Showing that in a triangle $ABC$, $\angle ABC = 60°$ , the following intersection points, and points $B$, and $C$ lie on the same circle

Question

The problem:

In an acute-angled triangle $ABC$, point $H$ is the orthocenter, point $O$ is the center of the circumscribed circle, point $J$ is the center of the inscribed circle, $\angle BAC = 60°$. Prove that points $B, H, O, J, C$ lie on the same circle.*

The figure:

I didn't know how to even approach this problem, so I used a hint from the textbook, that says the following:

Prove that the points $H, O, J$ belong to the points locus, from which the segment $BC$ is seen from an angle of $120°$.

At first I saw that $\angle BOC = 120°$, since it's a central angle that rests on the same arc as the inscribed $\angle BAC$ of the given circle. After quite some time of looking at the figure, I also saw that $\angle BHC = 120°$, since it's simply adjacent to the angle that's measure is $60°$. I got stuck at the point where needs to be proven that $\angle BJC = 120°$ as well. I can't see how to prove that.

So I've tried to prove what's initially asked, but at that point I did stuck completely, I don't understand how the given in the textbook hint proves that they are on the same circle.
How to prove that ?

P.s. Please, note, that this problem is from an 8th grade school textbook, so any angle functions or similar, more advanced tools cannot be used here. The problem itself is on the matter "Inscribed and central angles".

Answer 1

You have shown that $BOHC$ is a cyclic quadrilateral, since $\angle BOC=\angle BHC$.
It is easy to show that $$\angle BJC=90+\frac{\angle{A}}{{2}}=120^{\circ}$$ Therefore, $B,O,J,H,C$ all lie on the same circle.

Answer 2

J is center of inscribed circle that means it is the intersection of bisectors of angles of triangle. Now we use this theorem that says :

$\widehat {BJC}= 90 +\frac {\angle BAC}2=90+30=120^o$

So J must also be on the circle.