Answer this ques.

Dear Student,
Given - In triangle ABC,  D is the midpoint of BC and DQ are drawn parallel to BA.

To Prove - FD=1/3FE
Proof -  In triangle ABC,  D is the midpoint of BC and DQ are drawn parallel to BA.

therefore, Q is the mid-point of AC.

AQ = QC

Now, FA DQ PC and AQC is transversal such that AQ = QC and FDP is the another transversal on them.

FD = DP  .........(i) [by intercept theorem]

EC= (1/2) AC = Qc

in tr. EQD, 

C is the midpoint of EQ and CP DQ.

P must be the midpoint of DE.

DP = PE.....(ii)

thus, FD = DP = PE [from (i) and (ii)]

hence, FD = (1/3) FE.

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