Question

Ray OC bisects $\angle$AOB
Prove that:
$\angle$COD = $\frac{1}{2}$ [$\angle$BOD-$\angle$AOD]

Answer 1

Given: $\mathrm{Ray}\mathrm{OC}\mathrm{bisects}\angle \mathrm{AOB},\mathrm{i}.\mathrm{e}.,\angle \mathrm{BOC}=\angle \mathrm{AOC}$
To prove : $\angle \mathrm{COD}=\frac{1}{2}\left[\angle \mathrm{BOD}-\angle \mathrm{AOD}\right]$
We have,
$$\begin{gathered} \angle \mathrm{COD}=\angle \mathrm{COA}-\angle \mathrm{AOD} \\ \Rightarrow \angle \mathrm{COD}=\angle \mathrm{BOC}-\angle \mathrm{AOD}(\because \angle \mathrm{COA}=\angle \mathrm{BOC}) \\ \mathrm{Adding}\angle \mathrm{COD}\mathrm{to}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{have}, \\ 2\angle \mathrm{COD}=\angle \mathrm{BOC}+\angle \mathrm{COD}-\angle \mathrm{AOD} \\ \Rightarrow 2\angle \mathrm{COD}=\angle \mathrm{BOD}-\angle \mathrm{AOD} \\ \Rightarrow \angle \mathrm{COD}=\frac{1}{2}\left(\angle \mathrm{BOD}-\angle \mathrm{AOD}\right) \\ \mathrm{Hence},\mathrm{proved}. \end{gathered}$$