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

Given 

P, Q, R and S are the mid points of Rectangle ABCD.

Construction: Join AC and BD

To Prove: PQRS is a rhombus.

Proof

We will use the theorem: The line segment joining the mid-points of two sides of a triangle is parallel to the third side.

Considering taking ∆ACD

We observe that S and R are the mid points of side AD and DC respectively. [Given]

Hence, SR || AC and SR = ½ AC (NCERT Theorem 8.9)……………………….(1)

Now, taking ∆ACB

We observe that P and Q are the mid points of side AB and BC respectively. [Given]

Hence, PQ || AC and PQ = ½ AC (NCERT Theorem 8.9)…………………………(2)

So from (1) and (2), we conclude that

SR || PQ and SR = PQ

Hence, PQRS is a parallelogram. (If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram)

Now in ∆QBP and ∆QCR

QC = QB (Q is the mid point of BC)

RC = PB (opposite sides are equal, hence half length is also equal)

∠QCR = ∠QBP (Each 90°)

∆QBP ≅ ∆QCR (By SAS congruency)

QR = QP (By C.P.C.T.)

As PQRS is a parallelogram and having adjacent sides equal

Hence, PQRS is a rhombus

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