Mathematics
S is any point on side QR of triangle PQR. Prove that :
PQ + QR + RP > 2PS

Triangles
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Answer
In △ PQS,
⇒ PQ + QS > PS ……..(1) [Sum of any two sides of a triangle is greater than the third side]
In △ PRS,
⇒ RP + RS > PS ……..(2) [Sum of any two sides of a triangle is greater than the third side]
Adding equations (1) and (2), we get :
⇒ PQ + QS + RP + RS > PS + PS
⇒ PQ + (QS + RS) + RP > 2PS
⇒ PQ + QR + RP > 2PS.
Hence, proved that PQ + QR + RP > 2PS.
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