In a square \\\\(ABCD,\\\\) \\\\(P,Q,R\\\\) and \\\\(S\\\\) are the mid-points of \\\\(AB,BC,CD\\\\) and \\\\(DA\\\\) respectively.\\\\(\\Rightarrow\\\\) \\\\(AB=BC=CD=AD\\\\) [ Sides of square are equal ]In \\\\(\ riangle ADC,\\\\)\\\\(SR\\parallel AC\\\\) and \\\\(SR=\\dfrac{1}{2}AC\\\\) [ By mid-point theorem ] ---- ( 1 )In \\\\(\ riangle ABC,\\\\)\\\\(PQ\\parallel AC\\\\) and \\\\(PQ=\\dfrac{1}{2}AC\\\\) [ By mid-point theorem ] ---- ( 2 )From equation ( 1 ) and ( 2 ),\\\\(SR\\parallel PQ\\\\) and \\\\(SR=PQ=\\dfrac{1}{2}AC\\\\) ---- ( 3 )Similarly, \\\\(SP\\parallel BD\\\\) and \\\\(BD\\parallel RQ\\\\)\\\\(\ herefore\\\\) \\\\(SP\\parallel RQ\\\\) and \\\\(SP=\\dfrac{1}{2}BD\\\\)and \\\\(RQ=\\dfrac{1}{2}BD\\\\)\\\\(\ herefore\\\\) \\\\(SP=RQ=\\dfrac{1}{2}BD\\\\)Since, diagonals of a square bisect each other at right angle.\\\\(\ herefore\\\\) \\\\(AC=BD\\\\)\\\\(\\Rightarrow\\\\) \\\\(SP=RQ=\\dfrac{1}{2}AC\\\\) ----- ( 4 )From ( 3 ) and ( 4 )\\\\(SR=PQ=SP=RQ\\\\)We know that the diagonals of a square bisect each other at right angles.\\\\(\\angle EOF=90^o.\\\\)Now, \\\\(RQ\\parallel DB\\\\)\\\\(RE\\parallel FO\\\\)Also, \\\\(SR \\parallel AC\\\\)\\\\(\\Rightarrow\\\\) \\\\(FR \\parallel OE\\\\)\\\\(\ herefore\\\\) \\\\(OERF\\\\) is a parallelogram.So, \\\\(\\angle FRE =\\angle EOF=90^o\\\\) (Opposite angles are equal)Thus, \\\\(PQRS\\\\) is a parallelogram with \\\\(\\angle R=90^o\\\\) and \\\\(SR=PQ=SP=RQ\\\\)\\\\(\ herefore\\\\) \\\\(PQRS\\\\) is a square.