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

Let \[ \frac{BD}{CD} = x. \] Step 1: Convert the given ratios into segment ratios. From $AD : AT = 4 : 3$, \[ \frac{AD}{AT} = \frac{4}{3},\quad \Rightarrow\quad \frac{AT}{AD} = \frac{3}{4},\quad \frac{AT}{TD} = \frac{3}{1},\quad\Rightarrow\quad \frac{DT}{TA} = \frac{1}{3}. \] From $BE : BT = 5 : 4$, \[ \frac{BE}{BT} = \frac{5}{4},\quad \Rightarrow\quad \frac{BT}{BE} = \frac{4}{5},\quad \frac{BT}{TE} = \frac{4}{1},\quad\Rightarrow\quad \frac{ET}{TB} = \frac{1}{4}. \]
Step 2: Apply Menelaus’ Theorem in $\triangle ADC$ with transversal $B\!-\!T\!-\!E$. Line $BTE$ intersects: - $AD$ at $T$, - $AC$ at $E$, - and the extension of $CD$ at $B$. Menelaus’ Theorem for $\triangle ADC$ with transversal $BTE$: \[ \frac{AE}{EC} \cdot \frac{CB}{BD} \cdot \frac{DT}{TA} = 1. \] Now, \[ \frac{CB}{BD} = \frac{CD + BD}{BD} = \frac{CD}{BD} + 1 = \frac{1}{x} + 1 = \frac{1 + x}{x}, \quad \frac{DT}{TA} = \frac{1}{3}. \] Thus, \[ \frac{AE}{EC} \cdot \frac{1 + x}{x} \cdot \frac{1}{3} = 1 \quad\Rightarrow\quad \frac{AE}{EC} = 3 \cdot \frac{x}{1 + x} = \frac{3x}{1 + x}. \] So \[ \frac{EC}{AE} = \frac{1 + x}{3x}. \]
Step 3: Apply Menelaus’ Theorem in $\triangle CBE$ with transversal $A\!-\!T\!-\!D$. Line $ATD$ intersects: - $CE$ (extended) at $A$, - $BE$ at $T$, - and $BC$ at $D$. Menelaus’ Theorem for $\triangle CBE$ with transversal $ATD$: \[ \frac{CA}{AE} \cdot \frac{ET}{TB} \cdot \frac{BD}{DC} = 1. \] We have: \[ \frac{CA}{AE} = \frac{CE + AE}{AE} = \frac{CE}{AE} + 1 = \frac{1 + x}{3x} + 1 = \frac{1 + x + 3x}{3x} = \frac{4x + 1}{3x}, \] \[ \frac{ET}{TB} = \frac{1}{4},\qquad \frac{BD}{DC} = x. \] Substitute: \[ \left(\frac{4x + 1}{3x}\right) \cdot \frac{1}{4} \cdot x = 1. \]
Step 4: Solve for $x$. \[ \frac{x(4x + 1)}{12x} = 1 \quad\Rightarrow\quad \frac{4x + 1}{12} = 1 \quad\Rightarrow\quad 4x + 1 = 12 \quad\Rightarrow\quad 4x = 11 \quad\Rightarrow\quad x = \frac{11}{4}. \] Therefore, \[ \frac{BD}{CD} = \frac{11}{4} \quad\Rightarrow\quad BD : CD = 11 : 4. \]