Q. 424.6( 5 Votes )
Given: S is any point on the side QR
To prove: PQ+QR+RP>2PS.
Since in a triangle, sum of any two sides is always greater than the third side.
So in ∆PQS, we have,
PQ + QS > PS …(1)
Similarly, ∆PSR, we have,
PR + SR > PS …(2)
Adding 1 and 2
PQ + QS + PR + SR > 2PS
PQ + PR + QR > 2PS …as PR = QS +SR
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