Show that the bisectors of angles of a parallelogram form a rectangle.

Explanation

Let a parallelogram ABCD

As we know that `anglePAS=angleRAB=1/2anglePAB`____1

Similarly,`angleCBQ=angleQBA=1/2angleCBA`_____2

So, AD||BC then `angleOAB+angleCAB=180^@`

`1/2angleDAB+1/2angleCAB=90^@`

`angleRAB+angleRBA=180^@`

Now, In `triangleRAB`

`angleRAB+angleARB=180^@`

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