RMO!

$\color{#D61F06}{\text{Motivation}}:$ Since we have been asked to find the value of an angle we will focus on angles and not sides.The first thing that comes to mind is the sine-rule.So we use that, $\text{In triangle ABC,we have,}\\ \dfrac{BC}{\sin A}=\dfrac{AB}{\sin B}\\ \Longrightarrow \dfrac{BC}{AB}=\dfrac{\sin A}{\sin B}$ .Let us keep that aside for now,and focus on the piece of information given to us, $BC=AB+AI\\ \Longrightarrow \dfrac{BC}{AB}=1+\dfrac{AI}{AB}$ ,we have divided by $AB$ since we know $\dfrac{BC}{AB}$ in terms of angles,now,if we can find the value of $\dfrac{AI}{AB}$ in terms of angles we can have an equation in terms of angles.Now,how do we find the value of $\dfrac{AI}{AB}$ ,that is the question.That can also be done with the help of sine-rule!just join $B$ and $I$ and apply it in the $\bigtriangleup{ABI}$ we have, $\dfrac{AI}{AB}=\dfrac{\sin{\dfrac{B}{2}}}{\cos{\dfrac{C}{2}}}=\tan{\dfrac{B}{2}}$ ,now,substituting all the values in the equation we get, $\dfrac{\sin{A}}{\sin{B}}=1+\tan{\dfrac{B}{2}}$ .Now,it would certainly be easier if the equation had just one variable which we can do by substituting $180^{\circ}-2B=A \Longrightarrow \sin{A}=\sin{2B}=2\sin B\cos B$ ,hence we have, $2\cos B=1+\tan{\dfrac{B}{2}}$ ,now,let us try and convert $\cos B$ into $\tan{\dfrac{B}{2}}$ ,we have, $\tan^2{\dfrac{B}{2}}=\dfrac{1-\cos{B}}{1+\cos{B}}\\ \text{Applying componendo and dividendo we have,}\\ \dfrac{1+\tan^2{\dfrac{B}{2}}}{(1-\tan{\dfrac{B}{2}})(1+\tan{\dfrac{B}{2}})}=\dfrac{1}{\cos{B}}\\ \Longrightarrow \dfrac{1+\tan^2{\dfrac{B}{2}}}{(1-\tan{\dfrac{B}{2}})(2\cos{B})}=\dfrac{1}{\cos{B}}\\ \Longrightarrow 1+\tan^2{\dfrac{B}{2}}=2-2(\tan{\dfrac{B}{2}})\\ \Longrightarrow 2(\tan{\dfrac{B}{2}})=1-\tan^2{\dfrac{B}{2}}\\ \Longrightarrow \tan{B}=1!\\ \Longrightarrow \angle B=45^{\circ}\\ \Longrightarrow \angle A=90^{\circ}$ .And done!

Since $\Delta ABC$ is isosceles , $\angle B = \angle C \Rightarrow \pi=\angle A+2\angle C \Rightarrow \angle C=\dfrac{\pi- A}{2}$ .

We know that $BC=2R\sin A \ , \ AB=2R\sin C \ , \ AI=4R\sin\dfrac{B}{2}\sin\dfrac{C}{2}$ . Thus , by given condition ,

$\begin{aligned}&\Rightarrow 2R\sin A = 2R\sin C + 4R\sin\dfrac{B}{2}\sin\dfrac{C}{2} \\ &\Rightarrow \sin A = \sin C + 2\sin^2 \dfrac{C}{2} \\ &\Rightarrow \sin A =\sin\left(\dfrac{\pi- A}{2}\right) + \sin^2\left(\dfrac{\pi- A}{4}\right)\\ &\Rightarrow \sin A =\sin\left(\dfrac{\pi}{2}-\dfrac{ A}{2}\right) + \sin^2\left(\dfrac{\pi-A}{4}\right) \\ &\Rightarrow - \sin^2\left(\dfrac{\pi- A}{4}\right) = \cos\dfrac{ A}{2} - \sin A \\ &\Rightarrow 1 - \sin^2\left(\dfrac{\pi- A}{4}\right) = 1+\cos\dfrac{ A}{2} - \sin A \\ &\Rightarrow \cos\left(\dfrac{\pi- A}{2}\right) = 1+\cos\dfrac{ A}{2}-2\sin\dfrac{ A}{2}\cos\dfrac{ A}{2}\\ &\Rightarrow \sin\dfrac{ A}{2}= 1+\cos\dfrac{ A}{2}-2\sin\dfrac{ A}{2}\cos\dfrac{ A}{2}\\ &\Rightarrow \sin\dfrac{ A}{2} -\cos\dfrac{ A}{2}+2\sin\dfrac{ A}{2}\cos\dfrac{ A}{2}=1 \\ &\Rightarrow \sin\dfrac{ A}{2} -\cos\dfrac{ A}{2}+2\sin\dfrac{ A}{2}\cos\dfrac{ A}{2} = \sin^2\dfrac{ A}{2} +\cos^2\dfrac{ A}{2} \\ &\Rightarrow \sin\dfrac{ A}{2} -\cos\dfrac{ A}{2}= \left(\sin\dfrac{ A}{2} -\cos\dfrac{ A}{2}\right)^2 \\ &\Rightarrow \left(\sin\dfrac{ A}{2} -\cos\dfrac{ A}{2}\right)\left(\sin\dfrac{ A}{2} -\cos\dfrac{ A}{2}-1\right) =0 \end{aligned}$

Case 1:

$\sin\dfrac{ A}{2} -\cos\dfrac{ A}{2}=1\Rightarrow \left(\sin\dfrac{ A}{2} -\cos\dfrac{ A}{2}\right)^2=1 \Rightarrow 1-2\sin\dfrac{A}{2}\sin\dfrac{A}{2}=1 \Rightarrow \sin A=0 \Rightarrow A=180^\circ n \ \forall n \in \mathbb{Z}$

This case must be discarded since , there is no solution in $0^\circ < A < 180^\circ$

Case 2:

$\sin\dfrac{ A}{2} -\cos\dfrac{ A}{2}=0\Rightarrow \sin\dfrac{ A}{2} = \cos\dfrac{ A}{2} \Rightarrow \dfrac{A}{2}=45^\circ \Rightarrow A=90^\circ \quad \because 0^\circ< A < 180^\circ$

Considering all the cases we have $\boxed{\angle A=\angle BAC=90^\circ}$

Let M the midpoint of BC. Then, AI/IM =2c/a (Angle Bisector Theorem) or AI/AM=2c/(a+2c). But AM=√(c²-a²/4); hence, AI=(√(c²-a²/4))(2c)/(a+2c) or AI= a -c =(√(c²-a²/4))(2c)/(a+2c) which simplifies to a²=2c². But b=c, hence a²=b²+c² or tr. ABC is a rt. angled triangle with angle A=90°

$Since\quad we\quad are\quad asked\quad for\quad the\quad angle\quad not\quad the\quad sides,\quad \\ WLOG,\quad let\quad AB=AC=1,\quad AI=x\\ \\ Then,\quad let\quad the\quad inscribed\quad circle\quad centred\quad at\quad I\quad with\quad radius\quad y\quad touch\quad AB,\quad AC\quad and\quad BC\quad at\quad M,\quad N\quad and\quad R\quad respectively.\\ Given\quad that\quad BC=1+x,\quad by\quad properties\quad of\quad isos.\quad triangle,\quad we\quad have\quad \\ BR=RC=\frac { 1+x }{ 2 } \\ Besides,\quad by\quad tangents\quad from\quad external\quad point,\quad we\quad have\\ BM=BR=\frac { 1+x }{ 2 } \quad and\quad AM=1-\frac { 1+x }{ 2 } =\frac { 1-x }{ 2 } \\ Consider\quad \triangle AMI,\quad using\quad Pyth.\quad Theorem,\quad we\quad have\\ { y }^{ 2 }+{ \left( \frac { 1-x }{ 2 } \right) }^{ 2 }={ x }^{ 2 }\quad \cdots \cdots (1)\\ \\ Consider\quad \triangle ARB,\quad using\quad Pyth.\quad Theorem,\quad we\quad have\\ { \left( \frac { 1+x }{ 2 } \right) }^{ 2 }+{ \left( x+y \right) }^{ 2 }=1\quad \cdots \cdots (2)\\ From\quad (1)\quad and\quad (2),\quad we\quad have\quad { x }^{ 2 }+2x-1=0\\ \\ In\quad \triangle ABC,\quad { AB }^{ 2 }+AC^{ 2 }=2\quad and\quad { BC }^{ 2 }=\left( 1+x \right) ^{ 2 }=1+2x+{ x }^{ 2 }\\ By\quad converse\quad of\quad Pyth\quad Theorem,\quad the\quad required\quad angle\quad is\quad { 90 }^{ \circ }.$

I thought of posting a solution, but Rohit Camfar's solution is exactly the same.

$\color{#D61F06}{\text{Huh?}}$ No one actually got respect for the problem proposer. I know he just picked up a $right$ $\Delta ABC$ and an $incenter$ to find some results on $\color{#20A900}{\text{inscribed}}$ and $\color{#20A900}{\text{circumscribed}}$ figures and $fortunately$ got a simple result. But at least respect the $\color{#D61F06}{\text{situation}}$ . $\large{ { \color{#20A900}{K.I.P.K.I.G.}}}$ $Const$ : Produce $BA$ to $D$ such that $BD$ = $BI$ . Join $DC$ and $IC$ . Let $\angle ABC$ = $2θ$ Now $AB$ = $BC$ => $\angle A$ = $90-θ$ . => $\angle BIC$ = $90$ + $angle A$ / 2 = $135$ - $θ/2$ But also, $AB$ + $AI$ = $BD$ = $BC$ . => $\angle D$ = $45$ + $θ/2$ Then $BDCI$ is a cyclic quadrilateral. Also, $\angle ADI$ = $θ/2$ => $\angle CDI$ = $45$ + $θ/2$ - $θ/2$ = $45$ . But Since $BDCI$ is cyclic. $\angle CAI$ = $\angle CDI$ = $45$ . => $\angle A$ = 90 $degrees.$