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Question

# State true or false:The side AC of a triangle ABC is produced to point E so that CE=12AC. 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. Hence, 3DF=EF

A
True
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B
False
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Solution

## The correct option is A TrueGiven: In △ABC, D is mid point of BC and DP∥BA∥CRTo Prove: 3 DF = EFIn △ABC, D is the midpoint of BC and DP are drawn parallel to BA. Therefore, P is the midpoint of AC.AP=PCNow, FA∥DP∥RC and APC is transversal such that AP = PC and FDR is the another transversal Hence, FD=DR .........(I) (by intercept theorem)EC=12AC=PCIn △EPD,C is the midpoint of EP and CR∥DP . R must be the midpoint of DE. Thus, DR=RE .....(II)Hence, FD=DR=RE (from (I) and (II))FE=3DF

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