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.