Question

The angle between the internal bisector and the external bisector of an angle is:

• A. ${60}^{\circ }$
• B. ${90}^{\circ }$
• C. ${120}^{\circ }$
• D. ${135}^{\circ }$

Answer 1

The correct option is B ${90}^{\circ }$

In $△ABC$, let $\angle ABD$ be the exterior angle.

Let ray $BE$ be bisector of interior $\angle ABC$ and ray $BF$ be bisector of exterior $\angle ABD$.

$\therefore \angle ABE=\angle EBC$ and $\angle ABF=\angle FBD$ let $\angle ABE=\angle EBC=x$ ...(1)

let $\angle ABF=\angle FBD=x$ ...(2)

We know, $\angle ABC+\angle ABD={180}^o$ ....angles in linear pair

$\therefore (\angle ABE+\angle EBC)+(\angle ABF+\angle FBD)={180}^o$ ....angle addition property

$\therefore (x+x)+(y+y)={180}^o$ ...from (1) & (2)

$\therefore 2x+2y={180}^o$

Dividing by $2$ on both sides we get,

$x+y={90}^o$

$\therefore \angle ABE+\angle ABF={90}^o$ ....from (1) & (2)