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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 $\triangle$ACD,

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 $\triangle$ABC,

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 $\triangle$BCD,

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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