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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.

Q : 2 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: ABCD is a rhombus in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA . AC, BD are  diagonals.

To prove: the quadrilateral PQRS is a rectangle.

Proof: In \triangleACD,     

S is midpoint of DA.                (Given)

 R  is midpoint of DC.          (Given)

 By midpoint theorem,

  \small SR\parallel AC  and   \small SR=\frac{1}{2}AC...................................1

 In \triangleABC,

 P is midpoint of AB.            (Given)

 Q  is mid point of BC.           (Given)

 By mid point theorem,

 \small PQ\parallel AC  and   \small PQ=\frac{1}{2}AC.................................2

From 1 and 2,we get

 \small PQ\parallel SR  and   \small PQ=SR=\frac{1}{2}AC

Thus, \small PQ=SR and \small PQ\parallel SR

So,the quadrilateral PQRS is a parallelogram.

Similarly, in \triangleBCD,

Q is mid point of BC.                (Given)

 R  is mid point of DC.               (Given)

 By mid point theorem,

 \small QR\parallel BD 

So,  QN || LM ...........5

LQ || MN ..........6  (Since, PQ || AC)

From 5 and 6, we get

LMPQ is a parallelogram.

Hence, \small \angleLMN=\small \angleLQN   (opposite angles of the parallelogram)

But, \small \angleLMN= 90             (Diagonals of a rhombus are perpendicular)

so, \small \angleLQN=90

Thus, a parallelogram whose one angle is right angle,ia a rectangle. Hence,PQRS is a rectangle.

 

 

 

 

 

 

 

   

 

 

 

 

 

Posted by

seema garhwal

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