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

Explanation

Solution
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 `/_\BPQ`,
`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 = `PQ` by `CPCT` --- (i)
In `QCR` and `/_SAP`,
`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, `/_\QCR ≅ /_\SAP` 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.