# 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 $\angle C = 60°$

• For the OP: many users here tend to downvote so-called PSQs (Problem Statement Questions) like yours, and answers, too (some god will forgive them, but the poor human Jack won't). So it is in your (and community's) best interest to improve your actual question by adding some context (your attempts, why this question is relevant to you, something along these lines). Cheers. Oct 4 '16 at 14:55
• Yes, indeed, basically if we are going to spend some time (and effort) writing solutions or hits to your homework problem, you should invest effort and time to add a motivating statement and give people an incentive. Oct 4 '16 at 14:59

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 $\widehat{C}=60^\circ$.
$$\angle AOB=2\angle ACB$$ $$\angle AJB=90^{\circ}+\frac{\angle ACB}2$$ and $$\angle AOB=\angle AJB$$ Let $$\angle ACB=x$$. Then $$2x=90+\frac x2$$ $$x=60$$
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 $\angle \, BOL$?