Math, asked by shikha9gndubey, 5 months ago

ABCD is a rhombus and P,Q,R,S are midpoints of the sides AB,BC,CD and DA respectively.Show that the quadrilateral PQRS is a rhombus.

Answers

Answered by aishitaaish2020
1

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.

Answered By

toppr

8345 Views

How satisfied are you with the answer?

This will help us to improve better

answr

Get Instant Solutions, 24x7

No Signup required

girl

More Questions by difficulty

EASY

MEDIUM

HARD

Related Questions to study

In the given figure , point T is in the interior of rectangle PQRS, prove that, TS

2

+TQ

2

=TP

2

+TR

2

Similar questions