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: Two cevians \(AD\) and \(BE\) meet at \(T\) with known split ratios. This is a textbook mass-point setup: load masses so \(T\) balances on both cevians, then \(BD:DC\) falls out as a mass ratio.

Step 1: Convert the given data to the splits at \(T\). \(AD:AT=4:3\) means \(AT:TD=3:1\). \(BE:BT=5:4\) means \(BT:TE=4:1\).

Step 2: Balance on cevian \(AD\): masses satisfy \(m_A\cdot AT=m_D\cdot TD\), so \(\dfrac{m_A}{m_D}=\dfrac{TD}{AT}=\dfrac14\) \(\cdots\) wait, use \(\dfrac{m_A}{m_D}=\dfrac{TD}{AT}=\dfrac{1}{3}\). Take \(m_A=\alpha\Rightarrow m_D=3\alpha\).

Step 3: Balance on cevian \(BE\): \(\dfrac{m_B}{m_E}=\dfrac{TE}{BT}=\dfrac14\Rightarrow m_E=4m_B\).

Step 4: Use that \(D\) lies on \(BC\) and \(E\) on \(AC\): \[ m_D=m_B+m_C,\qquad m_E=m_A+m_C. \] So \(3\alpha=m_B+m_C\) and \(4m_B=\alpha+m_C\). Subtracting, \(4m_B-3\alpha=\alpha-m_B\Rightarrow 5m_B=4\alpha\Rightarrow m_B=\tfrac{4\alpha}{5}\), then \(m_C=3\alpha-\tfrac{4\alpha}{5}=\tfrac{11\alpha}{5}\).

Step 5: On segment \(BC\), \(D\) divides it inversely to the masses: \[ \frac{BD}{DC}=\frac{m_C}{m_B}=\frac{11\alpha/5}{4\alpha/5}=\frac{11}{4}. \] (The parallel line \(DF\) is the geometric reason these splits stay consistent.)

Answer: \(BD:CD = 11:4\) (option 2).

Answer 2

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, \[ \frac{BD}{CD} = \frac{11}{4} \quad\Rightarrow\quad BD : CD = 11 : 4. \]