Question

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

Answer 1

Answer:

Step-by-step explanation:

I will be using 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.

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

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.

Answer 2

Answer:

QR

Step-by-step explanation:

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