In △ABC , points D and E are on the sides BC and AC, respectively. BE and AD intrested at point T such that AD:AT=4:3, and BE:BT=5:4. Point F lies on AC such that DF is parallel to BE. Then, BD:CD is

The correct option is 11:4.

Step-by-Step Explanation:

Step 1: Understand the given ratios
We are given that point $T$ is the intersection of $AD$ and $BE$ in $\Delta ABC$.
The ratio of $AD:AT=4:3$. This implies:

$\frac{AT}{TD}=\frac{3}{4-3}=\frac{3}{1}$

Similarly, the ratio $BE:BT=5:4$. This implies:

$\frac{BT}{TE}=\frac{4}{5-4}=\frac{4}{1}$

Step 2: Apply Menelaus's Theorem on $\Delta ADC$
Consider $\Delta ADC$ intersected by the transversal line $B-T-E$.
By Menelaus's Theorem:

$\left(\frac{CB}{BD}\right)\times \left(\frac{DT}{TA}\right)\times \left(\frac{AE}{EC}\right)=1$

Note that $CB=CD+BD$, so:

$\frac{CB}{BD}=1+\frac{CD}{BD}$

Substituting $\frac{DT}{TA}=\frac{1}{3}$ into Menelaus's equation:

$\left(1+\frac{CD}{BD}\right)\times \frac{1}{3}\times \frac{1}{\left(EC/AE\right)}=1$

Rearranging this gives:

$\frac{EC}{AE}=\frac{1}{3}\left(1+\frac{CD}{BD}\right)$

Step 3: Apply Menelaus's Theorem on $\Delta BEC$
Now consider $\Delta BEC$ intersected by the transversal line $A-T-D$.
By Menelaus's Theorem:

$\left(\frac{EA}{AC}\right)\times \left(\frac{CD}{DB}\right)\times \left(\frac{BT}{TE}\right)=1$

Substitute $\frac{BT}{TE}=4$:

$\left(\frac{EA}{AC}\right)\times \left(\frac{CD}{DB}\right)\times 4=1$

$\frac{AC}{EA}=4\left(\frac{CD}{DB}\right)$

Since $AC=AE+EC$, we have $\frac{AC}{EA}=1+\frac{EC}{AE}$, which gives:

$\frac{EC}{AE}=4\left(\frac{CD}{DB}\right)-1$

Step 4: Solve for $BD:CD$
Equating the two expressions for $\frac{EC}{AE}$:

$\frac{1}{3}\left(1+\frac{CD}{BD}\right)=4\left(\frac{1}{BD/CD}\right)-1$

Let $x=\frac{CD}{BD}$. The equation becomes:

$\frac{1}{3}\left(1+x\right)=4x-1$

Multiply both sides by 3:

$1+x=12x-3$

$4=11x$

$x=\frac{4}{11}$

Therefore, the required ratio is:

$\frac{BD}{CD}=\frac{1}{x}=\frac{11}{4}$

Thus, $BD:CD=11:4$.