Question

$\text{In figure, ‘S’ is any point on the side}QR\text{of}△PQR.$
Prove that $PQ+QR+RP>2PS.$

Answer 1

$\text{In}△PQS,\text{we have}$
PQ + QS > PS …(i)
[$\because$ sum of any two sides of a triangle is greater than the third side]
$\text{In}△PRS,\text{we have}$
RP + RS > PS …(ii)
Adding (i) and (ii), we have
PQ + (QS + RS) + RP > 2PS
$\text{Hence,}PQ+QR+RP>2PS.[\because QS+RS=QR]$