S is any point on side QR of a ΔPQR. Show that: PQ + QR + RP > 2 PS.


Given: S is any point on side QR of a ΔPQR


In ΔPQS,


PQ + QS > PS (sum of two sides is greater than the third side) ...(1)


Similarly, In ΔPRS,


SR + RP > PS (sum of two sides is greater than the third side) ...(2)


Add (1) and (2)


PQ + QS + SR + RP > 2 PS


PQ + (QS + SR) + RP > 2 PS


PQ + QR + RP > 2 PS (as, QS + SR = QR)


Hence, proved.


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