Question

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

Answer 1

Given: $S$ is any oint in the interior of $\Delta PQR$
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 $\Delta$ is greater than the third side]
$\Rightarrow PQ+PT>QS+ST$ .. (i)
$[QT=QS+ST]$
In $\Delta RST$, we have
$ST+TR>SR$ ... (ii)
Adding (i) and (ii), we get
$PQ+PT+ST+TR>SQ+ST+SR$
$\Rightarrow PQ+(PT+TR)>SQ+SR$
$\Rightarrow PQ+PR>SQ+SR$
$\Rightarrow SQ+SR<PQ+PR$.