Question

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

Answer 1

∠AOB =  1/2 * (∠C + ∠D)

Explanation:

Given : AO and BO are the bisectors of ∠A and ∠B respectively.

The picture is missing in the question. Please refer the attached picture.

∠1 = ∠4 and ∠3 = ∠5 ----- (1)

In the quadrilateral ABCD, we know that sum of angles is 360°

∠A + ∠B + ∠C + ∠D = 360°

1/2 * (∠A + ∠B + ∠C + ∠D) = 180°  

1/2 ∠A  + 1/2 ∠B + 1/2 (∠C + ∠D) = 180°  

∠1 + ∠3 + 1/2 (∠C + ∠D) = 180° ------ (2)

In ∆AOB, sum of angles is 180°

∠1 + ∠2 + ∠3 = 180° ------ (3)

Equation 2 and 3 both equate to 180°. So we get:

∠1 + ∠2 + ∠3 = ∠A + ∠B + 1/2 * (∠C + ∠D)

∠1 + ∠2 + ∠3 = ∠1 + ∠3 +  1/2 * (∠C + ∠D)

∠2 =  1/2 * (∠C + ∠D)

∠AOB =  1/2 * (∠C + ∠D)

Hence proved.