Question

In a quadrilateral ABCD,AO and BO are the bisectors of angle A and angle B.prove that angle AOB =half of (angle C+angle D)

Answer 1

in triangleAOB,
      A+B+AOB=180           A=1,B=2
AOB=180-(A+B)
A+B+C+D=360
1/2(A+B)+1/2(C+D)=180
1+2+!/2(C+D)=180
1/2(C+D)=180-(1+2)
                  =AOB
1/2(C+D)=ANG AOB
HENCE PROVED

Answer 2

∠a + ∠b = 180°
1/2 (∠a + ∠b) = 90°
in tri. AOB, ∠oab + oba +∠aob = 180°
now, ∠aob + 90° = 180° (∵ ∠oab + ∠oba = 90° )
so ∠aob = 180° - 90° = 90°
now, ∠d + ∠c = 180°
so, 1/2 (∠d + ∠c) = 90°
therefore,∠aob = 1/2(∠d + ∠c) = 90°
hence, proved