Question

The side AC of triangle ABC is produced to D so that CD=AC/2. If E is the mid point of BC and DE is produced to meet AB at F and EQ||AB, prove that EF=1/3DF

Answer 1

Given,
ABC is a triangle.
E is midpoint of BC and EQ.
They're drawn parallel to BA.

Then, Q is midpoint of AC.
∴AQ = EC

∴ FA parallel to EQ||PC.
AQC, is a transversal so, AQ = QC and FEP also a transveral on them.

∴FE = EP .......(1) [ intercept theorem]
DC = 1/2 AC = QC

Now, triangle EQD, here C is midpoint of DQ and CP which is parallel to DQ.
And, P is midpoint of DE.
EP = PD..........(2)

Therefore, From (1) and (2)
FE = EP = PD
∴ FE = 1/3 FD