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.

Given:

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)

Thus, 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)

Thus, ΔQCR ≅ ΔSAP [SAS congruency]

RQ = SP [CPCT]———————- (ii)

Now,

In ΔCDB

R and Q are the midpoints of CD and BC, respectively.

⇒ QR || BD

also,

P and S are the midpoints of AD and AB, respectively.

⇒ PS || BD

⇒ QR || PS

Therefore, PQRS is a parallelogram.

Also, ∠PQR = 90°

In PQRS,

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

∠Q = 90°

Therefore, PQRS is a rectangle.