if triangle QRS has incenter, what is the incenter of triangle xyz

Question

if triangle ABC has incenter I how is it possible to show the circumcenter of triangle BIC lies on the circumcircle of triangle ABC?

[![ddiagram][1]][1]

D is the circumcenter of BIC

Answer 1

Join $AI$ and extend till it intersects circumcircle. If possible Ray$AI$ does not intersect $D$ but at $D'$ . Join $D'B$ and $D'C$ . By angle chasing, $\angle BID' = \angle IBD'$ And $\angle CID' = \angle ICD'$. Therefore, $BD'$ = $ID'$ = $CD'$. $\implies$ $D'$ is circumcentre of $\triangle BIC$ . As given $D$ is circumcentre of $\triangle BIC$ . Contradiction to considerations, $D$ coincides $D'$. Hence, proved.

Answer 2

In the standard notation $$\measuredangle BIC=180^{\circ}-\frac{\beta}{2}-\frac{\gamma}{2}=180^{\circ}-\frac{180^{\circ}-\alpha}{2}=90^{\circ}+\frac{\alpha}{2},$$ which says that in the circumcircle of $\Delta BIC$ the arc $BC$ is equal to $180^{\circ}+\alpha$.

Thus the arc $BIC$ is $$360^{\circ}-\left(180^{\circ}+\alpha\right)=180^{\circ}-\alpha$$ and since $$\measuredangle BDC+\measuredangle BAC=180^{\circ},$$we obtain that the circumcenter of $\Delta BIC$ is placed on the circumcircle of $\Delta ABC.$