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.

We know that,

The diagonals of a rectangle are equal.

⇒ BD = AC = x (let)

In △ ABC,

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

By mid-point theorem,

⇒ PQ || AC and PQ = $\dfrac{1}{2}AC = \dfrac{x}{2}$ ….(1)

In △ ADC,

S and R are the mid-points of AD and CD respectively.

SR || AC and SR = $\dfrac{1}{2}AC = \dfrac{x}{2}$ …..(2)

From equation (1) and (2), we get :

PQ || SR and PQ = SR

In quadrilateral PQRS, one pair of opposite sides are equal and parallel to each other.

∴ PQRS is a parallelogram.

In △ BCD, Q and R are the mid-points of side BC and CD respectively.

By mid-point theorem,

⇒ QR || BD and QR = $\dfrac{1}{2}BD = \dfrac{x}{2}$ ….(3)

In △ BAD, P and S are the mid-points of side AB and AD respectively.

By mid-point theorem,

PS || BD and PS = $\dfrac{1}{2}BD = \dfrac{x}{2}$ …….(4)

From equations (1), (2), (3), (4), we get :

PQ = QR = SR = PS

Hence, proved that the quadrilateral PQRS is a rhombus.