PROBLEM
Let $ABC$ be a triangle with incenter $I$. The circumscribed circle of triangle $AIC$ cuts the line $BC$ at point $D$, and the circumscribed circle of triangle $ABI$ cuts the line $BC$ at the point $E$. Show that $I$ is the center of the circumscribed circle of triangle $ADE$.
DRAWING
MY IDEAS
So, we can simply show that $AO_1IO_2$ is a rhombus. (idk if this is right)
If we want to show that $I$ is the centre of the circumscribed circle we have to show that triangles $IED$, $AIE$ and $AID$ are isosceles, am I right?
I don't really have much ideas of how can I show those and I don't know if my idea is right. Can you help me ? I will give a bounty to the one who answers. Thank you!
Since $I$ is the incentre of $\triangle ABC$, $\angle ICA=\angle ICD=\frac{\angle C}{2}$, $\angle IBA=\angle IBE=\frac{\angle B}{2}$
$\angle IDE= \frac{\angle C}{2} + \angle DIC =\frac{\angle C}{2} + \theta$, let $\angle DIC=\theta$
Since $AIDC$ is a cyclic quadrilateral$\angle IAD = \angle ICD=\frac{\angle A}{2}-\theta=\frac{\angle C}{2}$ and $\angle DIC= \angle DAC=\theta $
$\angle IED= \frac{\angle B}{2} + \angle BIE =\frac{\angle B}{2} + \alpha$, let $\angle BIE=\alpha$
Since $AIEB$ is a cyclic quadrilateral$\angle IAE= \angle IBE==\frac{\angle A}{2}-\alpha=\frac{\angle B}{2}$ and $\angle BIE= \angle BAE=\alpha $
$\angle IED=\frac{\angle B}{2} + \alpha = \frac{\angle A}{2}=\frac{\angle C}{2} + \theta = \angle IED $ (exterior $\angle$ of $\triangle$)
then $IE=ID=IA$ (sides are $=$ opposite $= \angle's$)($\angle$'s are $=$ subtended same segment$)