S is any point in the interior of ∆PQR. Show that SQ +SR<PQ + PR.
Given, S is any point in the interior of ∆PQR.
Produce QS to intersect PR at T.
In ∆PQT, we have PQ + PT > QT
[∵sum of any two sides of a triangle is greater than the third side]
=> PQ + PT >SQ+ST ,…(i)
In ∆TSR, we have
ST + TR >SR … .(ii)
On adding Eqs. (i) and (ii), we get
=>PQ + PT + ST + TR > SQ + ST + SR
=>PQ + PT + TR>SQ + SR
=>PQ + PR >SQ + SR Or
SQ+SR <PQ + PR