# We get a rhombus

Let ABCD is a rectangle such as AB = CD and BC = DA

P, Q, R and S are the mid points of the sides AB, BC, CD and DA respectively

Construction: Join AC and BD

In

P and Q are the mid-points of AB and BC respectively

Therefore,

PQ AC and PQ = AC (Mid-point theorem) (i)

Similarly,

In

SR AC and SR = AC (Mid-point theorem) (ii)

Clearly, from (i) and (ii)

PQ SR and PQ = SR

Since, in quadrilateral PQRS one pair of opposite sides is equal and parallel to each other, it is a parallelogram.

Therefore,

PS QR and PS = QR (Opposite sides of a parallelogram) (iii)

In

Q and R are the mid-points of side BC and CD respectively

Therefore,

QR BD and QR = BD (Mid-point theorem) (iv)

However, the diagonals of a rectangle are equal

Therefore,

AC = BD (v)

Now, by using equation (i), (ii), (iii), (iv), and (v), we obtain

PQ = QR = SR = PS

Therefore, PQRS is a rhombus.

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