In the adjoining figure, AD is a median of `triangle ABC` and E is the midpoint of AD. Also, BE produced meets AC in F. Prove that `AF=(1)/(3) AC.`

Explanation

Draw DP||EF.
In `triangle ADP`, E is the midpoint of AD and EF||DP.
`therefore ` F is the midpoint of AP, i.e., AF = FP.
In `triangle FBC`, D is the midpoint of BC and DP||BF. ltbr. `therefore ` P is the midpoint of FC, i.e., FP = PC.
Thus, AF = FP = PC.
Hence, AF = `(1)/(3)`AC.