In a quadrilateral `A B C D ,\ C O\ a n d\ D O` are the bisectors of `/_C\ a n d\ /_D` respectively. Prove that `/_C O D=1/2(/_A+/_B)dot`

Explanation

To prove that \( \angle COD = \frac{1}{2}(\angle A + \angle B) \) in quadrilateral \( ABCD \) where \( CO \) and \( DO \) are the bisectors of angles \( C \) and \( D \) respectively, we can follow these steps:

### Step 1: Understand the Angles in Triangle COD
In triangle \( COD \), the sum of the angles is equal to \( 180^\circ \). Therefore, we can write:
\[ \angle COD + \angle 1 + \angle 2 = 180^\circ \]
where \( \angle 1 \) is half of angle \( D \) and \( \angle 2 \) is half of angle \( C \).
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