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Maharashtra State BoardSSC (English Medium) 9th Standard

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 PMPRPMPR=13. [Hint: DN || QM] - Geometry

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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]

Sum
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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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Chapter 5: Quadrilaterals - Practice Set 5.5 [Page 73]

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Balbharati Mathematics 2 [English] Standard 9 Maharashtra State Board
Chapter 5 Quadrilaterals
Practice Set 5.5 | Q 4 | Page 73

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