Question

ABCD is a rhombus and P, Q, R, S are the mid-
points of AB, BC, CD and DA respectively. Prove
that OPQRS is a rectangle​

Answer 1

Step-by-step explanation:

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 ∆DRS and ∆BPO,

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, ∆DRS = ∆BPQ by SAS congruence

condition.

RS PO by CPCT--- (i)

In ∆QCR and ∆SAP,

RC PA (Halves of the opposite sides of

the rhombus)

RCQ =4PAS (Opposite angles of the

rhombus)

CQ = AS (Halves of the opposite sides of

the rhombus)

Thus, ∆QCR = ∆SAP by SAS congruence

condition.

RQ SP by CPCT- -- (i)

Now,

In ∆CDB,

R and Q are the mid points of CD and BC

respectively.

QR II BD

also,

Pand S are the mid points of AD and AB

respectively.

PS II BD

QR II PSS

Thus, PGRS is a parallelogram.

also, PQR = 90°

Now,

In PQRS,

RS PO and RQ = SP from (i) and (i)

<Q = 90°

Thus, PQRS is a rectangle.

(<) is the angle