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

Construction: Join AC

Proof: In ΔABC, P and Q are the mid points of AB and BC respectively.

Midpoint theorem: The line segment joining the mid-points of two sides of a triangle is parallel to the third side and equal to half the third side.

Therefore,

By Mid-point theorem

PQ || AC and PQ = AC (i)

Similarly,

By Mid-point theorem

RS || AC and RS = AC (ii)

From (i) and (ii), we get

PQ ‖ RS and PQ = RS

As we know a quadrilateral is a parallelogram, if one pair of opposite sides is parallel and equal.

Thus, PQRS is a parallelogram

AB = BC (Given)

Therefore,

P and Q are mid points of AB and BC respectively

PB = BQ

In ΔPBQ,

PB = BQ

Therefore,

As Equal sides have equal angles opposite to them.

∠BQP = ∠BPQ (iii)

In ΔAPS and ΔCQR,

1/2 AB = 1/2 BC

⇒ AP = CQ

1/2 AD = 1/2 CD

AS = CR

PS = RQ (Opposite sides of parallelogram are equal)

Therefore By SSS congruence rule,

APS CQR

(By c.p.c.t)

∠APS = ∠CQR (iv)

Now,

∠BPQ + ∠SPQ + ∠APS = 180°

∠BQP + ∠PQR + ∠CQR = 180°

Therefore,

∠BPQ + ∠SPQ + ∠APS = ∠BQP + ∠PQR + ∠CQR

∠SPQ = ∠PQR (v) [From (iii) and (iv)]

PS || QR and PQ is the transversal,

Therefore,

∠SPQ + ∠PQR = 180^o (Sum of adjacent interior an angles is 180^o)

∠SPQ + ∠SPQ = 180° [From (v)]

2 ∠SPQ = 180°

∠SPQ = 90°

Thus, PQRS is a parallelogram such that ∠SPQ = 90^o

Hence, PQRS is a rectangle.