**Given :** A rectangle ABCD in which P, Q, R, S are the mid-points of AB, BC, CD and DA respectively, PQ, QR, RS and SP are joined.

**To Prove :** PQRS is a rhombus.

**Construction :** Join AC

**Proof :** In ∆ABC, P and Q are the mid-points of the sides AB and BC.

∴ PQ || AC and PQ = 1 2 AC …(i) [Mid point theorem]

Similarly, in ∆ADC,

SR || AC and SR = 1 2 AC …(ii)

From (i) and (ii), we get

PQ || SR and PQ = SR …(iii)

Now in quadrilateral PQRS, its one pair of opposite sides PQ and SR is parallel and equal [From (iii)]

∴PQRS is a parallelogram.

Now AD = BC …(iv) [Opposite sides of a rectangle ABCD]

∴ 1/ 2 AD = 1/ 2 BC

⇒ AS = BQ

In ∆APS and ∆BPQ

AP = BP [∵ P is the mid-point of AB]

AS = BQ [Proved above]

∠PAS = ∠PBQ [Each = 90°]

∆APS ≅ ∆BPQ [SAS axiom]

∴ PS = PQ …(v)

From (iii) and (v), we have

PQRS is a rhombus **Proved.**