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 $\frac{CD}{BD}$?
This question is from https://artofproblemsolving.com/wiki/index.php/2004_AMC_10B_Problems/Problem_20 and although I do understand the solutions I wanted to attempt one by applying menelaus to both $\triangle BCE$ and $\triangle CAD$. I almost reach the original solution but end up with a quadratic that gives me a wrong solution. Can someone kindly complete it with specifically applying menelaus to both that will give 2 equations containing the answer ratio and $\frac{CE}{EA}$.
P.S: The answer will contain BC and AC which can be simplified to known terms.
Best Answer
No quadratic equation arises. By Menelaus’ Theorem, we get:
$$\frac{AE}{EC}\cdot\frac{BC}{BD}\cdot \frac{DT}{AT}=1$$ $$\frac{BD}{DC}\cdot \frac{CA}{AE}\cdot \frac{TE}{BT}=1$$ By substituting values and assuming the required ratio as $p$, we get: $$\frac{1}{p}\cdot \frac{CE+EA}{AE} = 4$$ $$\frac{1}{p} \cdot (\frac{CE}{EA}+1)=4$$ $$\frac{AE}{EC} \cdot\frac{BD+CD}{BD} = 3$$ $$\frac{AE}{EC}(1+p)=3$$ This yields, $$\frac{1+p}{4p-1}=3$$ Hence $$1+p=12p-3$$ And $$p=\frac{4}{11}$$