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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.

 

 

 

 

 

 

 

   

 

 

 

 

 

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