In a quadrilateral ABCD, CO and DO are the bisectors of ∠C and ∠D respectively.
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.