In the adjoining quadrilateral abcd , ao and bo are the bisectors of
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Answer:
Step-by-step explanation:
First way :-
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
Second Way: -
∠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
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