Question

In a $\Delta ABC$, AD is the median of $\angle BAC$ and E is the midpoint of AD. Prove that $AF=\left(\frac{1}{3}\right)AC$

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 $AF=FG\to \left(1\right)$
Similarly, in $\Delta BCF$
D is the midpoint of BC and DG || BF
G is midpoint of CF and $FG=GC\to \left(2\right)$
From equations (1) and (2), we get
$AF=FG=GC\to \left(3\right)$
From the figure we have, AF + FG + GC = AC
AF + AF + AF = AC [from (3)]
3 AF = AC
$AF=\left(\frac{1}{3}\right)AC$