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

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

• Angle chase. Extend AI to meet the circumcircle at D, the midpoint of arc BC, as in your diagram. Chase the angle DBC and hence DBI and BID, Similarly switch B and C. Jan 15 '19 at 0:27

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.
• Let $\angle AIB=x$and $\angle ABI=y$. $\angle BID'=x+y$ . $\angle CBI=y$ and $\angle IBD'=x$ as $\square ABCD'$ is cyclic. Thus $\angle BID'=\angle IBD'= x+y$. Repeat the process and get $\angle CID'=\angle ICD'$. Jan 22 '19 at 10:00
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 $$\Delta ABC.$$
• @user604720 They are angles of the triangle. $\measuredangle BAC=\alpha$, $\measuredangle ABC=\beta$ and $\measuredangle ACB=\gamma.$ Jan 16 '19 at 6:00