Q. 54.3( 34 Votes )

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Answer :

Given: S is any point on side QR of ΔPQR.

To Prove: PQ + QR + RP > 2PS

Proof:

We know, sum of two sides of triangle is greater than the third side

∴ In ΔPQS

PQ + QS > PS …………(1)

In Δ PSR

PR + SR > PS ……..(2)

Adding (1) and (2)

PQ + QS + PR + SR > PS + PS

⇒ PQ + QS + SR + PR > 2PS

⇒ PQ + QR + PR > 2PS [QR = QS + SR]

Hence, proved.

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