Answer:

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,

AC and BD are joined.

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 by SAS congruence condition.

RS = PQ by 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 by SAS congruence condition.

RQ = SP by CPCT --- (ii)

Now,

In ΔCDB,

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

⇒ QR || BD

also,

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

⇒ PS || BD

⇒ QR || PS

Thus, PQRS is a parallelogram.

also, ∠PQR = 90°

Now,

In PQRS,

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

∠Q = 90°

Thus, PQRS is a rectangle.

Verified answer

Ste

Given In quadrilateral ABCD, P, Q, R and S are the mid-points of the sides AB, BC, CD and DA, respectively.Also, AC⊥BDTo prove PQRS is a rectangle.Proof Since, AC⊥BD∴ ∠COD=∠AOD=∠AOB=∠COB=90∘ ltBrgt In ΔADC, S and R are the mid-points of AD and DC respectively, then by mid-point theorem SR∣∣ACandSR=12AC ...(i)In ΔABC, P and Q are the mid-points of AB and BC respectively, then by mid-point theorem PQ∣∣ACandPQ=12AC ...(ii)From Eqs. (i) and (ii), PQ∣∣SRandPQ=SR=12AC ...(iii)Similarly, SP∣∣RQandSP=RQ=12BD ...(iv)Now, in quadrilateral EOFR, OE||FR,OF||ER∴ ∠EOF=∠ERF=90∘ [∵∠COD=90∘⇒∠EOF=90∘]...(v)So, PQRS is a rectangle. Hence proved.