Given, a ABC in which AD is a median and E is mid-point of AD. Now, drawing DP||EF.
In ΔADP, E is mid-point of AD and EF||DP.
F is mid-point of AP.
In ΔFBC, D is mid-point of BC and DP||BF.
P is mid-point of FC.
Thus, AF = FP = PC
Hence, AF = 1/3 AC
Hence, proved.