Question

# Show that the bisector of angles of a parallelogram from a rectangle.

Solution

## Let a parallelogram $$ABCD$$we know that$$\angle PAS=\angle RAB=\dfrac{1}{2}< PAB$$..............(1)Similarly$$\angle CBQ=\angle QBA=\dfrac{1}{2}< CBA$$..............(2)So$$AD||BC$$ then$$\angle OAB+\angle CAB=180^0$$$$\dfrac{1}{2}\angle DAB+\dfrac{1}{2}\angle CAB=90^0$$$$\angle RAB+\angle RBA=180^0$$Now In $$\Delta le$$ $$RAB$$$$\angle RAB+\angle RBA_\angle ARB=180^0$$$$90^0+\angle ARB=180^0$$$$\boxed{\angle ARB=90^0}$$Similarly$$\boxed{\angle DPC=90^0}$$So opposite angles are $$90^0$$, soIt is a rectangle.The first line of solution should be$$\angle DAS = \angle RAB = \dfrac{1}{2} = \angle DAB$$Maths

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