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Question
S is any point on side QR of a ∆PQR. Show that: PQ + QR + RP > 2PS.
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Solution
Given: In ∆PQR, S is any point on side QR.
To show: PQ + QR + RP > 2PS
Proof: In ∆PQS,
PQ + QS > PS ...(i) [Sum of two sides of a triangle is greater than the third side]
Similarly, in ∆PRS,
SR + RP > PS ...(ii) [Sum of two sides of a triangle is greater than the third side]
On adding equations (i) and (ii), we get
PQ + QS + SR + RP > 2PS
⇒ PQ + (QS + SR) + RP > 2PS
⇒ PQ + QR + RP > 2PS ...[∵ QR = QS + SR]
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