ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. The quadrilateral PQRS is a rectangle. Is the given statement true or false?

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Given that, ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. The quadrilateral PQRS is a rectangleThe following figure depicts the given situation:The diagonals of the rhombus perpendicularly bisect each other.

⇒∠AOD=∠DOC=∠COB=∠BOA=
    
     90
    ∘

In

ΔADC
  , S is the midpoint of AD and R is the midpoint of DC. So, from midpoint theorem,

SR||AC and SR=
    1
    2
   
   AC
  ………. [1]In

ΔABC
  , P is the midpoint of AB and Q is the midpoint of BC. So, from midpoint theorem,

PQ||AC and PQ=
    1
    2
   
   AC
  ………. [2]From [1] and [2] we get,

PQ||SR and PQ=SR=
    1
    2
   
   AC
  Similarly,

PS||QR and PS=QR=
    1
    2
   
   BD
  Thus, we get that PQRS is a parallelogram. Now, the corresponding angles when SR||AC is,

∠KON=∠SND=
    
     90
    ∘

.From the figure linear pair is,

∠SNO=∠SND=
    
     90
    ∘

.Thus, sum of interior angles on the same side of transversal is,

∠SNO=∠KSN=
    
     90
    ∘

Hence,

∠PSR=
    
     90
    ∘

.A parallelogram with one of its angles as a right angle is a rectangle. Thus, PQRS is a rectangle. Hence, the statement is true.Therefore, the correct option is 1.

ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. The quadrilateral PQRS is a rectangle. Is the given statement true or false?

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