Given: seg PD is a median of ∆PQR. Point T is the midpoint of seg PD.
To Prove: PM/PR = 1/3
Construction: Draw seg DN ||seg QM such that P-M-N 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-QT-M]
∴ PR = PM + PM + PM [From (iii) ]
∴ PR = 3 PM
PM/PR = 1/3
