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A B C D is a rhombus and P ,\ Q ,\ R ,\ ...

` A B C D` is a rhombus and `P ,\ Q ,\ R ,\ S` are the mid-points of `A B ,\ B C ,\ C D ,\ D A` respectively. Prove that ` P Q R S` is a rectangle.

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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 `/_\CDB,
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.
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