Question

Let $\overrightarrow{OX}$ be a ray and let$\overrightarrow{OA}$ and $\overrightarrow{OB}$ be two rays on the same side of $\overrightarrow{OX}$, with $\overrightarrow{OA}$ between $\overrightarrow{OX}$ and $\overrightarrow{OB}$. Let $\overrightarrow{OC}$ be the bisector of $\angle AOB$. Prove that $\angle XOA+\angle XOB=2\angle XOC$.

Answer 1

$OC$ is bisector of $AOB$

$\therefore \angle BOC=\angle AOC=x°$

To proof $\angle XOA+\angle XOB=2\angle XOC$

$L.H.S$

$\angle XOA+\angle XOA+\angle AOB$

$⟹2\angle XOA+2\angle AOC$

$⟹2(\angle XOA+\angle AOC)$

$⟹2\angle XOC$ $(R.H.S)$ $\left(Proved\right)$