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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
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Solution
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.
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