Question

Side AC of a ABC is produced to point E so that CE = `(1)/(2)"AC"`. D is the mid-point of BC and ED produced meets AB at F. Lines through D and C are drawn parallel to AB which meets AC at point P and EF at point R respectively. Prove that: 3DF = EF

Answer 1

In ΔBDF and ΔDRC,
BD = DC    ...(D is the mid-point of BC)
CR || PD || AB
∠BFD = DRC  ...(alternate angles)
∠BDF = RDC  ...(vertivally opposite angles)
Therefore,
ΔBDF ≅ ΔDRC
⇒ DF = DR    .....(i)

In ΔABC,
D is the mid-point of BC and DP || AB
Therefore, P is the mid-point of AC.

In ΔDEP,
C is the mid-point of PE and DP || RC || AB   ...(CE = `(1)/(2)"AC"` and P is the mid-point of AC)
Therefore, R is the mid-point of DE.
⇒ DR = RE ......(ii)
But EF = DF + DR + RE
EF = DF + DF + DF
EF = 3DF.