⚡ AZUCATION SHORTCUT:
Using $DF \parallel TE$: In $\triangle ADF$, $\frac{AT}{AD} = \frac{AE}{AF} = \frac{3}{4} \implies AF = \frac{4}{3}AE$.
In $\triangle CBE$ with $DF \parallel BE$, $\frac{CD}{CB} = \frac{CF}{CE}$.
By Mass Point Geometry or Similarity, $BD : CD = \mathbf{11 : 4}$.

Step 1: Analyze Ratio of AD and AT
Given $AD : AT = 4 : 3$. Since $T$ lies on $AD$, $TD = AD - AT$.
Therefore, $AT : TD = 3 : 1$.

Step 2: Use Similarity in $\triangle ADF$
In $\triangle ADF$, $TE \parallel DF$ (since $BE \parallel DF$). By Basic Proportionality Theorem:
$\frac{AE}{EF} = \frac{AT}{TD} = \frac{3}{1} \implies AE = 3x, EF = x$.
Also, $\frac{TE}{DF} = \frac{AT}{AD} = \frac{3}{4}$.

Step 3: Analyze ratio of BE and BT
Given $BE : BT = 5 : 4$. Let $BT = 4y$ and $BE = 5y$, then $TE = y$.
From Step 2, $DF = \frac{4}{3} TE = \frac{4}{3}y$.

Step 4: Use Similarity in $\triangle CBE$
In $\triangle CBE$, $DF \parallel BE$. Therefore, $\triangle CDF \sim \triangle CBE$:
$\frac{CD}{CB} = \frac{DF}{BE} = \frac{\frac{4}{3}y}{5y} = \frac{4}{15}$.
This means if $CD = 4$ units, then $CB = 15$ units.

Step 5: Find the final ratio
$BD = CB - CD = 15 - 4 = 11$ units.
Thus, $BD : CD = \mathbf{11 : 4}$.

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