Show that the quadrilateral formed by joining the mid-points of the pairs of adjacent sides of a rectangle is a rhombus.

Join BD.

By mid-point theorem,

The line segment joining the mid points of any two sides of a triangle is parallel to the third side and is equal to half of it.

In △ABC,

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

By mid-point theorem,

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

In △ADC,

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

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

From eq.(1) and (2), we have :

PQ = SR and PQ || SR …(3)

In △ABD,

Since, P and S are the mid-points of AB and AD respectively.

PS || BD

⇒ PS = $\dfrac{1}{2}$ BD …(4)

In △BCD,

Since, Q and R are the mid-points of BC and CD respectively.

QR || BD

⇒ QR = $\dfrac{1}{2}$ BD …(5)

From eq.(4) and (5), we have:

PS = QR and PS || QR …(6)

From eq.(3) and (6), we have:

In quadrilateral PQRS opposite sides are parallel and equal.

∴ PQRS is a parallelogram.

Given,

ABCD is a rectangle.

AB = CD and AD = BC

In △ASP and △BQP,

⇒ AP = BP (P is the mid-point of AB)

⇒ AS = BQ (As, S and Q are mid-points of equal sides AD and BC respectively)

⇒ ∠SAP = ∠QBP (Both equal to 90°)

∴ △ASP ≅ △BQP (By S.A.S axiom)

⇒ PS = PQ ….(7) (Corresponding parts of congruent traingles are equal)

From eq.(3), (6) and (7), we have:

PS = PQ = QR = SR

Since, all sides are equal and opposite sides are parallel,

∴ PQRS is a rhombus.

Hence, the quadrilateral formed by joining the mid-points of the pairs of adjacent sides of a rectangle is a rhombus.