Prove that the quadrilateral formed by the bisectors of the angles of a parallelogram is a rectangle.

Explanation

Given Let ABCD be a parallelogram and AP, BR, CR, be are the bisectors of `angleA, angleB, angleC and angleD,` respectively.
To prove Quadrilateral PQRS is a rectangle.
Proof Since, ABCD is a parallelogram, then DC||AB and DA is a transversal.
We have, `" "angleA+angle D= 180^(@)`
`" "` [sum of cointerior angles of parallelogram is `180^(@)`]
`rArr" "(1)/(2)angleA+(1)/(2)angleD=90^(@)" "`[dividing both sides by 2]
`rArr" "anglePAD+anglePDA=90^(@)` ltBrgt `rArr" "angleAPD=90^(@)" "` [since, sum of all angles of a triangle is `180^(@)`]
`therefore" "angleSPQ=90^(@)" "` [vertically opposite angles]
Similarly, `" "anglePQR=90^(@)`
`" "angleQRS=90^(@)`
and `" "anglePSR=90^(@)`
Thus, PQRS is a quadrilateral whose each angle is `90^(@)`.
Hence, PQRS is a rectangle. `" "` Hence proved.