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

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.

PQ || AC

⇒ PQ = $\dfrac{1}{2}$ AC …(1)

In △ADC,

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

SR || AC

⇒ SR = $\dfrac{1}{2}$ AC …(2)

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

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 …(6)

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

∴ PQRS is a parallelogram.

We know that,

Diagonals of rhombus intersect at right angles.

⇒ ∠EOF = 90°

In quadrilateral OERF,

ER || OF and EO || RF

∴ OERF is a parallelogram.

Opposite angles of a parallelogram are equal.

⇒ ∠EOF = ∠ERF = 90°

In parallelogram PQRS,

⇒ ∠QRS = ∠QPS = 90°

⇒ ∠PSR = ∠PQR = x (let)

∠QRS + ∠QPS + ∠PSR + ∠PQR = 360°

⇒ 90° + 90° + x + x = 360°

⇒ 180° + 2x = 360°

⇒ 2x = 360° - 180°

⇒ 2x = 180°

⇒ x = $\dfrac{180°}{2}$

⇒ x = 90°

⇒ ∠PSR = ∠PQR = 90°

Since, in parallelogram PQRS, opposite sides are equal and parallel and all the interior angles equal to 90°.

∴ PQRS is a rectangle.

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