Question

In the adjoining figure, $ABC$ is a triangle having a median $AD$. If $E$ is mid-point of $AD$ and $BE$ meets $AC$ at $F$, then prove that $AF=\frac{1}{3}AC$.

Answer 1

Consider the problem

Given, $AD$ is the medium of $\Delta ABC.E$ is the mid-point of $AD.BE$ produced meets $AD$ at $F$.

To prove :

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

Construction :

From point $D$,draw $DG\parallel BF$

Proof :

In $\Delta ADG,E$ is the mid-point of $AD$ and $EF\parallel DG$.

$F$ is the mid-point of $AG$

$AF=FG$ ---- $\left(i\right)$

In $\Delta BCF,D$ is the mid-point of $BC$ of $BC$ and $DG\parallel BF$

$G$ is the mid-point of $CF$

$FG=GC$ $\left(ii\right)$

from $\left(i\right)$ and $\left(ii\right)$, we get

$AF=FG=GC$ ---- $\left(iii\right)$

Now,

$AF+FG+GC=AC$

$AF+AF+AF=AC$ using $\left(iii\right)$

$3AF=AC$

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

Hence prove,