
1. Let $A B C$ be a triangle in which $A B=A C$ and let $I$ be its in-centre. Suppose $B C=A B+A I$. Find $\angle B A C$.


## Solution:

![](https://cdn.mathpix.com/cropped/2024_06_05_6f4d28fa51cab68f2750g-1.jpg?height=535&width=532&top_left_y=652&top_left_x=819)

We observe that $\angle A I B=90^{\circ}+(C / 2)$. Extend $C A$ to $D$ such that $A D=A I$. Then $C D=C B$ by the hypothesis. Hence $\angle C D B=\angle C B D=90^{\circ}-(C / 2)$. Thus

$$
\angle A I B+\angle A D B=90^{\circ}+(C / 2)+90^{\circ}-(C / 2)=180^{\circ}
$$

Hence $A D B I$ is a cyclic quadrilateral. This implies that

$$
\angle A D I=\angle A B I=\frac{B}{2}
$$

But $A D I$ is isosceles, since $A D=A I$. This gives

$$
\angle D A I=180^{\circ}-2(\angle A D I)=180^{\circ}-B
$$

Thus $\angle C A I=B$ and this gives $A=2 B$. Since $C=B$, we obtain $4 B=180^{\circ}$ and hence $B=45^{\circ}$. We thus get $A=2 B=90^{\circ}$.
