Question

In the adjoining figure, the side AC of ∆ABC is produced to E such that CE = 1/2 AC. If D is the midpoint of BC and ED produced meets AB at F, and CP, DQ are drawn parallel to BA, then FD is

• A. FD = 1/3 FE
• B. FD = 2/3 FE
• C. FD = 1/2 FE
• D. FD = 1/4 FE

Answer 1

The correct option is A

FD = 1/3 FE

In ∆ABC, D is the midpoint of BC and DQ || BA.

Therefore Q is the midpoint of AC.

Therefore AQ = AC.

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

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

Now EC = 1/2 AC = QC.

Therefore in ∆EQD, C is the midpoint of EQ and CP || DQ.

Therefore P must be the midpoint of DE.

Therefore DP = PE (ii)

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

Hence FD = 1/3 FE.