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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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Answer:

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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