Question

In a quadrilateral ABCD, CO and DO are the bisectors of ∠C and ∠D respectively. Prove that $\angle COD=\frac{1}{2}(\angle A+\angle B).$

Answer 1

$$\begin{gathered} \angle \mathrm{COD}=180°-\left(\angle \mathrm{OCD}+\angle \mathrm{ODC}\right) \\ =180°-\frac{1}{2}\left(\angle \mathrm{C}+\angle \mathrm{D}\right) \\ =180°-\frac{1}{2}\left[360°-\left(\angle \mathrm{A}+\angle \mathrm{B}\right)\right] \\ =180°-180°+\frac{1}{2}\left(\angle \mathrm{A}+\angle \mathrm{B}\right) \\ =\frac{1}{2}\left(\angle \mathrm{A}+\angle \mathrm{B}\right) \\ =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$