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Question
In the given figure, seg PD is a median of ΔPQR. Point T is the mid point of seg PD. Produced QT intersects PR at M. Show that `"PM"/"PR" = 1/3`.
[Hint: DN || QM]
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Solution
Given: seg PD is a median of ΔPQR. Point T is the mid point 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 ΔPDN,
Point T is the mid-point of seg PD. ...(Given)
seg TM || seg DN ...(Construction)
∴ Point M is the mid-point of seg PN. ...(Converse of mid-point theorem)[P-M-N]
∴ PM = MN ...(i)
In ΔQMR,
Point D is the mid-point of seg QR. ...(Given)
seg DN || seg QM. ...(Construction)
∴ Point N is the mid-point of seg MR. ...(Converse of mid-point theorem)[M-N-R]
∴ RN = MN ...(ii)
PM = MN = RN ...[From (i) and (ii)] ...(iii)
Now,
PR = PM + MN + RN
∴ PR = PM + PM + PM ...[From (iii)]
∴ PR = 3PM
∴ `"PM"/"PR" = 1/3`
Hence proved.
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