The side AC of a triangle ABC is produced to point E, so that CE = ½ AC. D is the
midpoint of BC and ED produced meets AB at F. Lines through D and C are drawn
parallel to AB which meet AC at point P and EF at point R respectively. Prove that:
(ii) 4CR = AB
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Answer:
The side AC of a triangle ABC is produced to point E so that CE=21AC. 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 meet AC at point R respectively. Prove that:
3DF=EF

Easy
Answer
Given that D is the midpoint of BC and DP is parallel to AB, therefore P is the midpoint of AC
PD=21AB
Again from the triangle AEF we have AE∥PD∥CR and AP=31AE
Therefore DF=31EF or we can say that 3DF=EF.
Hence it is shown.
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