Incenter and Circumcenter lie on a circle
$O$ is the center of the circumcircle of $\triangle ABC$ and $J$ is the center of the incircle. It is known that $A, B, O, J$ lie on a circle. Prove that $\angle C = 60° $
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$\begingroup$Approach 1. If $I,O$ are the incenter and circumcenter of $ABC$, we have $$ \widehat{AIB} = 90^\circ+\tfrac{1}{2}\widehat{C},\qquad \widehat{AOB}=2\widehat{C} $$ by angle chasing. If $A,I,O,B$ lie on the same circle, $\widehat{AIB}=\widehat{AOB}$ must hold, and the claim readily follows.
Approach 2. Let $D$ be the point of intersection between the internal angle bisector through $C$ and the circumcircle of $ABC$. By a well-known lemma, $DA=DI=DB$. If $O$ lies on the circumcircle of $AIB$, we have that both $ADO$ and $BDO$ are equilateral triangles, hence $\widehat{C}=60^\circ$.
$\endgroup$ 2 $\begingroup$Hint:
$$\angle AOB=2\angle ACB$$$$\angle AJB=90^{\circ}+\frac{\angle ACB}2$$and $$\angle AOB=\angle AJB$$Let $\angle ACB=x$. Then $$2x=90+\frac x2$$$$x=60$$
$\endgroup$ $\begingroup$Another hint: Let line $CJ$ intersect the circumcircle at point $L \neq C$. Then show that $LA = LJ = LB$ by proving that triangles $ALJ$ and $BLJ$ are isosceles. Consider the circle $c_L$ centered at $L$ and of radius $LA = LJ = LB$. Then points $A, J, B$ lie on circle $c_L$. Now, in the case of this problem, point $O$ lies on the same circle as $A, J, B$ which is circle $c_L$ with center $L$. What is then the length of $LO = ?$ What about the lengths $BO$ and $AO$? What kind of triangles are $AOL$ and $BOL$? What can you conclude about angles $\angle\, AOL$ and $\angle \, BOL$?
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