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In a quadrilateral $A B C D$, $C O$ and $D O$ are the bisectors of $\angle C$ and $\angle D$ respectively. Prove that $\angle C O D=\frac{1}{2}(\angle A+\angle B)$
Given :
In quadrilateral ABCD, CO and DO are the bisectors of $\angle C$ and $\angle D$ respectively.
To do :
We have to prove that, $\angle C O D=\frac{1}{2}(\angle A+\angle B)$
Solution :
We know that,
Sum of the angles in a quadrilateral is 360°.
Therefore,
Form the figure
$∠COD + \frac{1}{2} ∠C + \frac{1}{2} ∠D = 180°$ (Sum of the angles in a triangle is 180°)
$∠COD = 180° -\frac{1}{2} ∠C - \frac{1}{2} ∠D$
$∠COD = 180° - \frac{1}{2} (∠C+∠D)$----------(1)
$∠A+∠B+∠C+∠D = 360°$
$∠C + ∠D = 360° - ∠A - ∠B$--------------(2)
Substituting (2) in (1)
$∠COD = 180° - \frac{1}{2}(360° - ∠A - ∠B)$
$∠COD = 180° - 180° + \frac{1}{2} ∠A + \frac{1}{2} ∠B$
$∠COD = \frac{1}{2}(∠A + ∠B)$.
Hence proved.
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