In a quadrilateral ABCD, AO and BO are bisectors of angle A and angle B respectively meeting at O. Prove that
angle AOB = half of [angle C+ angle D]

ABCD is a quadrilateral

To prove : ∠AOB=

2

1

(∠C+∠D)

AO and BO is bisector of A and B

∠1=∠2∠3=∠4...(1)

∠A+∠B+∠C+∠D=360

(Angle sum property)

2

1

(∠A+∠B+∠C+∠D)=180...(2)

In △AOB

∠1+∠3+∠5=

2

1

(∠A+∠B+∠C+∠D)

∠1+∠3+∠5=∠1+∠3+

2

1

(∠C+∠D)

∠AOB=

2

1

(∠C+∠D)

Explanation: In a quadrilateral ABCD. AO and BO are bisectors of angle A and angle B respectively. Prove that ∠AOB=

2

1

{∠C+∠D}.

won't that be plagrisim thought?

step-by-step explanation: