Question

In the given figure, $OD$ is the bisector of $\angle AOC,OE$ is the bisector of $\angle BOC$ and $OD\perp OE$. Then the points $A,O$ and $B$ are collinear. Also $\angle AOC+\angle BOC={180}^o$.

• A. True
• B. False

Answer 1

The correct option is A True

Given:

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

$OE$ is the bisector of $\angle BOC$

$OD\perp OE$

Proof:

From given we have $\angle AOD=\angle DOC$ .... $\left(1\right)$ and $\angle BOE=\angle EOC$ .... $\left(2\right)$

$\angle DOC+\angle EOC={90}^o$ ...... $\left(a\right)$

Substituting from $1$ and $2$

$\angle AOD+\angle BOE={90}^o$ ...... $\left(b\right)$

$\left(a\right)+\left(b\right)$

$\angle AOD+\angle DOC+\angle EOC+\angle BOE={90}^o+{90}^o$

$\angle AOD+\angle DOC+\angle EOC+\angle BOE={180}^o$

$\therefore$ $\angle AOB={180}^0$ and $A,O$ and $B$ are Collinear.

$\angle AOD+\angle DOC=\angle AOC$ and $\angle BOE+\angle EOC=\angle BOC$

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