Question

$E$ is the mid-point of the median $AD$ of $\Delta ABC$ and $BE$ is produced to meet $AC$ at $F$ .Show that $AF\frac{1}{3}AC.$

Answer 1

Given: A $\Delta ABC$ in which $E$ is the mid point of median $AB$ and $BE$ is produced to meet $AC$ at $F$.

Refer image,

To prove: $AF=\frac{1}{3}AC$

Construction: Draw $DG||BF$ intersecting $AC$ at $G$.

Proof: In $\Delta ADG,E$ is the mid-point of $AD$ and $EF||DG$

$\therefore AF=FG$........(1) [Converse of mid-point theorem]

In $\Delta FBC,D$ is the midpoint if $BC$ and $DG||BF$.

$\therefore FG=GC$...........(2) [Converse of mid-point theorem]

From equation (1) and (2), we get,

$AF=FG=GC$.........(3)

But, $AC=AF+FG+GC$

$\Rightarrow AC=AF+AF+AF$ (Using (3))

$\Rightarrow AC=3AF$

$\Rightarrow AF=\frac{1}{3}AC$

Hence proved.