triangle BDT\sim \triangle BFE$, so $BF:BD=EF:TD=BE:BT=(BT+TE):BT=5:4$. Hence $BD:DF=4:1$, i.e. $FD:BD=1:4$.
Since $EF\parallel AD$, $\triangle CEF\sim \triangle CAD$, so $CF:CD=EF:AD=(EF:TD)\cdot (TD:AD)=\frac54\cdot \frac{TD}{AT+DT}=\frac54\cdot \frac14=\frac5{16}$.
In $\triangle ABC$ points $D$ and $E$ lie on $BC$ and $AC$, respectively. If $AD$ and $BE$ intersect at $T$ so that $AT/DT=2$ and $BT/ET=3$, what is $CD/BD$?
$\mathrm{(A) \ } \frac{1}{8} \qquad$ $ \mathrm{(B) \ } \frac{2}{9} \qquad$ $ \mathrm{(C) \ } \frac{3}{10} \qquad$ $ \mathrm{(D) \ } \frac{4}{11} \qquad$ $ \mathrm{(E) \ } \frac{5}{12}$ 
