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# ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA, respectively. Show that the quadrilateral PQRS is a rhombus. Given: ABCD is a rectangle.
$$\Rightarrow$$ $$\angle{A}$$ = $$\angle{B}$$ = $$\angle{C}$$ = $$\angle{D}$$ = $$90^\circ$$ and
AB = CD and BC = AD.

Also, P, Q, R and S are mid-points of the sides AB, BC, CD and DA, respectively.

Therefore, by midpoint theorem,

PQ || BD and PQ = $$\frac{1}{2}$$ BD and
SR || AC and SR = $$\frac{1}{2}$$ AC
In rectangle ABCD,
AC = BD
$$\therefore$$ PQ = SR ...(i)

Now, in $$\triangle{ASP}$$ and $$\triangle{BQP}$$, we have,
AP = BP ...(Given)
AS = BQ ...(Given)
$$\angle{A}$$ = $$\angle{B}$$ ...(Given)
$$\therefore$$ $$\triangle{ASP}$$ $$\displaystyle \cong$$ $$\triangle{BQP}$$ ...(By SAS rule)
$$\therefore$$ SP = BQ ...(ii)(By CPCT)

Similarly, in $$\triangle{RDS}$$ and $$\triangle{RCQ}$$, we have,
SD = CQ ...(Given)
DR = RC ...(Given)
$$\angle{C}$$ = $$\angle{D}$$ ...(Given)
$$\therefore$$ $$\triangle{RDS}$$ $$\displaystyle \cong$$ $$\triangle{RCQ}$$ ...(By SAS rule)
$$\therefore$$ SR = RQ ...(iii)(By CPCT)

Thus, from eq. (i), (ii) and (iii), we can say that,
The quadrilateral PQRS is a rhombus is proved.