Question:
In a quadrilateral $A B C D, C O$ and $D O$ are the bisectors of $\angle C$ and $\angle D$ respectively. Prove that $\angle C O D=\frac{1}{2}(\angle A+\angle B)$.
Solution:
$\angle \mathrm{COD}=180^{\circ}-(\angle \mathrm{OCD}+\angle \mathrm{ODC})$
$=180^{\circ}-\frac{1}{2}(\angle \mathrm{C}+\angle \mathrm{D})$
$=180^{\circ}-\frac{1}{2}\left[360^{\circ}-(\angle \mathrm{A}+\angle \mathrm{B})\right]$
$=180^{\circ}-180^{\circ}+\frac{1}{2}(\angle \mathrm{A}+\angle \mathrm{B})$
$=\frac{1}{2}(\angle \mathrm{A}+\angle \mathrm{B})$
= RHS
HEnce proved.
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