# ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle.

**Solution:**

We will use the mid-point theorem here. It states that the line segment joining the mid-points of any two sides of the triangle is parallel to the third side and is half of it.

In ΔABC, P and Q are the mid-points of sides AB and BC respectively.

∴ PQ || AC and PQ = 1/2 AC (Using mid-point theorem) ... (1)

In ΔADC,

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

∴ RS || AC and RS = 1/2 AC (Using mid-point theorem) ... (2)

From Equations (1) and (2), we obtain

PQ || RS and PQ = RS

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

Since the sides of a rhombus are equal, AB = BC

1/2 × AB = 1/2 × BC

PB = BQ (P and Q are the mid-points of sides AB and BC respectively)

∠QPB = ∠PQB (Sides opposite to equal angles are equal) ..... (3)

In ΔAPS and ΔCQR,

AP = CQ (P and Q are the mid-points of sides AB and BC respectively)

AS = CR (S and R are the mid-points of sides AD and CD respectively)

PS = QR (Opposite sides of a parallelogram are equal)

BY SSS congruency, ΔAPS ≅ ΔCQR

So, ∠APS = ∠CQR (By CPCT) .... (4)

Since AB is a straight line, ∠APS + ∠SPQ + ∠QPB = 180^{°}

Since BC is a straight line, ∠PQB + ∠PQR + ∠CQR = 180^{°}

∠APS + ∠SPQ + ∠QPB = ∠PQB + ∠PQR + ∠CQR

∠APS + ∠SPQ + ∠QPB = ∠QPB + ∠PQR + ∠APS (By equations (3) and (4))

∠SPQ = ∠PQR .... (5)

Since ∠SPQ and ∠PQR are interior angles on the same side of the transversal PQ, they form a pair of supplementary angles.

∠SPQ + ∠PQR = 180^{°}

2∠SPQ = 180^{°} [From (5)]

∠SPQ = 90^{° }

Clearly, PQRS is a parallelogram having one of its interior angles as 90°.

Hence, PQRS is a rectangle.

**☛ Check: **Class 9 Maths NCERT Solutions Chapter 8

**Video Solution:**

## ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle.

NCERT Maths Solutions Class 9 Chapter 8 Exercise 8.2 Question 2

**Summary:**

If ABCD is a rhombus and P, Q, R, S are mid-points of the sides AB, BC, CD, and DA respectively, then the quadrilateral PQRS is a rectangle.

**☛ Related Questions:**

- 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.
- ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see Fig. 8.30). Show that F is the mid-point of BC
- In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see the given figure). Show that the line segments AF and EC trisect the diagonal BD.
- Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.

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