Question

Prove that the bisectors of a pair of vertically opposite angles are on the same straight line.

Answer 1

Let $AB$ and $CD$ be straight lines intersecting at $O$.

Also, let $OX$ be the bisector of $\angle AOC$ and $OY$ be the bisector of $\angle BOD$

$OY$ is the bisector of $\angle BOD$

$\therefore \angle 1=\angle 6$....(i)

$OX$ is the bisector of $\angle AOC$

$\therefore \angle 3=\angle 4$.....(ii)

$\angle 2$ and $\angle 5$ are vertically opposite angles.

$\therefore \angle 2=\angle 5$.....(iii)

We know that, sum of all angles $={360}^o$

$\therefore \angle 1+\angle 2+\angle 3+\angle 4+\angle 5+\angle 6={360}^o$

Using the relations from (i), (ii) and (iii), we get:

$\angle 1+\angle 2+\angle 3+\angle 3+\angle 2+\angle 1={360}^o$

$2\angle 1+2\angle 2+2\angle 3={360}^o$

$\Rightarrow \angle DOY+\angle AOD+\angle AOX={180}^o$

But, $\angle DOY+\angle AOD+\angle AOX=\angle XOY$

$\therefore \angle XOY={180}^o$

Since, $\angle XOY={180}^o$, both $OX$ and $OY$ are on the same straight line.$\left[Henceproved\right]$