Question

If ABCD is a rectangle and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively, then quadrilateral PQRS is a rhombus.

Answer 1

ABCD is the given rectangle. Join AC. In ∆ABC, P and Q are the mid-points of sides AB and BC respectively.

∴ By mid-point theorem, 

PQ || AC and PQ = \(\frac12\) AC                         ...(i)

In ∆ADC, R and S are the mid-points of CD and AD respectively

∴ SR || AC and SR = \(\frac12\)AC                         ...(ii) 

From (i) and (ii), we get 

PQ || SR and PQ = SR ⇒ PQRS is a parallelogram 

ABCD is a rectangle, AD = BC ⇒ \(\frac12\)AD = \(\frac12\)BC ⇒ AS = BQ 

∴ In ∆s APS and BPQ 

AS = BQ                              (Proved above) 

AP = PB                              (P is mid-point of AB) 

∠SAP = ∠QBP = 90°

∴ ∆APS ≅ ∆BPQ                (SAS) 

⇒ PS = PQ 

∴ PQRS is a parallelogram with adjacent sides equal 

⇒ PQRS is a rhombus.