`A B C D` is a kite having `A B=A D` and `B C=C D` . Prove that the figure formed by joining the mid-points of the sides, in order, is a rectangle.

Explanation

Given: A kite `ABCD` having `AB=AD` and `BC=CD`
`P`, `Q`, `R`, `S` are the midpoint of sides `AB`, `BC`, `CD` and `DA` respectively.
`PQ`, `QR`, `RS` and spare joined.
To prove: `PQRS` is a rectangle.
Proof: In `/_\ABC`, `P` and `Q` are the midpoints of `AB` and `BC` respectively.
`:.PQ` parallel `AC` and `PQ=1/2AC` ....(1)
In `/_\ADC`,
`R` and `S` are the midpoint of `CD` and `AD` respectively.
`:.RS` parallel `AC` and `RS=1/2AC` ....(2)
From (1) and (2) we have
`PQ` parallel `RS` and `PQ=RS`
Thus, in quadrilateral `PQRS`, a pair of opposite sides are equal and parallel.
So `PQRS` is a parallelogram.
Now, we shall prove that one angle of parallelogram `PQRS` it is a right angle.
Since `AB=AD`
`=>1/2AB=1/2AD`
`=>AP=AS` ........(3)
`=>∠1=∠2` ......(4)
Now, in `/_\PBO` and `/_\SDR` we have
`PB=SD` since `AD=AB`
`=>1/2AD=1/2AB`
`BQ=DR`
`:.PB=SD`
And `PQ=SR` since `PQRS` is a parallelogram.
So by `SSS` criterion of congruence, we have
`/_\PBQ~=/_\SOR`
`=>/_3=/_4` by `CPCT`
Now, `/_3+/_SPQ+/_2=180^@`
And `/_1+/_PSR+/_4=180^@`
`:./_3+/_SPQ+/_2=/_1+/_PSR+/_4`
`=>/_SPQ=/_PSR` since `/_1=/_2` and `/_3=/_4`
Now, transversal `PS` cuts parallel lines `SR` and `PQ` at `S` and `P` respectively.
`:./_SPQ+/_PSR=180^@`
`=>2/_SPQ=180^@`
`=>/_SPQ=180^@/2=90^@`
Thus, `PQRS` is a parallelogram such that `/_SPQ=90^@`
Hence, PQRS is a parallelogram.