Question

The side AC of a triangle ABC is produced to point E so that CE = 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:

1. 3DF = EF
2. 4CR = AB

Answer 1

Consider the figure :

Here D is the midpoint of BC and DP is parallel to AB, therefore P is the midpoint of AC and 

PD = `[1]/[2]`AB

(i) Again from the triangle AEF, we have AF || PD || CR and AP = `[1]/[3]`AE

Therefore, DF = `[1]/[3]` EF or we can say that 3DF = EF.

Hence, it is shown.

(ii) From the triangle PED, we have PD || CR and C is the midpoint of PE, therefore, CR = `[1]/[2]`PD

Now,

PD = `1/2` AB

`1/2"PD" = 1/4`AB

CR = `1/4`AB

4CR = AB

Hence, it is shown.