In the given figure, AD is a median of ABC and E is the mid-points of AD. Also, BE produced meets AC in F. Then AF =
13AC
Draw DP∥EF
In ΔADP, E is the mid-point of AD and EF∥DP
∴ F is the mid-point of AP.
In ΔFBC, D is the mid-point of BC and DP∥BF.
∴ P is the mid-point of FC
Thus, AF = FP = PC
Hence, AF =13AC