In ` A B C ,A D` is the median through `A` and `E` is the mid-point of `A D` . `B E` produced meets `A C` in `F` (Figure). Prove that `A F=1/3A Cdot`

Explanation

AD is the median of `/_\ABC \and\ E` is the midpoint of AD
Through `D \draw\ DG∣∣BF`
In `/_\ ADG` E is the midpoint of AD
`EF∣∣DG`
By converse of midpoint theorem we have,F is midpoint of AG and `AF=FG`...1
Similarly, in `/_\BCF`D is the midpoint of BC
`DG∣∣BF`
G is midpoint of CF
`FG=GC` ..............2
From equations 1 and 2
`AF=FG=GC`........3

`AF+FG+GC=AC`
`AF+AF+AF=AC`
`AF=1/3AC`