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 ?