The correct option is B 13FE
In ΔABC, D is the mid - point of BC and DQ∥BA.
∴ Q is the mid - point of AC.
[Converse of mid-point theorem]
⇒AQ=QC
Now, FA∥DQ∥PC, and AQC is the transversal such that AQ =QC and FDP is the other transversal on them.
∴ FD =DP ….(1) [By intercept theorem]
Now, EC=12AC=QC
∴ In EQD, C is the mid-point of EQ and CP∥DQ
∴ P must be the mid-point of DE.
∴ DP = PE
Thus, FD = DP = PE [From (1) and (2)]
Hence, FD=13FE