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

To Prove,

PQRS is a rectangle.

Construction,

Join AC and BD.

Proof:

In ΔDRS and ΔBPQ,

DS = BQ (Halves of the opposite sides of the rhombus)

∠SDR = ∠QBP (Opposite angles of the rhombus)

DR = BP (Halves of the opposite sides of the rhombus)

ΔDRS ≅ ΔBPQ [SAS congruency]

RS = PQ [CPCT] ............ (i)

In ΔQCR and ΔSAP,

RC = PA (Halves of the opposite sides of the rhombus)

∠RCQ = ∠PAS (Opposite angles of the rhombus)

CQ = AS (Halves of the opposite sides of the rhombus)

ΔQCR ≅ ΔSAP [SAS congruency]

RQ = SP [CPCT] ...........(ii)

Now,

In ΔCDB,

R and Q are the mid points of CD and BC respectively.

⇒ QR || BD

P and S are the mid points of AD and AB respectively.

⇒ PS || BD

⇒ QR || PS

PQRS is a parallelogram.

∠PQR = 90°

In PQRS,

RS = PQ and RQ = SP from (i) and (ii)

∠Q = 90°

PQRS is a rectangle.