Question

In the following figure, S is any point on the side QR of Δ PQR. Prove that PQ + QR + RP > 2 PS.

Answer 1

The sum of any two sides of a triangle is grater than the third side.
In $∆PQS\mathrm{and}∆PSR$, we have
PQ + QS > PS ..... (i)
PR + SR > PS ..... (ii)
Adding (i) and (ii), we get
PQ + PR + (QS + SR) > 2PS
Since QS + SR = QR, therefore
PQ + PR + QR > 2PS
Hence, PQ + QR + RP > 2PS.