ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA, respectively. Show that the quadrilateral PQRS is a rectangle.

Given: ABCD is a rhombus and P, Q, R and S are mid-points of AB, BC, CD and DA.

By midpoint theorm, Now, in \(\triangle{ADC}\), we have,
Thus, SR || AC ...(i)
and SR = (\(\frac{1}{2} \) ) AC ...(ii)
Similarly, in \(\triangle{ABC}\), we have,
PQ || AC ...(iii)
and PQ = (\(\frac{1}{2} \) ) AC ...(iv)
from (i), (ii), (iii) and (iv), we get,
SR = PQ = (\(\frac{1}{2} \) ) AC and PQ || SR
\(\therefore \) Quadrilateral PQRS is a parallelogram.

Now, we know that diagonals of a rhombus bisect each other at right angles.
\(\therefore \) \(\angle{EOF}\) = \(90^\circ\).

Now, By midpoint theorem, we have,
RQ || BD
Thus, RE || OF
As SR || AC ...(from (i))
Thus, FR || OE

\(\therefore \) OERF is a parallelogram
So, \(\angle{ERF}\) = \(\angle{EOF} = 90° \)
(since, Opposite angle of a quadrilateral is equal)

Thus, PQRS is a parallelogram with \(\angle{R}= 90° \)
Hence, it is proved that PQRS is a rectangle.