Q5 of 45 Page 49

In figure 3.60, point S is any point on side QR of ΔPQR.

Prove that: PQ + QR + RP > 2PS


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.


More from this chapter

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