In a \triangle A B C , points D and E are on the sides BC and AC , respectively. BE and AD intersect at point T such that \mathrm { AD } : A T = 4 : 3 , and \mathrm { BE } : \mathrm { BT } = 5 : 4 . Point F lies on AC such that DF is parallel to BE . Then, \mathrm { BD } : \mathrm { CD } is

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CAT 2025 Slot 2 QA · Question 21

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

A.
$9 : 4$
B.
$15 : 4$
C.
$11 : 4$
D.
$7 : 4$

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