P, Q, R and S are the mid points of Rectangle ABCD.
Construction: Join AC and BD
To Prove: PQRS is a rhombus.
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