Question

Let $\overleftrightarrow{AB}$ and $\overleftrightarrow{CD}$ be straight lines intersecting at O. Let $\overrightarrow{OP}$ be the bisector of $\angle BOD$ and $\overrightarrow{OQ}$ be the bisector of $\angle AOC$. Prove that $Q,O,P$ are collinear.

Answer 1

$\overleftrightarrow{AB}$ and $\overleftrightarrow{CD}$ be straight lines intersecting at O.

$\angle AOC=\angle BOD$ ....... [Opposite angles] ........ $\left(i\right)$

$\overrightarrow{OP}$ bisects $\angle BOD$ ........ [Given]

$⟹\angle BOP=\angle POD$

$⟹\angle BOP=\frac{1}{2}\angle BOD$ ......... $\left(ii\right)$

$\overrightarrow{OQ}$ bisects $\angle AOC$ ....... [Given]

$⟹\angle COQ=\frac{1}{2}AOC$ ......... $\left(iii\right)$

Now, $\angle BOP+\angle BOC+\angle COQ$

$=\frac{1}{2}\angle BOD+\angle BOC+\frac{1}{2}\angle AOC$

$=\frac{1}{2}(\angle AOC+BOD)+\angle BOC$

$=\angle AOC+\angle BOC$ ........ From $\left(i\right)$

$={180}^o$ ..... [Since $AB$ is a straight line and $O$ is a point on $AB$]

$\therefore \angle BOP+\angle BOC+\angle COQ={180}^o$

$⟹\angle BOP,\angle BOC,\angle COQ$ form a linear pair.

Thus, $\angle POQ={180}^o$

Hence, $Q,O,P$ are collinear