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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

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  • $\begingroup$ 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. $\endgroup$ Jan 15 '19 at 0:27
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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.

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  • $\begingroup$ Hi, how would I go about angle chasing to prove ∠BID′=∠IBD′ And ∠CID′=∠ICD' $\endgroup$
    – user604720
    Jan 20 '19 at 19:13
  • $\begingroup$ 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 $\angle CID'=\angle ICD'$. $\endgroup$
    – RB MCPE
    Jan 22 '19 at 10:00
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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 $\Delta ABC.$

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  • $\begingroup$ hi, quick question, what do β,γ, and α represent? $\endgroup$
    – user604720
    Jan 16 '19 at 1:01
  • $\begingroup$ @user604720 They are angles of the triangle. $\measuredangle BAC=\alpha$, $\measuredangle ABC=\beta$ and $\measuredangle ACB=\gamma.$ $\endgroup$ Jan 16 '19 at 6:00

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