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Question
In a quadrilateral ABCD, AO and BO are bisectors of angle A and angle B respectively. Show that:
∠AOB = (∠C + ∠D)
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Solution
In a quadrilateral ABCD, AO and BO are the bisectors of ∠A and ∠B, respectively. We need to prove that:
∠AOB = ∠C + ∠D.
The sum of the interior angles of a quadrilateral is: ∠A + ∠B + ∠C + ∠D = 360∘.
Since AO and BO are the bisectors of ∠A\ and ∠B\, we can express: `angleAOB=(angleA)/2+(angleB)/2`
From the sum of the interior angles of the quadrilateral, rearrange to find ∠A+∠B
∠A + ∠B = 360∘ − (∠C + ∠D).
Now substitute ∠A+∠B into the expression for ∠AOB:
`angleAOB= (angleA)/2+(angleB)/2=(angleA+angleB)/2`
Replace ∠A + ∠B with 360∘ − (∠C + ∠D)
`angleAOB=(360°-(angleC+angleD))/2`
Simplify: `angleAOB = 180°-(angleC+angleD)/2`
∠AOB = ∠C + ∠D.
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