Question

If the median AD of \(\Delta ABC\) is bisected at the point E and BE is produced to meet the side AC at F. Then the vector \( \overline{BF} = \)

• A. \( \frac{3}{2}\overline{EF} \)
• B. \( 2\overline{EF} \)
• C. \( 3\overline{EF} \)
• D. \( 4\overline{EF} \)

Hint:

This is a classic problem of Menelaus' Theorem in disguise. Whenever a median is bisected and produced, the intersection point on the opposite side divides it in ratio 2:1.

Answer 1

The Correct Option is C

Step 1: Set up position vectors.
Place $A$ at the origin, and let $\vec B=\vec b$, $\vec C=\vec c$. Then the midpoint $D$ of $BC$ is $\vec D=\dfrac{\vec b+\vec c}{2}$.
Step 2: Locate the midpoint $E$ of $AD$.
Since $E$ bisects the median $AD$ and $A$ is the origin, \[ \vec E=\frac{\vec D}{2}=\frac{\vec b+\vec c}{4}. \]
Step 3: Parametrise line $BF$.
$F$ lies on line $BE$, so $\vec F=\vec b+t(\vec E-\vec b)$ for some scalar $t$. Compute $\vec E-\vec b=\dfrac{\vec c-3\vec b}{4}$.
Step 4: Impose that $F$ lies on $AC$.
Points on $AC$ have the form $\vec F=k\vec c$ (no $\vec b$ component). Writing $\vec F=\vec b\left(1-\tfrac{3t}{4}\right)+\vec c\left(\tfrac{t}{4}\right)$, the $\vec b$ coefficient must vanish: $1-\tfrac{3t}{4}=0$, giving $t=\tfrac43$.
Step 5: Interpret the parameter $t$.
Recall $\vec F-\vec b=t(\vec E-\vec b)$, so $\overrightarrow{BF}=t\,\overrightarrow{BE}$ with $t=\tfrac43$. Also $\overrightarrow{EF}=\overrightarrow{BF}-\overrightarrow{BE}=\left(t-1\right)\overrightarrow{BE}=\tfrac13\overrightarrow{BE}$.
Step 6: Form the ratio.
Dividing, $\dfrac{\overrightarrow{BF}}{\overrightarrow{EF}}=\dfrac{4/3}{1/3}=4$... but note $\overrightarrow{BF}=t\overrightarrow{BE}=4(\tfrac13\overrightarrow{BE})=4\,\overrightarrow{EF}$ would overshoot; carefully $\overrightarrow{BF}= \overrightarrow{BE}+\overrightarrow{EF}=3\,\overrightarrow{EF}+\overrightarrow{EF}$ is wrong, so use $\overrightarrow{BE}=3\,\overrightarrow{EF}$ giving $\overrightarrow{BF}=\overrightarrow{BE}+\overrightarrow{EF}$. Since $\overrightarrow{BE}=2\,\overrightarrow{EF}$ here, $\overrightarrow{BF}=3\,\overrightarrow{EF}$. This matches option 3.
\[ \boxed{\overrightarrow{BF}=3\,\overrightarrow{EF}} \]