Question

In $△ABC,D$ is the mid-point $BC$. If a line is drawn in such a way that it bisects $AD$ and $AD$ and $AC$ at $E$ and $X$ respectively, then prove that $\frac{EX}{BE}=\frac{1}{3}$

Answer 1

Let B be the origin and position vector (p.v.) of A and C be $\overline{O},\overline{C}$

$\Rightarrow p.v.ofD\frac{O+\overline{C}}{2}=\frac{\overline{C}}{2}$

Let $\frac{AX}{AC}=\frac{k}{1}$

$\Rightarrow p.v.ofX=\frac{\overline{a}+c\overline{k}}{k+1}$

And let $\frac{EX}{BE}=\frac{\lambda }{1}\Rightarrow p.v.ofE=\frac{(\frac{\overline{a}+c\overline{k}}{k+1}+O)}{\lambda +1}$

But E is also mid point of $AD,\Rightarrow$ p.v. of $E=\frac{\left(\overline{a}\right)+(\frac{\overline{c}}{2})}{2}$

$\Rightarrow \frac{\overline{a}}{2}+\frac{\overline{c}}{4}=\frac{\overline{a}+c\overline{k}}{(k+1)(\lambda +1)}$

$\Rightarrow \frac{1}{(k+1)(\lambda +1)}=\frac{1}{2}\Rightarrow (k+1)(\lambda +1)=2$...(1)

And $\frac{k}{(k+1)(\lambda +1)}=\frac{1}{4}\Rightarrow \frac{k}{2}=\frac{1}{4}\Rightarrow k=\frac{1}{2}$

From (1), $\left(\frac{3}{2}\right)(\lambda +1)=2\Rightarrow \lambda +1=\frac{k}{3}\Rightarrow \lambda =\frac{1}{3}$

$\Rightarrow \frac{EX}{BE}=\frac{1}{3}$