Question

In Fig.6.9, OD is the bisector of $\angle AOC,OE$ is the bisector of $\angle BOC$ and $OD\perp OE$ Show that the points A, O and B are collinear.

Answer 1

Given: In figure, $OD\perp OE$ (i.e $\angle DOE={90}^0$), $OD$ and $OE$ are the bisector of $\angle AOC$ and $\angle BOC$.

To prove: points $A,O$ and $B$ are collinear i,e., $AOB$ is a straight line.

Proof: Since, $OD$ and $OE$ bisect angles $\angle AOC$ and

$\angle BOC$ respectively.

$\therefore \angle AOC=2\angle DOC$...........(1)

And $\angle COB=2\angle COE$............(2)

On adding equations (1) and (2), we get

$\angle AOC+\angle COB=2\angle DOC+2\angle COE$

$\Rightarrow \angle AOC+\angle COB=2(\angle DOC+\angle COE)$

$\Rightarrow \angle AOC+\angle COB=2\angle DOE$

$\Rightarrow \angle AOC+\angle COB=2\times {90}^0[\because OD\perp OE]$

$\Rightarrow \angle AOC+\angle COB={180}^0$

$\therefore \angle AOB={180}^0$

So, $\angle AOC+\angle COB$ are for forming linear pair or $AOB$ is a straight line. Hence, points $A,O$ and $B$ are collinear.