RMO 2009

The first thing which comes in the mind is: Trigonometry.. No, Nah, Nope.

$\large{ { \color{#20A900}{K.I.P.K.I.G.}}}$ $Const$ : 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. $\angle CAI$ = $\angle CDI$ = $45$ . => $\angle A$ = $90^{\circ.}$