If $$S$$ is any point in the interior of $$\Delta PQR$$, prove that $$(SQ + SR) < (PQ + PR)$$.

Solution



Let's project $$QS$$ to intresect $$PR$$ at $$T$$
From
 $$\triangle PQT$$, we have
$$PQ+PT>QT$$

$$PQ+PT>SQ+ST$$          ----------(1)
$$ST+TR > SR$$                     ----------(2)

Adding equation(1) and (2),
$$PQ+PT+ST+TR>SQ+ST+SR$$
$$PQ+PT+TR>SQ+SR$$
$$PQ+PR>SQ+SR$$

$$\Rightarrow SQ+SR<PQ+PR$$

Hence proved.