Question

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Answer 1

Step-by-step explanation:

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Answer 2

✪AnSwEr

${\tt{\red{\underline{\large{Given}}}}}$

• Quadrilateral ABCD
• AO and BO are yhe the bisector of angle A and Angle B

${\sf{\green{\underline{\large{To\:Prove}}}}}$

• $\tt\angle{AOB}=\frac{1}{2}(\angle{c}+\angle{D})$

${\sf{\pink{\underline{\Large{Explanation}}}}}$

We know that sum of the angle of quadrilateral is 360°

Therefore

A+B+C+D=360°

Now taking 1/2 at both side

=>1/2(A+B+C+D)=360° × 1/2

=>1/2(A+B+C+D)=180°

=>1/2(A+B)=180° - 1/2 (C+D)___(1)

Coming to the ∆ AOB

We know that sum of the angle of ∆s is 180°

Therefore

1/2(A+B)+AOB=180

¶utting the value of equation 1

=>1/2(A+B)+AOB=180°

=>180° - 1/2 (C+D)+AOB=180°

$\implies\tt\angle{AOB}=\frac{1}{2}(\angle{c}+\angle{D})$

Hence , proved