** Given : **ABCD is a rhombus in which P, Q, R and S are mid points of sides AB, BC, CD and DA respectively :

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

**Construction : **Join AC, PR and SQ.

**Proof :** In ∆ABC

P is mid point of AB [Given]

Q is mid point of BC [Given]

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

Similarly, in ∆DAC,

SR || AC and SR = 1 2 AC …(ii)

From (i) and (ii), we have PQ||SR and PQ = SR

⇒ PQRS is a parallelogram [One pair of opposite sides is parallel and equal]

Since ABQS is a parallelogram

⇒ AB = SQ [Opposite sides of a || gm]

Similarly, since PBCR is a parallelogram.

⇒ BC = PR

Thus, SQ = PR [AB = BC]

Since SQ and PR are diagonals of parallelogram PQRS, which are equal.

⇒ PQRS is a rectangle.** Proved.**