in a quadrilateral ABCD , AO and BO are the bisectors of angle A and angle B respectively. Prove that angle AOB = 1/2 (angle C+ angle D)

thnx..i got another way for finding the ans..but this one is easy

your welcome

Given, AO and BO are the bisectors of angle A and angle B respectively.

∴ ∠1 = ∠4 and ∠3 = ∠5 ... (1)

To prove: ∠2 =  (∠C + ∠D)

Proof:

In quadrilateral ABCD

∠A + ∠B + ∠C + ∠D = 360�

 (∠A + ∠B + ∠C + ∠D) = 180� ... (2)

Now in ΔAOB

∠1 + ∠2 + ∠3 = 180� ... (3)

equating (2) and (3), we get

∠1 + ∠2 + ∠3 =  ∠A +  ∠B +  (∠C + ∠D)

∠1 + ∠2 + ∠3 = ∠1 + ∠3 +  (∠C + ∠D)

∴ ∠2 = [∠C + ∠D]

Hence proved