Math, asked by poorvichaudhary004, 5 hours ago

please help me solve this Q

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Answered by IbrahimKhalil1

1

Given:In ∆PQR, S is any point on side QR

To show PQ+QR+RP>2PS

Proof:in ∆PQS, PQ+QS>PS....(1) [sum of two side of a triangle is greater than the third side]

Similarly,in ∆PRS, SR+RP>PS....(2)

add (1)+(2):

PQ+QS+SR+RP>2PS

PQ+(QS+SR)+RP>2PS [QR=QS=SR]

PQ+QR+RP>2PS

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