Question

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:

• A. \(15 : 4\)
• B. \(11 : 4\)
• C. \(9 : 4\)
• D. \(7 : 4\)

Hint:

When internal cevians intersect (like $AD$ and $BE$ meeting at $T$) and you need a side ratio, Menelaus’ Theorem on carefully chosen triangles with transversals through $T$ can be more direct than coordinate or mass-point geometry.

Answer 1

The Correct Option is B

Approach: Drop the synthetic geometry and use coordinates. Place the triangle conveniently, write $D$ and $E$ with unknown ratios, force $T$ to sit on both cevians with the given splits, and solve.

Step 1: Let $A=(0,0),\ B=(1,0),\ C=(0,1)$. Put $D$ on $BC$ with $BD:DC=t:(1-t)$, so $D=(1-t,\,t)$. Put $E$ on $AC$ with $AE:EC=s:(1-s)$, so $E=(0,\,s)$.

Step 2: $T$ is on $AD$ with $AT:TD=3:1$, i.e. $AT=\tfrac34 AD$: \[ T=A+\tfrac34(D-A)=\left(\tfrac34(1-t),\ \tfrac34 t\right). \]

Step 3: $T$ is also on $BE$ with $BT:TE=4:1$, i.e. $BT=\tfrac45 BE$: \[ T=B+\tfrac45(E-B)=\left(1-\tfrac45,\ \tfrac45 s\right)=\left(\tfrac15,\ \tfrac45 s\right). \]

Step 4: Match $x$-coordinates: $\tfrac34(1-t)=\tfrac15\Rightarrow 1-t=\tfrac{4}{15}\Rightarrow t=\tfrac{11}{15}$.

Step 5: Therefore $BD:DC=t:(1-t)=\tfrac{11}{15}:\tfrac{4}{15}=11:4$. (Matching $y$-coordinates gives $s=\tfrac{11}{16}$, consistent, but we already have the answer.)

Answer: $BD:CD = 11:4$.