Question

In $\Delta ABC$, $AD$, is a median and $E$ is the mid point of $AD$. If $BE$ is produced to meet $AC$ in $F$, show that $AF=\frac{1}{3}AC$.

Answer 1

$AD$ is median $\Rightarrow BD=CD$

$E$ is mid point of $AD$

Draw a line $DG$, such that $DG||BF$

In $△ADG$

$EF||DG$ and $E$ is mid point of $AD$

$\therefore F$ is midpoint of $AG$ [converse of midpoint theorem]

$\Rightarrow AF=FG---1$

In $△BCF$

$D$ is midpoint of $BC$ and $DG||BF$

$\therefore FG=GC---2$

From $1$ and $2$

$\therefore AF=FG=GC$

$AC=AF+FG+GC$

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