Question

In the figure the side AC of triangle ABC is produced to E such that CE =1/2AC. If D is the midpoint of BC and ED produced to meet AB at F and CP and DQ are parallel to AB prove that FD = 1/3 FE

Answer 1

Given,
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

Answer 2

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Alistair Dias asked in Math

The side AC of triangle ABC is produced to E such that CE=1/2 AC. If D is the midpoint of BC and ED produced to meet AB at F. CP and DQ are drawn parallel to BA. Prove that FD=I/3 FE. 

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Varun Rawat answered this
21668 helpful votes in Math, Class XII-Science

Dear Student,
Please find below the solution to the asked query :

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

To Prove - FD=1/3FEProof -  In triangle ABC,  D is the midpoint of BC and DQ are drawn parallel to BA.

therefore, Q is the mid-point of AC.

AQ = QC

Now, FA  DQ PC and AQC is transversal such that AQ = QC and FDP is the another transversal on them.

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

EC= (1/2) AC = Qc

in tr. EQD, 

C is the midpoint of EQ and CP  DQ.

P must be the midpoint of DE.

DP = PE.....(ii)

thus, FD = DP = PE [from (i) and (ii)]

hence, FD = (1/3) FE.