In the adjoining figure, the side AC of `triangle`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, prove that `FD =(1)/(3)FE.`

Explanation

In `triangle `ABC, D is the midpoint of BC and DQ||BA.
`therefore ` Q is the midpoint of AC.
`therefore AQ=QC.`
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 " triangle 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 " " `[form (i) and (ii) ].
Hence, FD` =(1)/(3)FE.`