Show that the quadrilateral formed by joining the mid-points of the consecutive sides of a rectangle is a rhombus.

Explanation

Solution
A rectangle `ABCD` in which `P,Q,R` and `S` are the mid-points of sides `AB,BC,CD` and `DA` respectively.
`PQ,QR,RS` and `SP` are joined
Join `AC`. In `/_\ABC,P` and `Q` are the mid-points of sides `AB` and `BC` = respectively.
∴`PQ∣∣AC` and `PQ= 1/2 AC` .....(i) (Mid pt. Theorem)
In `'/_\ADC,R'` and '`S`' are the mid-points of sides `CD` and `AD` respectively.
∴`'SR∣∣AC'` and `'SR`= `1/2`
`AC` ....(ii) (Mid. pt. Theorem)
From (i) and (ii), we have `PQ∣∣SR` and `'PQ=SR'`
⇒`'PQRS'` is a parallelogram.
Now `'ABCD'` is a rectangle.
∴`AD=BC`
⇒ `1/2 AD = 1/2 BC`
⇒`AS=BQ ` .....(iii)'
In `/_\S,APS` and `'BPQ'`, we have
`AP=BP` (`P` is the mid-point of `AB`)
`/_PAS=/_PBQ` (Each is equal to `90^@`)
and `AS=BQ` (From (iii))
∴`/_\APQ≅/_\BPQ` (SAS)
⇒`PS=PQ`
∴`PQRS` is a parallelogram whose adjacent sides are equal.
⇒`PQRS` is a rhombus.