CAT 2025

Slot 2

QA

Question & SolutionQues & Sol

In $\triangle ABC$, 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 the question, it is given

$$AD:AT=4:3$$

So,

$$TD:AT=1:3$$

Let,

$$TD=x,\ AT=3x$$

Also,

$$BE:BT=5:4$$

So,

$$BT:TE=4:1$$

Let,

$$BT=4y,\ TE=y$$

Given, DF is parallel to BE

So, DF is parallel to TE

So,

$$\triangle\ ATE\sim\triangle\ ADF$$

(by A.A., as $\angle ATE=\angle ADF$ and $\angle AET=\angle AFD$)

So,

$$\dfrac{AT}{AD}=\dfrac{TE}{DF}$$

So,

$$\dfrac{3x}{x+3x}=\dfrac{y}{DF}$$

So,

$$DF=\dfrac{4y}{3}$$

Also, since DF is parallel to BE,

$$\triangle\ CFD\sim\triangle\ CBE$$

(by A.A., as $\angle CFD=\angle CEB$ and $\angle CDF=\angle CBE$)

So,

$$\dfrac{CD}{BC}=\dfrac{DF}{BE}$$

So,

$$\dfrac{CD}{BC}=\dfrac{\dfrac{4y}{3}}{5y}=\dfrac{4y}{15y}=\dfrac{4}{15}$$

So,

$$\dfrac{CD}{BD}=\dfrac{4}{11}$$

So,

Final Answer: BD : CD = 11 : 4