Question

Let $ABC$ be a triangle in which $AB=AC$ and let $I$ be its in-centre. Suppose $BC=AB+AI$. Find $\angle BAC$.

• A. ${60}^o$
• B. ${45}^o$
• C. ${75}^o$
• D. ${90}^o$

Answer 1

The correct option is D ${90}^o$

From figure, $\angle AIB={90}^o+\left(C/2\right)$.
In figure, extend $CA$ to $D$ such that $AD=AI$.
Let $CD=CB$ by hypothesis.
$\therefore \angle CDB=\angle CBD={90}^o-\left(C/2\right)$.
$\angle AIB=\angle ADB={90}^o+\left(C/2\right)+{90}^o-\left(C/2\right)={180}^o$
Hence $ADBI$ is a cyclic quadrilateral.
$\therefore \angle ADI=\angle ABI=\frac{B}{2}$
But $ADI$ is isosceles triangle,
since $C=B$
$4B={180}^o$
$B={45}^o$
$\therefore A=2B={90}^o$