# 1. The side AC of triangle 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 meet AC at point P and EF at point R respectively. Prove that: 3DF=EF and 4CR=AB

The figure is:

In triangle ABC, D is the midpoint of BC and DP are drawn parallel to BA.

therefore, P is the mid-point of AC.

AP = PC

Now, FA DP RC and APC is transversal such that AP = PC and FDR is the another transversal on them.

FD = DR .........(i) [by intercept theorem]

EC= (1/2) AC = PC

in tr. EPD,

C is the midpoint of EP and CR DP.

R must be the midpoint of DE.

DR = RE.....(ii)

thus, FD = DR = RE [from (i) and (ii)]

hence, FD = (1/3) FE

or 3FD= FE.

2)

From the triangle PED we have PD||CR and C is the midpoint of PE therefore CR = PD/2

Now PD = AB/2 so PD/2 = AB/4 so CR = AB/4

so 4CR = AB

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