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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

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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.

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