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.
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
ABC is a triangle.
D is midpoint of BC and DQ.
They're drawn parallel to BA.
Then, Q is midpoint of AC.
∴AQ = DC
∴ FA parallel to DQ||PC.
AQC, is a transversal so, AQ = QC and FDP also a transveral on them.
∴FD = DP .......(1) [ intercept theorem]
EC = 1/2 AC = QC
Now, triangle EQD, here C is midpoint of EQ and CP which is parallel to DQ.
And, P is midpoint of DE.
DP = PE..........(2)
Therefore, From (1) and (2)
FD = DP = PE
FD = 1/3 FE
(ii) From triangle PED we have PDIICR and C is the midpoint of PE therefore CR=1/2PD(1)
CR=1/4PB( from 1)