Triangle incenter relation

Question

Let $ABC$ be a triangle in which $AB = AC$ and let $I$ be its in-centre. Suppose $BC = AB + AI$. Find $∠BAC$. I do not see how to start even, please help.

Answer 1

hint: let $BA=BF \to AI=FI=FC$

Answer 2

We observe that $\angle AIB = 90^{\circ} + (C/2)$. Extend $CA$ to $D$ such that $AD = AI$. Then, $CD = CB$ by the hypothesis. Hence, $\angle CDB = \angle CBD = 90^{\circ} - (C/2)$.

Thus $\angle AIB + \angle ADB = 180^{\circ}$. Hence $ADBI$ is a cyclic quadrilateral. This implies that $\angle ADI = \angle ABI = B/2$.

But $ADI$ is isosceles, since $AD = AI$. This gives $\angle DAI = 180^{\circ} - 2(\angle ADI) = 180^{\circ} - \angle B$.

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

Thus we get $A = 2B = 90^{\circ}$.

Sorry can't use latex.

This question is from RMO - 2009: http://www.artofproblemsolving.com/Forum/resources.php?c=78&cid=48&year=2009&sid=48c6849690355f862898aeff37d1965c

Here are some solution other than mine: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1695934&sid=48c6849690355f862898aeff37d1965c#p1695934