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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 quadrilataral PQRS is a rhombus.

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 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.

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