Question

in a quadrilateral ABCD. AO and BO are the bisector in a quadrilateral ABCD A and B O are the bisector of angle A and Angle B respectively prove that angle AOB is equal to one upon two angle C + Angle B

Answer 1

According to the question,
Angle DAO= angle BAO= 1/2×angle A............1 and
Angle CBO = angle ABO= 1/2×angle B.............2

In triangle AOB
Angle BAO+ angle ABO +angle AOB = 180°
180° - (Angle BAO+ angle ABO)= angle AOB
180° -(1/2+angle A+ 1/2× angle B)= angle AOB...........3

In quadrilateral ABCD
Angle C +angle D= 360° - (angle A +angle B)
On dividing the equation by 2, we get
1/2( angle C + angle D) =180 - (1/2×angle A +1/2× angle B)
1/2( angle C + angle D) = angle AOB (from 3)

Hence proved.