Question

The median $\overrightarrow{AD}$ of $△ABC$ is bisected at $E$ and $\overrightarrow{BE}$ is producted to meet the side $\overrightarrow{AC}$ in $F.$ Then the ratio $\overrightarrow{AF}:\overrightarrow{FC}$ is

• A. $1:3$
• B. $1:4$
• C. $1:2$
• D. $3:1$

Answer 1

The correct option is C $1:2$

Let point $A$ be taken as origin and $\overrightarrow{AB}=\overrightarrow{b},\overrightarrow{AC}=\overrightarrow{c}$
$\therefore D=\frac{\overrightarrow{b}+\overrightarrow{c}}{2},E=\frac{\overrightarrow{b}+\overrightarrow{c}}{4}$
The equation of the line $\overrightarrow{AC}$ is $\overrightarrow{r}=t\overrightarrow{c}\cdots \left(i\right)$
and the equation of line $\overrightarrow{BE}$ is $\overrightarrow{r}=(1-s)\overrightarrow{b}+s\left(\frac{\overrightarrow{b}+\overrightarrow{c}}{4}\right)\cdots \left(ii\right)$
At the common point $F,$
$1-s+\frac{s}{4}=0$ and $t=\frac{s}{4}$
$\therefore s=\frac{4}{3},t=\frac{1}{3}$
Putting the value of $t$ in $\left(i\right),$ we get
$\overrightarrow{AF}=\frac{1}{3}\overrightarrow{AC}$
$\therefore \overrightarrow{AF}:\overrightarrow{AC}=1:2$