Question

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

Answer 1

Refer image

Given : $ABCD$ a parallelogram

To Prove : $PQRS$ is a rectangle

$DC||AB$ and $DA$ is a traversal

$\therefore \angle A+\angle D={180}^0$

$\frac{1}{2}\angle A+\frac{1}{2}\angle D={90}^0$

$\Rightarrow \angle SAD+\angle SDA={90}^0$

$\therefore \angle ASD={90}^0$ (by angle sum property)

$\therefore$ By concept of vertically opposite angles:

$\angle ASD=\angle PSR={90}^0$

Similarly, we will get

$\angle PQR=\angle QPS=\angle QRS={90}^0$

$\therefore PQRS$ is a rectangle