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Question
In the given figure, point S is any point on side QR of ΔPQR Prove that: PQ + QR + RP > 2PS
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Solution
In △PQS
PQ + QS > PS ...(1) (Sum of two sides of a traingle is greater than the third side)
In △PRS
RP + RS > PS ...(2) (Sum of two sides of a traingle is greater than the third side)
Adding (1) and (2), we get
PQ + QS + RP + RS > PS + PS
∴ PQ + (QS + SR) + PR > 2PS
∴ PQ + QR + RP > 2PS ...(Q - S - R)
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