Question

In the adjoining quadrilateral ABCD, AO and BO are the bisectors of ZA and D
B respectively. Prove that angle AOB 1 / 2 (<C +< D).
​

Answer 1

Answer:

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)

Hence,Proved.

Answer 2

Step-by-step explanation:

ANSWER

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)

Hence,Proved.