Question

The median AD of the $\Delta ABC$ is bisected at E, BE meets AC in F, then AF: AC is equal to

• A. $\frac{3}{4}$
• B. $\frac{1}{3}$
• C. $\frac{1}{2}$
• D. $\frac{1}{4}$

Answer 1

The correct option is B $\frac{1}{3}$

Taking A as the origin let the position vectors of B and C be $\overrightarrow{b}$ and $\overrightarrow{c}$ respectively. Equations of lines BF and AC are $\overrightarrow{r}=\overrightarrow{b}+\lambda (\frac{\overrightarrow{b}+\overrightarrow{c}}{4}-\overrightarrow{b})$ and $\overrightarrow{r}=\overrightarrow{0}+\mu \overrightarrow{c}$ respectively. For the point of intersection F. We have
$\overrightarrow{b}+\lambda \left(\frac{\overrightarrow{c}-3\overrightarrow{b}}{4}\right)=\mu \overrightarrow{c}$
$\Rightarrow 1-\frac{3\lambda }{4}=0$ and $\frac{\lambda }{4}=\mu$
$\Rightarrow \lambda =\frac{4}{3}$ and $\mu =\frac{1}{3}$
Therefore, the position vector of $\overrightarrow{F}$ is $\overrightarrow{r}=\frac{1}{3}\overrightarrow{c}$
Now, $\overrightarrow{AF}=\frac{\overrightarrow{c}}{3}\Rightarrow \overrightarrow{AF}=\frac{1}{3}\overrightarrow{AC}$
Hence, $AF:AC=\frac{1}{3}:1=\frac{1}{3}$