Question

In the adjoining figure, seg PD is a median of ∆PQR. Point T is the midpoint of seg PD. Produced QT intersects PR at M. Show that \(\frac{PM}{PR}\) = .\(\frac{1}{3}\) [Draw DN || QM]

Answer 1

Given: seg PD is a median of ∆PQR. Point T is the midpoint of seg PD. 

To Prove: = \(\frac{PM}{PR}\) =  \(\frac{1}{3}\)

Construction: Draw seg DN ||seg QM such that P-MN and M-N-R.

Proof: 

In ∆PDN,

 Point T is the midpoint of seg PD and seg TM || seg DN [Given] 

∴ Point M is the midpoint of seg PN. [Construction and Q-T-M] 

∴ PM = MN [Converse of midpoint theorem] In ∆QMR,

 Point D is the midpoint of seg QR and seg DN || seg QM [Construction]

∴ Point N is the midpoint of seg MR. [Converse of midpoint theorem] 

∴ RN = MN …..(ii) 

∴ PM = MN = RN …..(iii) [From (i) and (ii)] 

Now, PR = PM + MN + RN [ P-M-R-Q-T-M]

∴ PR = PM + PM + PM [From (iii) ] 

∴ PR = 3PM 

 \(\frac{PM}{PR}\) = \(\frac{1}{3}\)