In Δ ABC,

P and Q are mid points of AB and BC

By mid-point theorem,

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

In Δ ADC,

S and R are mid points of AD and DC.

By mid-point theorem,

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

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

⇒ SR = PQ and SR || PQ.

Since, one of the opposite pairs of PQRS are parallel and equal.

∴ PQRS is a parallelogram.

Let the diagonals AC and BD intersect at O.

In Δ BAC,

P and Q are mid points of AB and BC.

By mid-point theorem,

PQ || AC and PQ = $\dfrac{1}{2}AC$

∴ MQ || ON

In Δ BCD,

Q and R are mid points of BC and CD.

By mid-point theorem,

QR || BD and QR = $\dfrac{1}{2}BD$

∴ QN || OM

Since, opposite sides of quadrilateral OMQN are parallel.

∴ OMQN is a parallelogram

⇒ ∠MQN = ∠NOM (Opposite angles of parallelogram are equal)

We know that,

Diagonals of rhombus intersect at 90°

∴ ∠NOM = 90°

⇒ ∠PQR = ∠NOM = 90°

So, ∠PQR = 90°

In PQRS,

One pair of opposite side is parallel and equal and one of its interior angle is 90°.

Hence, proved that PQRS is a rectangle.

BRIGHT TUTORIALS

CBSE Class IX | Academic Year 2026-2027

9403781999

Excellence in Education