Question

In Fig. PQR is a triangle and S is any point in its interior, show that SQ + SR < PQ +PR

Answer 1

Given : S is any point in the interior of $△$PQR
To Prove : SQ + SR < PQ + PR
Construction : Produce QS to meet PR in T.
Proof : In PQT, we have
PQ + PT > QT
[Sum of two sides of a triangle is greater than third side]
$\Rightarrow$ PQ + PT > QS + ST ....(1)
[QT = QS + ST]
In $△$RST, we have
ST + TR > SR ...(2)
Adding (1) and (2), we get
PQ+PT+ST+TR>SQ+ST+SR
$\Rightarrow$ PQ+(PT+TR)>SQ+SR
$\Rightarrow$ PQ + PR>SQ +SR $\Rightarrow$SQ+SR <PQ +PR.