Question

If bisectors of angle A and B of a quadrilateral ABCD meet at O, Prove that Angle AOB = ½(Angle C+Angle D)

Answer 1

In any quadrilateral, sum of its four angles = 360° 

As such here in the quadrilateral ABCD also, <A + <B + <C + <D = 360° 

==> <C + <D = 360° - (<A + <B) 

2) Dividing the above by 2, 

(1/2)(<C + <D) = 180° - (1/2)(<A + <B) ------- (i) 

3) In the triangle AOB, <AOB = 180° - (1/2)*(<A + <B) ------- (ii) [Since given AO & Bo are bisectors of angles A & B respectively] 

4) Thus from (i) & (ii) above, 

<AOB = (1/2)(<C + <D) Proved

Answer 2

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)