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Question
In a quadrilateral ABCD, CO and DO are the bisectors of ∠C and ∠D respectively. Prove that \[∠COD = \frac{1}{2}(∠A + ∠B) .\]
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Solution
\[ ∠COD = 180° - \left(∠OCD + ∠ODC \right)\]
\[ = 180°- \frac{1}{2}\left(∠C + ∠D \right)\]
\[ = 180°- \frac{1}{2}\left[ 360° - \left(∠A + ∠B \right) \right]\]
\[ = 180°- 180°+ \frac{1}{2}\left( ∠A + ∠B \right)\]
\[ = \frac{1}{2}\left( ∠A +∠B \right)\]
\[ = RHS\]
\[\text{ Hence proved } .\]
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