Question

AD is a median of triangle ABC and E is the midpoint of AD. BE produced meets AC in F, Prove that AF 1/3 AC

Answer 1

given:-

$AD$ is the median of $\Delta ABC$ and $E$ is the midpoint of $AD$

Through $D$ draw $DG||BF$

In $\Delta ADG$

$E$ is the midpoint of $AD$ and $EF||DG$

By converse of midpoint theorem we have

$F$ is midpoint of $AG$ and $AF=FG$ ..............$1$

Similarly, in $\Delta BCF$

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

$G$ is midpoint of $CF$ and $FG=GC$ ..............$2$

From equations $1$ and $2$

we will get

$AF=FG=GC........3$

$AF+FG+GC=AC$

$AF+AF+AF=AC$ ......... from eq $3$

$AF=AC$

$AF=\left(1/3\right)AC$