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

Q : 3       ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.
 

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Given: ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively.

To prove: the quadrilateral PQRS is a rhombus.

Proof : 

In \triangleACD,     

S is the midpoint of DA.                (Given)

      R  is the midpoint of DC.               (Given)

  By midpoint theorem,

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

   In \triangleABC,

      P is the midpoint of AB.                (Given)

      Q  is the midpoint of BC.               (Given)

  By midpoint 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 the midpoint of BC.                (Given)

      R  is the midpoint of DC.               (Given)

  By midpoint theorem,

                        \small QR\parallel BD                  and      \small QR=\frac{1}{2}BD...................5

                         AC = BD.......................6(diagonals )

                            

From 2,  5 and 6, we get

                     PQ=QR 

Thus, a parallelogram whose adjacent sides are equal is a rhombus. Hence, PQRS is a rhombus.

 

 

 

 

 

 

 

   

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