Question

$D$ is the midpoint of the $lineAD$.

If a line is drawn through$B$, bisecting $AD$, which cuts $AD$ and $AC$ at $E$ and $X$ respectively,

then prove that $\frac{EX}{BE}=\frac{1}{3}$

Answer 1

Construct (Refer attached image)

A line through D parallel to BX such that it cuts AC at F.

Proof

Consider $△ADF$,

$\because AE=EDandEX\parallel DF$

By mid-point theorem, $2EX=DF$

$\therefore LetEX=a\&DF=2a$

Now consider $△BCX$,

$\because$ D is the mid-point of BC and $DF\parallel BX$,

$\therefore$ by mid-point theorem,

$\frac{BX}{DF}=2$

$\Rightarrow BX=2DF=4a$

$\therefore BF=BX-EX=3a$

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

Hence Proved