In a $\triangle ABC$, points $D$ and $E$ are on the sides $BC$ and $AC$, respectively. $BE$ and $AD$ intersect 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

In $\triangle ABC$, we have $D$ on $BC$, $E$ on $AC$, and their cevians $AD$ and $BE$ intersect at $T$.

From $AD : AT = 4 : 3$, we get $AT : TD = 3 : 1$ since $TD = AD - AT = 4 - 3 = 1$.

From $BE : BT = 5 : 4$, we get $BT : TE = 4 : 1$ since $TE = BE - BT = 5 - 4 = 1$.

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Let the position vectors of $A$, $B$, $C$ be $\vec{A}$, $\vec{B}$, $\vec{C}$. Let $BD : DC = p : q$ and $AE : EC = s : r$.

Since $T$ divides $AD$ in the ratio $3:1$, and $D$ divides $BC$ in the ratio $p:q$:

$\vec{D} = \dfrac{q\vec{B} + p\vec{C}}{p + q}$

$\vec{T} = \dfrac{\vec{A} + 3\vec{D}}{4} = \dfrac{(p+q)\vec{A} + 3q\vec{B} + 3p\vec{C}}{4(p+q)}$

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Since $T$ divides $BE$ in the ratio $4:1$, and $E$ divides $AC$ in the ratio $s:r$:

$\vec{E} = \dfrac{r\vec{A} + s\vec{C}}{s + r}$

$\vec{T} = \dfrac{\vec{B} + 4\vec{E}}{5} = \dfrac{4r\vec{A} + (s+r)\vec{B} + 4s\vec{C}}{5(s+r)}$

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Since both expressions represent the same point $T$, and any point in the plane of $\triangle ABC$ has a unique representation as a weighted combination of $\vec{A}, \vec{B}, \vec{C}$ (with weights summing to $1$), the corresponding coefficients must be equal.

Matching the $\vec{A}$ coefficient:

$\dfrac{1}{4} = \dfrac{4r}{5(s+r)}$

$5(s+r) = 16r \implies 5s = 11r \implies \dfrac{s}{r} = \dfrac{11}{5}$

Matching the $\vec{B}$ coefficient:

$\dfrac{3q}{4(p+q)} = \dfrac{1}{5}$

$15q = 4p + 4q \implies 11q = 4p \implies \dfrac{p}{q} = \dfrac{11}{4}$

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Verifying with the $\vec{C}$ coefficient using $p = 11, q = 4, s = 11, r = 5$:

$\dfrac{3(11)}{4(15)} = \dfrac{33}{60} = \dfrac{11}{20}$

$\dfrac{4(11)}{5(16)} = \dfrac{44}{80} = \dfrac{11}{20} \quad \checkmark$

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$BD : CD = 11 : 4$