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In a quadrilateral $ABCD, CO$ and $DO$ are the bisectors of $\angle C$ and $\angle D$ respectively. Prove that $\angle COD = \frac{1}{2}(\angle A + \angle B)$.
Given:
In a quadrilateral $ABCD, CO$ and $DO$ are the bisectors of $\angle C$ and $\angle D$ respectively.
To do:
We have to prove that $\angle COD = \frac{1}{2}(\angle A + \angle B)$.
Solution:
In $\Delta C O D$,
$\angle \mathrm{DCO}+\angle \mathrm{CDO}+\angle \mathrm{COD}=180^{\circ}$ (Sum of the angles in a triangle is $180^o$)
This implies,
$\frac{1}{2} \angle C+\frac{1}{2} \angle D+\angle C O D=180^{\circ}$
$\angle C O D=180^{\circ}-(\frac{1}{2} \angle C+\frac{1}{2} \angle D)$
$\angle COD =180^{\circ}-\frac{1}{2}(\angle C+\angle D)$
In quadrilateral $ABCD$,
$\angle A+\angle B+\angle C+\angle D=360^{\circ}$
$\angle C+\angle D=360^{\circ}-(\angle A+\angle B)$
$\angle C O D=180^{\circ}-\frac{1}{2}[360^{\circ}-\angle A+\angle B]$
$\angle C O D =180^{\circ}-180^{\circ}+\frac{1}{2}(\angle A+\angle B)$
$\angle C O D =\frac{1}{2}(\angle A+\angle B)$
Hence proved.
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