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.
