Question

In $\Delta ABC$, AD is median through A and E is mid -point of AD.BE is produced to meet AC at F. Then prove that $AF=\frac{1}{3}AC$.

Answer 1

Given : AD is the median 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||BF.
Proof: In $\Delta ADG$, E is the mid-point of AD and EF||DG
$\therefore$ F is the mid point of AG [Converse of the mid point theorem]
$\Rightarrow AF=FG...\left(i\right)$
In $\Delta BCF,$D is the mid point of BC and DG||BF
$\therefore$ G is the mid point of CF
$\Rightarrow FG=GC...\left(ii\right)$
Form (i) and (ii), we get,
AF=FG=GC...(iii)
Now, AF+FG+GC=AC
$\Rightarrow AF+AF+AF=AC$[Using (iii)]
$\Rightarrow 3AF=AC$
$\Rightarrow AF=\frac{1}{3}AC$