Question

In the given figure, if the bisectors of exterior angle $\angle BCE$ and $\angle CBD$ at I, then $\angle BIC=$___

• A. ${90}^{\circ }-\frac{\angle A}{2}$
• B. ${90}^{\circ }+\frac{\angle A}{2}$
• C. $-{90}^{\circ }+\frac{\angle A}{2}$
• D. $-{90}^{\circ }-\frac{\angle A}{2}$

Answer 1

The correct option is A

${90}^{\circ }-\frac{\angle A}{2}$

We know, $\angle A+\angle B+\angle C={180}^{\circ }$...... (Sum of $\angle s$ of a $\Delta$)

$\therefore \angle B+\angle c=180-\angle A$.......(i)

Also, $\angle BCE={180}^{\circ }-\angle C$ and $\angle CBD=180-\angle B$

$\therefore \angle BCI=\frac{180-\angle c}{2}={90}^{\circ }-\frac{\angle c}{2}...$ (ii)

$\angle CBI=\frac{180-\angle B}{2}=90-\frac{\angle B}{2}...$ (iii)

$\underset{–}{In\Delta BCI}$

$\angle BIC+\angle BCI+\angle CIB={180}^{\circ }...$ (Sum of $\angle s$ of a $\Delta$)

$\therefore \angle BCI=180-(90-\frac{\angle c}{2})-(90-\frac{\angle B}{2})$... (from (ii) and (iii))

$=180-(180-\frac{\angle B}{2}-\frac{\angle c}{2})$

$\angle BIC=180-180+\left(\frac{\angle B+\angle C}{2}\right)$

$=\frac{{180}^{\circ }-\angle A}{2}$......from (i)

$=90-\frac{\angle A}{2}$