The side AC of a triangle ABC is produced to point E so that CE = ½ 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 meet AC at point P and EF at point R respectively. Prove that: (i) 3DF = EF(ii) 4CR = AB.
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Step-by-step explanation:
Given that D is the midpoint of BC and DP is parallel to AB, therefore P is the midpoint of AC
PD=
2
1
AB
From the triangle PED we have PD∥CR and C is the midpoint of PE therefore CR=
2
1
PD
PD=
2
1
AB
2
1
PD=
4
1
AB
CR=
4
1
AB
4CR=AB
Hence it is shown.
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