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.

In ΔABC,
P is mid-point to AB [∵ Given]
Q is mid-point of BC [∵ Given]
Hence, PQ ∥ AC and PQ = (1/2)AC … (1) [∵ Mid-point Theorem]

Similarly, in ΔACD,
S is mid-point of AD [∵ Given]
R is mid-point of CD [∵ Given]
Hence, SR ∥ AC and SR = (1/2) AC …(2) [∵ Mid-Point Theorem]

From (1) and (2), we have
PQ ∥ SR …(3) [∵ PQ ∥ AC and SR ∥ AC]
and PQ = SR …(4) [∵ SR = (1/2)AC and PQ = (1/2)AC]

Hence, PQRS is a parallelogram.

Similarly, in ΔBCD,
Q is mid-point of BC [∵ Given]
R is mid-point of CD [∵ Given]
Hence, QR = (1/2) BD …(5) [∵ Mid-point theorem]
Given that: AC = BD …(6) [∵ Diagonals of a rectangle are equal]
From (1), (5) and (6), we have
PQ = QR
A parallelogram whose adjacent sides are equal, is a rhombus. Hence, PQRS is a rhombus.