1 The side AC of triangle ABC is produced to point E so that CE=1/2 AC D is the - Maths - Triangles

Question

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

Arpita D Pal · Accepted Answer

Dear Student, 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 $âˆ¥$ DP $âˆ¥$ 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 $âˆ¥$ 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 Regards

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